Math 1022 - Precalculus · Math 1022 - Precalculus: Table of Contents Chapter 1: unctionsF 1.1...

Math 1022 - Precalculus

c⃝ Boris A. Datskovsky, Marilena Downing: Mathematics Department, Temple University

Math 1022 - Precalculus: Table of Contents

Chapter 1: Functions

1. 1 Functions pp. 2-11

1. 2 Properties of Functions and Their Graphs pp. 12-20

1. 3 Basic Graphs and Their Transformations pp. 21-28

1. 4 Rational Functions and Asymptotic Behavior pp. 29-34

1. 5 Composition of Functions pp. 35-37

1. 6 Inverse Functions pp. 38-43

Chapter 1 Additional Practice Problems pp. 44-46

Chapter 2: Exponential and Logarithmic Functions

2. 1 Exponential Functions and Their Graphs pp. 47-54

2. 2 Logarithmic Functions and Their Graphs pp. 55-60

2. 3 Properties of Logarithms pp. 61-64

2. 4 Exponential and Logarithmic Equations pp. 65-69

2. 5 Applications pp. 70-73


Chapter 3: Trigonometric Functions

3. 1 Angles and Right Triangular Trigonometry pp. 76-83

3. 2 Trigonometric Functions of General Angles; Basic Identities pp. 84-91

3. 3 Graphs of Trigonometric Functions pp. 92-101

3. 4 Inverse Trigonometric Functions pp. 102-107

3. 5 Trigonometric Identities pp. 108-113

3. 6 Trigonometric Equations pp. 114-118

3. 7 Polar Coordinates pp. 119-121

3. 8 Supplementary Exercises Involving Trigonometric Functions pp. 122-123


Answer Key pp. 126-146

1

1. 1 Functions 2

1. 1 Functions

Introduction to Functions

De�nition: A function f is a rule that assigns to each element x in oneset one and only one element y in another set. The elementy is often denoted by f(x), read "f of x" or "the value off at x".

You can think of a function as a rule that takes as an input the value x and outputs thevalue y. So

f(input) = output or simply f(x) = y.

In the equation, y = f(x), x is called the independent variable and y is called the dependentvariable.

De�nition: The set of input values is called the domain of f .The set of output values is called the range of f .

Functions can be given by a verbal description, by a formula, or by a graph.

Example 1

(a) The rule that assigns to each Temple student their unique Temple Id number is a functionfrom the set of Temple students to the set of 9-digit numbers. The domain of the functionis the set of all Temple students and the range of the function is the set of all TempleId numbers currently in use.

(b) The equation y = x2 assigns to each real number its square. Since a number has oneand only one square, this equation de�nes a function, f(x) = x2. The domain of f isthe set of all real numbers and the range of f is the set of all real numbers ≥ 0.

Most of the functions we will consider will be described using formulas. Their domainsand ranges will be subsets of the set of real numbers.

(c) Below is a function represented graphically.

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

6

By examining the graph, we can see that this is the graph of the straight line with

slope 1/2 and y-intercept y = 1. So its equation is given by y =1

2x + 1. In functional

notation, this graph is the graph of the function f(x) =1

2x+ 1.

1. 1 Functions 3

Example 2 Not every equation de�nes a function. Consider the equation y2 = x2 + 1.Solving for y, we get

y = ±√x2 + 1.

So every input value x leads to two output values, y =√x2 + 1 and y = −

√x2 + 1. This

means that this equation does not de�ne a function.

Example 3 Not every graph de�nes a function either. In order to determine if a graphrepresents a function, we can apply the following test.

Vertical Line Test: A graph represents a function if and only if any vertical lineintersects the graph in at most one point.

By applying the vertical line test, determine whether the graphs shown below representfunctions.

(a) (b) (c) (d)

Solution: We can see from the graphs that any vertical line intersects the graphs (a) and(d) in precisely one point and that a vertical line intersects the graphs (b) and (c) in morethan one point. Hence, the graphs (a) and (d) are graphs of functions, while the graphs (b)and (c) are not.

Note on Functional Notation: Up to this point when discussing functions, we have usedx and y = f(x) to denote the variables. However, other letters can be used. For example,we may write s(t) = −16t2 + 50t + 6 to describe a certain motion problem or G(θ) = cos θto represent a certain trigonometric function. A variety of other letters are routinely usedto denote variables and functions.

Types of Functions

Below are several types of functions that occur frequently in mathematics. In this chapter,we will concentrate on these functions and some of their graphs.

• Polynomials: A polynomial function of degree n is a function of the form

p(x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0,

where an, an−1, . . . , a0 are real numbers, an ̸= 0, and all the exponents are nonnegativeintegers. The chart below contains names and examples of polynomial functions of lowdegrees:

1. 1 Functions 4

Degree Name Example

0 constant function f(x) = −4

1 linear function g(x) = 2x+ 3

2 quadratic function h(x) = x2 − 5x+ 2

3 cubic function k(x) = x3 + 3x2 − 1

• Rational Functions: Rational functions are functions of the form

r(x) =p(x)

q(x),

where p(x) and q(x) are polynomial functions. For example, F (x) =2x− 1

x2 + xand

G(x) = x−2 =1

x2are rational functions.

• Algebraic Functions: Algebraic functions can be obtained from polynomial func-tions by addition, subtraction, multiplication, division and taking roots. For example,

f(x) =√1− x2 and g(x) =

(3√x+ 1

)21−

√x

are algebraic functions.

• Piecewise Functions: Piecewise functions are functions that are de�ned by morethan one formula. For example,

f(x) =

{−2x+ 3, if x < 0

x2 + 1, if x ≥ 0

is a piecewise function. For x < 0, this function is de�ned by the formula y = −2x+3and for x ≥ 0 by y = x2 + 1.

Evaluating Functions

For any function f(x), x can be thought of as a placeholder. To evaluate f(x) at x = a,simply replace x by a. More generally, we can replace x by any algebraic expression in theformula for f(x).

Example 4

(a) For f(x) = x2 − 3x+ 2, �nd f(2), f(3a), and f(a− 1).

(b) For g(x) =x

2− x, �nd g(0), g(−2), g(2), and g

(1

x

).

(c) For h(t) =√16− t2, �nd h(1), h(−4), h(5), and h(2s+ 1).

(d) Evaluate f(x) =

7 if x < 0

x2 + 1 if 0 ≤ x ≤ 2

3x− 1 if x > 2

at x = −3, x = 0, x = 1/2, x = 2, and x = 5 .

1. 1 Functions 5

Solution

(a) f(2) = 22 − 3 · 2 + 2 = 4− 6 + 2 = 0 f(3a) = (3a)2 − 3(3a) + 2 = 9a2 − 9a+ 2

f(a− 1) = (a− 1)2 − 3(a− 1) + 2 = a2 − 2a+ 1− 3a+ 3 + 2 = a2 − 5a+ 6

Note the use of parentheses when needed.

(b) g(0) =0

2− 0= 0 g(−2) =

−2

2− (−2)= −1

2

g(2) is unde�ned since evaluating g at x = 2 leads to division by zero, which is notpossible in the set of real numbers.

g

(1

x

)=

1

x

2− 1

x

=

1

x2x− 1

x

=1

2x− 1

(c) h(1) =√

16− (1)2 =√15 h(−4) =

√16− (−4)2 =

√16− 16 = 0

h(5) is unde�ned since√

16− (5)2 =√

− 9 is not a real number.

h(2s+ 1) =√16− (2s+ 1)2 =

√16− (4s2 + 4s+ 1) =

√− 4s2 − 4s+ 15

(d) We will use a pictorial aid to help explain how to evaluate this piecewise function. Welet a number line represent the domain of f(x). The formulas for f(x) change at x = 0and at x = 2. Plotting these two numbers on the number line splits the number lineinto three intervals. For each interval, f(x) is given by one of the three formulas in itsde�nition. We will use a bracket when the formula is to be used for an endpoint of aninterval and a parenthesis when it is not to be used.

For x < 0, f(x) = 7 and since −3 < 0, f(−3) = 7.

For all x in the interval 0 ≤ x ≤ 2, f(x) = x2 + 1. Hence f(0) = 02 + 1 = 1, f(1/2) =(1/2)2 + 1 = 5/4, and f(2) = 22 + 1 = 5.

Finally, for x > 2, f(x) = 3x− 1. So f(5) = 3(5)− 1 = 14.

Example 5

(a) The length of a rectangular plot is x meters and thearea of the plot is 100 m2. Express the perimeter asfunction of x.

(b) Find the perimeter when x = 20m.x

y

1. 1 Functions 6

Solution

(a) If x is the length of the rectangular plot and y is its width, then the area is given by

A = xy. Therefore, 100 = xy and y =100

x. The formula for the perimeter of a rectangle

is P = 2x+ 2y. Hence P (x) = 2x+200

x.

(b) P (20) = 2 · 20 + 200

20= 40 + 10 = 50m.

Di�erence Quotient

Let f be a function whose graph is shown below.

The points A and B shown on the graph have coordinates (x, f(x)) and (x + h, f(x + h)).The line through the points A and B is called a secant line to the graph y = f(x). The slopeof this line is given by

f(x+ h)− f(x)

(x+ h)− x=

f(x+ h)− f(x)

h.

The latter expression is called the di�erence quotient. This expression is very important incalculus. Its value approaches the derivative of f at x as h approaches 0.

Example 6 Findf(x+ h)− f(x)

h, h ̸= 0 for the following functions:

(a) f(x) = x2 − x

(b) g(x) =1

3x+ 2

1. 1 Functions 7

Solution

(a)f(x+ h)− f(x)

h=

(x+ h)2 − (x+ h)− (x2 − x)

h

=x2 + 2xh+ h2 − x− h− x2 + x

h

=2xh+ h2 − h

h

=h(2x+ h− 1)

h

= 2x+ h− 1

(b)g(x+ h)− g(x)

h=

1

3(x+ h) + 2− 1

3x+ 2

h=

3x+ 2− (3x+ 3h+ 2)

(3x+ 3h+ 2)(3x+ 2)

h

=−3h

h(3x+ 3h+ 2)(3x+ 2)=

−3

(3x+ 3h+ 2)(3x+ 2)

Finding Domains of Functions

In some cases, the domain of a function is stated. For example, consider f(x) = x2, x ≥ 0.Here the domain of f is explicitly speci�ed as the set of all nonnegative real numbers. Moreoften, however, the domain is implied by the formula(s) that determine the function. Morespeci�cally, the domain of a function is the set of all real numbers x for which f(x) is areal number. When x is in the domain of f , we say that f is de�ned at x.

Notation: We shall use Df to denote the domain of a function and Rf to denote the rangeof a function.

In �nding domains of algebraic functions, we need to eliminate from the domain any realnumbers that result in

a) division by 0

b) an even root (square root, fourth root, sixth root, etc.) of a negative number.

Example 7 Find the domains of the following functions: (a) g(x) = x3 + 2x − 1, (b)

f(x) =x

x− 2, (c) h(t) =

√16− t2, (d) G(s) = (2− s)−3/4.

Solution

(a) g(x) = x3+2x−1 is a polynomial. This function can be evaluated for any x. Therefore,the domain of g is all real numbers. Equivalently, in interval notation, Dg = (−∞,∞).In general, the domain of any polynomial is all real numbers.

1. 1 Functions 8

(b) For f(x) =x

x− 2, the only domain issue is division by zero. Setting the denominator

equal to 0, we get x− 2 = 0 and so x = 2. So we must exclude x = 2 from the domain.Hence, the domain of f is x ̸= 2 or, in interval notation, Df = (−∞, 2) ∪ (2,∞).

(c) h(t) =√16− t2 is only de�ned when 16− t2 ≥ 0. So to �nd the domain of h, we solve

this quadratic inequality using sign diagrams:

16− t2 ≥ 0

(4− t)(4 + t) ≥ 0

Therefore, the domain of h is −4 ≤ t ≤ 4, or equivalently, Dh = [−4, 4].

(d) G(s) = (2 − s)−3/4 =1

(2− s)3/4=

14√

(2− s)3. The denominator is only de�ned if

(2 − s)3 ≥ 0, but we need to exclude all s's that give division by 0. So we must have(2 − s)3 > 0. (2 − s)3 > 0 if and only if 2 − s > 0 and this occurs if and only if s < 2.So DG = (−∞, 2).

Notice that there is no need to use a sign diagram for this inequality. Since (2− s)3 =(2 − s)2(2 − s) and (2 − s)2 is either zero or positive, we only need to determine when2 − s > 0. When considering factors, only multiplying or dividing by a negative canchange the sign of an expression.

The Algebra of Functions

As with real numbers, we can add, subtract, multiply, and divide functions.

De�nition: Let f and g be two functions. Then the functions f + g, f − g, fg, and f/gare de�ned as follows:

(f + g)(x) = f(x) + g(x) (f − g)(x) = f(x)− g(x)

(fg)(x) = f(x)g(x) (f/g)(x) =f(x)

g(x)

For any given x, f(x) + g(x), f(x) − g(x), and f(x)g(x) exist if and only if both f(x) andg(x) exist. So x must belong to the domains of both f and g. Hence,

Df+g = Df−g = Dfg = Df ∩Dg.

For f(x)/g(x) to be de�ned, not only must x be in the domains of both f and g but, inaddition, we must exclude any x for which g(x) is equal zero. Therefore

Df/g = Df ∩Dg − {x| g(x) = 0}.

1. 1 Functions 9

Example 8 Let f(x) = x2 + 1 and g(x) = 4−√x. Find f + g, f − g, fg, f/g, and their

domains.

Solution

(f + g)(x) = x2 + 1 + 4−√x = x2 + 5−

√x

(f − g)(x) = x2 + 1− (4−√x) = x2 − 3 +

√x

(fg)(x) = (x2 + 1)(4−√x)

(f/g)(x) =x2 + 1

4−√x

Df = (−∞,∞) and Dg = [0, ∞). Hence Df+g = Df−g = Dfg = Df ∩Dg = [0, ∞).

For f/g, there is an additional restriction that 4 −√x ̸= 0. 4 −

√x = 0 if and only if

4 =√x if and only if x = 16. So we must exclude x = 16 from Df ∩ Dg = [0, ∞).

Therefore, Df/g = [0, 16) ∪ (16,∞).

Notice that these domains can be determined without even computing the combinationsof the functions.

Example 9 Let f(x) = x +2

xand g(x) = x2 − 6. Evaluate f + g, f − g, fg, and f/g at

x = 2.

Solution

f(2) = 2 +2

2= 3 and g(2) = 22 − 6 = −2.

Therefore, using the de�nitions given above, we obtain

(f + g)(2) = 3 + (−2) = 1, (f − g)(2) = 3 − (−2) = 5, (fg)(2) = 3 · (−2) = −6, and(f/g)(2) = 3/(−2) = −3/2.

1. 1 Functions 10

Exercises 1. 1

1. Determine which of the following graphs represent a function.

(a) (b) (c) (d)

Evaluate the following functions at the indicated values. Simplify where possible.

2. f(x) = 2x2 − 2x+ 3 3. g(x) =2− x

2 + x

f(0), f(√2), f(x+ 1),

f(x)− f(1)

x− 1g(1), g(−3), g(3/2), g(2a+ 1), g(2− a)

4. h(x) =

2x+ 1, if x < −2

x2, if −2 ≤ x < 2

5, if x ≥ 2

5. F (x) =

√1− x, if x < 0

2, if x = 0

3− 2x, if x > 0

h(−4), h(−2), h(0), h(2), h(4) F (−3), F (0), F (3)

6. G(x) = x2 − 1 7. H(x) = x3 +1

x

G(x+ 2), G(x) +G(2) H(2x), 2H(x)

Findf(x+ h)− f(x)

h, h ̸= 0 for the following functions. Simplify fully.

8. f(x) = 3x+ 2 9. f(x) = −x2 + 1 10. f(x) =1

x

11. f(x) =1

2x+ 112. f(x) = 2x2 − x+ 5

Find the domains of the following functions. State your answers in interval notation.

13. f(x) = 3x+ 1 14. g(x) =1

2x+ 615. F (s) =

s− 1

s2 − 4

16. h(x) =√x− 10 17.G(t) = 3

√t− 1 18.H(x) =

x

x2 + 1

19. f(x) =√6− 4x 20. h(u) =

u2

√2− u

21. g(x) =

√x

2x2 + x− 1

22. F (x) =√x2 − 4 23.G(x) =

√x− 1

x+ 324.H(x) =

1√x− 3

25. f(x) =

√x− 2

x+ 126. g(x) = 4

√2x− x2

1. 1 Functions 11

For the pairs of functions in exercises 27 − 29, �nd f + g, f − g, fg, f/g, g/f , and theirdomains. Simplify your answers when possible and use interval notation for the domains.

27. f(x) = x2 − 4, g(x) = x+ 2 28. f(x) =x

x− 1, g(x) =

x

x+ 1

29. f(x) =√x+ 4, g(x) =

√2− x

30. Let f(x) =√8− x2 and g(x) =

x− 4

x. Evaluate f + g, f − g, fg, f/g, and g/f at

x = −2.

Applications. If necessary, draw sketches to help you solve these problems.

31. Express the area A of a rectangle as a function of its length l if the length of the rectangleis twice the width of the rectangle.

32. Express the perimeter P of a rectangle as a function of its length l if the length of therectangle is twice the width of the rectangle.

33. Express the area A of a circle as a function of its radius r.

34. Express the radius r of a circle as a function of its area A.

35. A balloon is rising vertically in the air. An observer is standing on the ground 100 mfrom the point on the ground directly below the balloon. Express the distance d fromthe observer to the balloon as a function of the height h of the balloon. See Figure 1.

36. An open box is to be constructed from a rectangular piece of cardboard of dimensions10 in x 12 in by cutting squares of side x from each of the corners of the cardboard andthen folding up the sides. Express the volume V as a function of x. See Figure 2.

Figure 1 Figure 2

1. 2 Properties of Functions and Their Graphs 12

1. 2 Properties of Functions and Their Graphs

Some Basic Properties

Many properties of a function can be obtained from its graph.

Example 1: Domain, Range, Intercepts and Sign of the Function

The graph of f(x) is given below. Determine (a) the domain and range of f , (b) the x- andy-intercepts, (c) f(−4) and f(4), (d) all values of x for which f(x) = 1 and f(x) = 2, (e)the interval(s) on which f(x) > 0 and the interval(s) on which f(x) < 0.

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

Solution:

(a) The domain of a function is the set of all x for which f(x) is de�ned, that is, the setof all x on the x-axis for which there is a point on the graph of the function. Here,Df = [−4, 4 ]. The range of f is the set of all y on the y-axis for which there is acorresponding point on the graph of f . Here, Rf = [−1, 3 ].

(b) The x-intercepts are (−1, 0 ) and ( 1, 0 ) and the y-intercept is ( 0,−1). Equivalently, thiscan be stated as the x-intercepts are x = −1 and x = 1 and the y-intercept is y = −1.

In terms of the function f , this means that f(1) = 0 and f(−1) = 0 and f(0) = −1.Note that a function can have at most one y-intercept, but may have more than onex-intercept.

(c) f(−4) is the y-coordinate of the point on the graph whose x-coordinate is equal to −4.So f(−4) = 3 and similarly, f(4) ≈ 1.5

(d) To �nd the values for which f(x) = 1, we must determine where the horizontal liney = 1 intersects the graph of f . This occurs at (−2, 1) and (2, 1). Therefore, f(x) = 1at x = −2 and at x = 2. Similarly, using the horizontal line y = 2, we see that f(x) = 2at x = −3.

(e) f(x) is positive when the graph of f lies above the x-axis and f(x) is negative whenthe graph of f lies below the x-axis. Hence, f(x) > 0 for x in [−4,−1 ) ∪ ( 1, 4 ] and


f(x) < 0 for x in (−1, 1). Notice that x = −1 and x = 1 are excluded in the answerssince f(x) = 0 at these values. We say that f(x) > 0 on [−4,−1 )∪ ( 1, 4 ] and f(x) < 0on (−1, 1).

When a function is given by a formula, most of the above properties can be found al-gebraically. Finding properties algebraically is particularly important when sketching thegraph of a function. However, ranges are generally di�cult to determine algebraically, so weexclude ranges from the discussion that follows.

Example 2: Let f(x) =x− 3

x2 − 4. Determine (a) the domain of f , (b) the x- and y-

intercepts, (c) f(−3), (d) the values of x for which f(x) = 1, (e) the interval(s) on whichf(x) > 0 and the interval(s) on which f(x) < 0.

Solution:

(a) The domain of f is the set of all x for which x2 − 4 ̸= 0. x2 − 4 = 0 when x = −2 andx = 2. Therefore, the domain of f is x ̸= ±2.

(b) To �nd x-intercepts, we must solve the equation f(x) = 0.x− 3

x2 − 4= 0 implies that

x − 3 = 0 and so x = 3 is the x-intercept. To �nd the y-intercept, we evaluate f atx = 0. f(0) = 3/4 and therefore the y-intercept is y = 3/4.

(c) f(−3) =−3− 3

(−3)2 − 4= −6

5.

(d) To �nd the value of x for which f(x) = 1, we solve the equationx− 3

x2 − 4= 1. Cross-

multiplying, we obtain x − 3 = x2 − 4 and collecting all the terms to one side gives

x2 − x− 1 = 0. Using the quadratic formula yields x =1−

√5

2and x =

1 +√5

2.

(e) Below is a number line with the zeros of the numerator and the denominator marked onit. (These are the only places where a rational function can change signs.) The signsbelow the diagram give the signs of the factors of the numerator and the denominatorof f , while the signs above the diagram are the signs of f(x) itself. f(x) > 0 on(−2, 2 ) ∪ ( 3,∞) and f(x) < 0 on (−∞,−2) ∪ ( 2, 3). Note that f is not de�ned atx = −2 and x = 2 and that f(x) = 0 at x = 3.


Increasing and Decreasing Functions

De�nition: A function f is increasing on an interval if for any x1 < x2

in the interval, f(x1) < f(x2).

A function f is decreasing on an interval if for any x1 < x2

in the interval, f(x1) > f(x2).

A function f is constant on an interval if for any x1, x2 inthe interval, f(x1) = f(x2).

f(x) is increasing if the graph of f rises as x increases, that is, as x moves from left to right.f(x) is decreasing if the graph of f falls as x increases. f(x) is constant on an interval if thegraph of f over the interval is a horizontal line segment.

Example 3 The graph of f(x) is given below. Determine (a) the domain and range of f ,(b) interval(s) on which f is increasing, (c) interval(s) on which f is decreasing, (d) interval(s)on which f is constant.

-4 -3 -2 -1 1 2 3 4

-5

-4

-3

-2

-1

1

2

3

4

Solution:

(a) Note that when there are no closed circles at the endpoints of a graph, it is assumed thatthe graph continues inde�nitely on both the left and the right in the same direction (upor down) as indicated by the graph. So for this function, Df = (−∞,∞), by convention.

As x goes to +∞, f(x) goes to −∞ and as x goes to −∞, f(x) goes to −∞. SoRf = (−∞, 4 ].

(b) f is increasing on (−∞,−1).

(c) f is decreasing on (−1, 0) and (2,∞).

(d) f is constant on (0, 2).

Note that all of these intervals are subsets of the domain, that is, intervals on the x-axis.


De�nition: We say that a function f

has a local maximum at x = a if f(a) ≥ f(x) for all x insome open interval containing a.

has a local minimum at x = a if f(a) ≤ f(x) for all x insome open interval containing a.

A local maximum occurs at a point where the function changes from increasing to decreasingand a local minimum occurs at a point where the function changes from decreasing toincreasing.

Example 4

The graph of f(x) is given below. Determine (a) the intervals on which the function isincreasing and the intervals on which the function is decreasing, (b) the values of x where fhas a local maximum, (c) the values of x where f has a local minimum.

-4 -3 -2 -1 1 2 3 4

-2

-1

1

2

3

4

Solution:

(a) f is increasing on (−2, 0) and (2,∞) and f is decreasing on (−∞,−2) and (0, 2).

(b) f has a local maximum at x = 0.

(c) f has a local minimum at x = −2 and at x = 2.

Even and Odd Functions

De�nition: A function f is called even if f(−x) = f(x) and odd if f(−x) = −f(x).

Most functions are neither even nor odd, but many of our basic functions happen to be ofthese two types. To determine if f is even, odd, or neither, �nd f(−x) and compare it tof(x) and −f(x).


Example 5 For each function, determine if the function is even, odd, or neither: (a)

f(x) = x4 − 2, (b) g(x) =x

x2 + 1, (c) h(x) = x3 − 1.

Solution:

(a) f(−x) = (−x)4 − 2 = x4 − 2 = f(x). Since f(−x) = f(x), f is an even function.

(b) g(−x) =−x

(−x)2 + 1=

−x

x2 + 1= − x

x2 + 1= −g(x). Since g(−x) = −g(x), g is an odd

function.

(c) h(−x) = (−x)3 − 1 = −x3 − 1. Since h(−x) ̸= h(x), h is not even. Furthermore,−h(x) = −x3 + 1 ̸= h(−x) and so h is not odd. Hence h is neither.

The graphs of even and odd functions have symmetry properties: The graph of an even func-

tion is symmetric with respect to the y-axis and the graph of an odd function is symmetric

with respect to the origin.

In the �gure below, the left graph is the graph of an even function. Given that f(−x) = f(x),if (x, y) lies on the graph, then so does (−x, y). This is why the part of the graph to the leftof the y-axis is the mirror-image of that to the right.

The right graph is the graph of an odd function. Given that f(−x) = −f(x), if (x, y) lies onthe graph, then so does (−x,−y). Therefore, the part of the graph to the left of the y-axis issame as the part to the right re�ected through the origin. Re�ection through the origin canbe thought of as a double re�ection, re�ection with respect to the x-axis and to the y-axis.

Even

Hx,yLH-x,yLHx,yL

H-x,-yL

Odd

Example 6 The graph of f for x ≥ 0 is given below. Complete the graph of f if (a) f isan even function, (b) f is an odd function.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3


Solution:

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

(a) even function (b) odd function

Points of Intersection

In calculus, you will need to �nd the points of intersection of the graphs of two functions,say f(x) and g(x). Any point of intersection (x, y) must lie on the graphs of both of thesefunctions and therefore must satisfy y = f(x) and also y = g(x). So in order to �nd anypoints of intersection, we set f(x) = g(x) and solve for x. Substituting the values of x thatwe �nd into either formula will give the y-coordinates of the points of intersection.

Example 7 Find the points of intersection of the graphs of y = 2x2 − 1 and y = x + 2.Compare the answer to the �gure provided.

Solution:

2x2 − 1 = x+ 2

2x2 − x− 3 = 0

(2x− 3)(x+ 1) = 0

x = 3/2 and x = −1

At x =3

2, y =

3

2+ 2 =

7

2.

At x = −1, y = −1 + 2 = 1.

Points of Intersection: (−1, 1)and (3/2, 7/2).

y = 2x2-1

y = x+2

-3 -2 -1 1 2 3

-2

-1

1

2

3

4


Exercises 1. 2

In Exercises 1 - 3, the graph of a function f is given. Determine

(a) the domain and range of f (e) the intervals on which f(x) > 0, if any

(b) the intercepts, if any (f) the intervals on which f(x) < 0, if any

(c) f(1) and f(−2) (g) if f is even, odd, or neither

(d) all x for which f(x) = 1 and for which f(x) = −3

1. 2. 3.

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

-2 -1 1 2

-2

-1

1

2

In Exercises 4 - 6, for each function determine

(a) the domain (d) all x for which f(x) = 3

(b) the intercepts, if any (e) the intervals on which f(x) > 0, if any

(c) f(5) (f) the intervals on which f(x) < 0, if any

4. f(x) = 3x2 + x 5. f(x) =x+ 3

1− x26. f(x) =

√x2 − 9

In Exercises 7 and 8, given the graph of a function f , determine

(a) the domain and range of f

(b) the intervals on which f(x) is increasing, if any

(c) the intervals on which f(x) is decreasing, if any

(d) the intervals on which f(x) is constant, if any

7. 8.

-6 -4 -2 2 4 6 8

-2

-1

1

2

3

-2 -1 1 2

-2

-1

1

2


In Exercises 9 - 11, the graph of f is given. Determine (a) the intervals on which the functionis increasing and the intervals on which the function is decreasing, (b) the values of x wheref has a local maximum, (c) the values of x where f has a local minimum, (d) if f is even,odd, or neither.

9. 10. 11.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

-4 -3 -2 -1 1 2 3 4

-2

-1

1

2

3

4

-3 -2 -1 1

-3

-2

-1

1

2

3

4

5

In Exercises 12 - 15, determine if the function is even, odd, or neither.

12. f(x) = x3 + x2 13. g(x) = 4x4 + 2 14. h(x) =x2 − 3

x

15. F (x) =x+ 1

x− 3

In Exercises 16 and 17, the graph of f for x ≤ 0 is given below. Complete the graph of f if(a) f is an even function, (b) f is an odd function.

16. 17.

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

In Exercises 18 and 19, �nd the points of intersection of the graphs of the two functionsgiven.

18. y = 2x+ 4 ; y = 3x− 1 19. y = x2 − x+ 5 ; y = −2x+ 7


20. The graphs of f and g are given.

(a) Which is larger, f(0) or g(0)? (d)On what interval(s) is f(x) < g(x)?

(b) Which is larger, f(2) or g(2)? (e) For which x is f(x) = g(x)?

(c) On what interval(s) is f(x) > g(x)?

f

g

-1 1 2 3 4 5

-2

-1

1

2

3

21. Sketch the graph of a function that satis�es the following conditions:

(a) f(0) = −2 and f(3) = 4

(b) f is increasing on (0, 3)

(c) f is decreasing on (−∞, 0) ∪ (3,∞)

1. 3 Basic Graphs and Their Transformations 21

1. 3 Basic Graphs and Their Transformations

y = x

1

1

y = ÈxÈ

1

1

D = (−∞,∞); R = (−∞,∞) D = (−∞,∞); R = [ 0,∞)

Odd function Even function

y = x2

1

1

y = x

1

1

D = (−∞,∞); R = [0,∞) D = [ 0,∞); R = [ 0,∞)

Even function

y = x3

1

1

y = x3

1

1

D = (−∞,∞); R = (−∞,∞) D = (−∞,∞); R = (−∞,∞)

Odd function Odd function


The previous page contains the graphs of some basic functions that you must know. Whensketching the graph of one of these functions, evaluate the function at a few integer valuesmentally, plot the corresponding points, and connect by using a graph of the appropriate

shape. Two more basic graph will be added in the next section, namely, y =1

xand y =

1

x2.

Vertical and Horizontal Shifts

VERTICAL SHIFTS:

To graph: Shift the graph of f(x):

y = f(x) + k, k > 0 Up k units.

y = f(x)− k, k > 0 Down k units.

Domains are not a�ected by vertical shifts, however ranges may change.

Example 1: Using a vertical shift of the graph of y = x2, graph f(x) = x2 + 2 along withthe graph of y = x2 on the same coordinate system. Determine the domain and range of fand any intercepts.

Solution:

y = x2

y = x2+ 2

1

1

Df = (−∞,∞), Rf = [ 2,∞)

x-intercept: None

y-intercept: y = 2

Shift the graph of y = x2 up by 2.

Note that the graph of f(x) = x2 + 2 is the graph of y = x2 shifted up by 2. Since everyy-value is increased by 2, the range of f(x) = x2 + 2 is [ 2,∞) instead of [ 0,∞) for y = x2.

An e�cient way to graph f(x) = x2 + 2 is to mentally shift the origin (0, 0) to (0, 2) andthen graph y = x2 starting at the "new" origin.

Example 2: Graph g(x) =√x−1. Determine the domain and range of g and any intercepts.

Solution:

y = x - 1

2

1 Dg = [ 0,∞), Rg = [−1,∞)

x-intercept: x = 1

y-intercept: y = −1

Shift the graph of y =√x down by 1.


Above, a procedure was given for graphing y = f(x) + k, k > 0. Now, consider the graphof y = f(x + k). A point (x0, y0) lies on the graph of y = f(x) if y0 = f(x0). But theny0 = f((x0 − k) + k). This implies that (x0 − k, y0) lies on the graph of y = f(x + k). Thepoint (x0 − k, y0) lies k units to the left of the point (x0, y0).

HORIZONTAL SHIFTS:

To graph: Shift the graph of f(x):

y = f(x+ k), k > 0 Left k units.

y = f(x− k), k > 0 Right k units.

Ranges are not a�ected by horizontal shifts, however domains may change.

Example 3: Using a horizontal shift of the graph of y = |x|, graph h(x) = |x − 2| alongwith the graph of y = |x| on the same coordinate system. Determine the domain and rangeof h and any intercepts.

Solution:

y = Èx -2Èy = ÈxÈ

1

1

Dh = (−∞,∞), Rh = [ 0,∞)

x-intercept: x = 2

y-intercept: y = 2

Shift the graph of y = |x| right by2.

Note that the x-coordinate of every point is increased by 2 with this horizontal shift.

Example 4: Graph F (x) = (x + 1)3. Determine the domain and range of F and anyintercepts.

Solution:

y = Hx+1L3

1

1

DF = (−∞,∞), RF = (−∞,∞)

x-intercept: x = −1

y-intercept: y = 1

Shift the graph of y = x3 left by 1.


Example 5: Graph G(x) =√x+ 3 − 2. Determine the domain and range of G and any

intercepts.

Solution:

y = x + 3 - 2

2

1DG = [−3,∞), RG = [−2,∞)

x-intercept: x = 1

y-intercept: y =√3− 2

Shift the graph of y =√x left by 3

and down by 2.

Re�ections

REFLECTIONS:

To graph:

y = −f(x) Re�ect the graph of f(x) across the x-axis.

y = f(−x) Re�ect the graph of f(x) across the y-axis.

Example 6: Graph f(x) =√x+ 1, g(x) = −

√x+ 1 and h(x) =

√−x+ 1.

Solution:

y = x + 1

1

1

y = - x + 11

1y = -x + 1

1

1

Example 7: Taking Absolute Values: Graph f(x) = |x2 − 4|. Determine the domain andrange of f and any intercepts.

Solution: |x2−4| = x2−4 if x2−4 ≥ 0 and |x2−4| = − (x2 − 4) if x2−4 < 0. Therefore,to graph f(x) = |x2 − 4|, start with the graph of y = x2 − 4 and re�ect across the x-axisonly the parts of the graph that lie below the x-axis. Df = (−∞,∞);Rf = [ 0,∞).

y = x2-4

1

1

y = Èx2-4È

1

1


Graphing Piecewise Functions

Recall from Section 1. 1 that piecewise functions are functions that are de�ned by more than

one formula. For example, |x| =

{x, if x ≥ 0

−x, if x < 0. One way to graph this function is to graph

the line y = x for x ≥ 0 and the line y = −x for x < 0.

In general, to graph a piecewise function, partition the domain into the pieces where each ofthe formulas apply. Then graph the appropriate formula on each piece. Be sure to evaluateeach formula at the boundary values (x-values where the formula changes) in order to beginand end each piece of the graph at the correct point. We will call these points anchor points.

Example 8: Graph and �nd the domain and range of f(x) =

{x2, if x ≤ 0

x+ 1, if x > 0.

Solution: Here the domain is divided into two pieces, x ≤ 0 and x > 0. We need to sketchthe graph of y = x2 for x ≤ 0 and y = x+ 1 for x > 0.

The boundary value of x = 0 leads to twoanchor points, (0, 0) from y = x2 and (0, 1)from y = x + 1. Since x = 0 belongs to the�rst piece of the domain, the point (0, 0) is in-cluded on its graph. This is indicated on thegraph by a closed circle at (0, 0). Since x = 0is not in the second piece of the domain,the point (0, 1) is not included on the graph.This is indicated by an open circle at (0, 1).Note that the graph passes the verticalline test. Remember this is the graph of afunction. Df = (−∞,∞), Rf = [0,∞).

Example 9: Graph and �nd the domain and range of g(x) =

−x− 2, if x ≤ −1

−x2, if −1 < x < 1.

1, if x ≥ 1

Solution:Here the domain is divided into three pieces,x ≤ −1, −1 < x < 1 and x ≥ 1. The an-chor points are (−1,−1) for the �rst piece,(−1,−1) and (1,−1) for the second piece,and (1, 1) for the third piece. Note that thereis no open or closed circle at (−1,−1) sincethe �rst and second piece come together atthat point and the closed circle from the left�lls in the open circle from the right. Notethat the graph passes the vertical linetest. Dg = (−∞,∞), Rg = [−1,∞).


Remark: Besides shifts and re�ections, there are other transformations that are sometimesused in graphing. For example, replacing y = f(x) by y = kf(x) , k > 0, results in thecompression or expansion of the graph of f(x). If 0 < k < 1, an vertical compression occurs(movement towards the x-axis). If k > 1, a vertical expansion occurs. In the �gure below,

compare the graphs of y = 2x2, y = x2 and y =1

2x2. While these transformations change

the size of the graph, the shape remains the same.

-3 -2 -1 1 2 3

-1

1

2

3

4

x2

2

x2

2 x2


Exercises 1. 3

1. The graph of f is given. Sketch the following transformations of the graph of f .

f

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

(a) y = f(x) + 1 (d) y = f(x− 2)− 1

(b) y = f(x+ 1) (e) y = |f(x)|(c) y = −f(x)

2. Each graph below is a transformation of the graph of one of the basic functions coveredin this section. Find an equation for the given graph.

(a) (b) (c)

1

11

1

1

1

(d) (e) (f)

1

1

1

1

1

1

In Exercises 3 - 10, graph the given function. Determine the domain and range of the functionand any intercepts.

3. f(x) = −|x|+ 3 4. g(x) = (x− 2)2 − 2 5. h(x) = (x+ 2)3

6. F (x) = 3√x+ 1 7. G(x) = −

√x+ 2 8. H(x) =

√x− 1 + 4

9. f(x) = |2x+ 3| 10. g(x) = 4− x2


In Exercises 11 - 14, sketch the piecewise function. State its domain and range.

11. f(x) =

1 if x ≤ −1

−x if −1 < x < 1√x if x ≥ 1

12. g(x) =

−2x− 6 if x < −2

−2 if −2 ≤ x < 3

6x− 20 if x ≥ 3

13. h(x) =

2x, if x ≤ −1

x2 − 1, if −1 < x ≤ 2

5− x, if x > 2

14. F (x) =

−1, if x < 0

0, if x = 0

1, if x > 0

1. 4 Rational Functions and Asymptotic Behavior 29

1. 4 Rational Functions and Asymptotic Behavior

Negative Powers of x

The most basic rational function besides polynomials is f(x) =1

x. The domain of this

function is x ̸= 0. Its behavior as x approaches 0 or as x approaches ±∞ is very di�erentfrom the behavior of polynomial functions as demonstrated by the following tables.

As x → 0+,1

x→ ∞

x 0.1 0.01 0.001 0.0001

1

x10 100 1000 10000

As x → ∞,1

x→ 0

x 10 100 1000 10000

1

x0.1 0.01 0.001 0.0001

Notation: x → 0+ means that x approaches 0 from the right.

f(x) → ∞ means that the values of f(x) are approaching positive in�nity,that is, the values become very large and positive.

f(x) =1

xis an odd function (why?). So its graph is symmetric with respect to the origin.

Therefore, as x → 0−, read "as x approaches 0 from the left",1

x→ −∞, and as x →

−∞,1

x→ 0. For example, f(−.0001) = −10000, whereas f(−10000) = −0.0001. This

behavior is exhibited in the graph in Figure 1.

-1 1

-1

1

Figure 1 Figure 2

Df = (−∞, 0) ∪ (0,∞) Dg = (−∞, 0) ∪ (0,∞)

Rf = (−∞, 0) ∪ (0,∞) Rg = (0,∞)

Figure 2 is the graph of g(x) =1

x2. It exhibits behavior similar to f(x) =

1

x: as x →

0+,1

x2→ ∞ and as x → ∞,

1

x2→ 0. However, there are some di�erences.

1

x2goes to


in�nity much faster than1

xas x → 0+ and

1

x2goes to 0 much faster than

1

xas x → ∞.

Also, since g(x) =1

x2is an even function (why?), as x → 0−, g(x) = 1

x2→ +∞ instead of

−∞.

In general, the graph of y =1

xn, n a positive integer, looks like the graph of

1

xif n is odd

and like1

x2if n is even.

Vertical Asymptotes

De�nition: The vertical line x = a is called a vertical asymptote of thegraph of a function y = f(x) if f(x) → +∞ or f(x) → −∞as x → a+ or x → a−.

For example, the vertical line x = 0 is a vertical asymptote for the graphs of both y =1

x

and y =1

x2. It is the only vertical asymptote for these functions.

Suppose f(x) =p(x)

q(x)is a rational function where p(x) and q(x) are polynomials and the

fraction is reduced. Then the vertical asymptotes of the graph of f are the lines x = awhere q(a) = 0. As x → a from either direction, the denominator q(x) → 0, whereas thenumerator, p(x) approaches a non-zero value. Dividing a non-zero number by a very smallnumber results in a very large answer, either positive or negative. Hence, x = a is a verticalasymptote of the graph of f .

To �nd vertical asymptotes : Reduce the rational expression if not already reduced. Thenset the denominator equal to 0 to �nd the values of a. Thevertical asymptotes are the lines x = a.

Example 1: Find the vertical asymptote(s) and intercepts of the f(x) =x2 + 1

x2 − 1.

Solution: This fraction is already reduced. So setting x2 − 1 = 0 yields x = −1 or x = 1.Hence, the graph of this function has two vertical asymptotes, x = −1 and x = 1. To �nd thex-intercept, set x2+1 = 0. But this equation has no real solutions, so there is no x-intercept.

To �nd the y-intercept, compute f(0) =0 + 1

0− 1= −1. So the y-intercept is (0,−1).


Behavior at In�nity and Horizontal Asymptotes

De�nition: The horizontal line y = b is called a horizontal asymptoteof the graph of a function y = f(x) if f(x) → b as x → +∞or x → −∞.

For example, the horizontal line y = 0 is a horizontal asymptote for the graphs of both y =1

x

and y =1

x2. It is the only horizontal asymptote for these functions (see Figures 1 and 2).

In order to understand behavior of rational functions at in�nity, we must �rst understandbehavior of polynomials. For example, consider p(x) = x3 − 3x2 +2x− 1. Factoring out theleading term, x3, we obtain

p(x) = x3

(1− 3

x+

2

x2− 1

x3

).

As x → ±∞, all the terms in the parentheses other than 1 approach 0 and so the expressionin the parentheses approaches 1. But this means that p(x) behaves like x3 for large x, bothpositive and negative. As x → +∞, x3 → +∞ and so does p(x), and as x → −∞, x3 → −∞and so does p(x). So this polynomial does not have any horizontal asymptotes. This is truefor any non-constant polynomial.

More generally, if a polynomial p(x) = anxn+an−1x

n−1+ · · ·+a0, thenp(x)

anxnis very close to

1 when |x| is large. This implies that p(x) behaves like the leading term anxn when x → +∞

or x → −∞.

We will write p(x) ∼ anxn to indicate that

p(x)

anxn→ 1 as x → +∞ or x → −∞.

Hence for any rational function f(x) = p(x)/q(x),

f(x) =p(x)

q(x)=

anxn + · · ·+ a0

bmxm + · · ·+ b0∼ anx

n

bmxm.

Thus, to understand asymptotic behavior of f(x) at in�nity, we simply need to look at thequotient of the leading terms of the numerator and the denominator.

Example 2: Find the horizontal asymptote(s) of the graphs of the following functions:

(a) f(x) =3x− 2

2x+ 1(b) g(x) =

x4 + 1

x2 − 3(c) h(x) =

6x+ 5

x2 + 1

Solution:

(a) f(x) =3x− 2

2x+ 1∼ 3x

2x=

3

2. This means that f(x) → 3/2 as x → ±∞. So the horizontal

asymptote is the line y = 3/2.


(b) g(x) =x4 + 1

x2 − 3∼ x4

x2= x2 → ∞ as x → ∞ and as x → −∞. Therefore, the graph of

g(x) has no horizontal asymptotes.

(c) h(x) =6x+ 5

x2 + 1∼ 6x

x2=

6

x→ 0 as x → ±∞. So the horizontal asymptote is the line

y = 0.

Transformations of y =1

x

Add the graph of y =1

xand y =

1

x2to the list of basic graphs that you should know. Here

we give an example of a transformation of y =1

x.

Example 3: Use transformations of y =1

xto sketch the graphs of (a) f(x) =

1

x− 1+ 2

and g(x) = −1

x− 1. Label any asymptotes. Find and plot any intercepts. Determine the

domain and range.

(a) HA: y = 2

VA: x = 1

-3 -2 -1 1 2 3 4 5

-1

1

2

3

4

This graph is obtained from the graph ofy = 1/x by shifting up by 2 and to the rightby 1. Notice the corresponding shifts in theasymptotes. The vertical asymptote is x = 1and the horizontal asymptote is y = 2. Theintercepts are (1/2, 0) and (0, 1).

Df = (−∞, 1) ∪ (1,∞)Rf = (−∞, 2) ∪ (2,∞).


(b)

HA: y = -1

VA: x = 0

-4 -3 -2 -1 1 2 3 4

-5

-4

-3

-2

-1

1

2

3

This graph is obtained from the graph of y =1/x by re�ecting across the x-axis and shiftingdown by 1. The vertical asymptote is x =0 and the horizontal asymptote is y = −1.The x-intercept is (−1, 0) and there is no y-intercept.

Df = (−∞, 0) ∪ (0,∞)Rf = (−∞,−1) ∪ (−1,∞).

Rational functions can have several vertical asymptotes (or possibly none at all) but atmost one horizontal asymptote. However, in general, a function can have zero, one, ortwo horizontal asymptotes. Two horizontal asymptotes occur when f(x) approaches twodi�erent values, one as x → ∞ and another as x → −∞. An example of such a function is

f(x) =x√

x2 + 1. f(x) → 1 as x → ∞ and f(x) → −1 as x → −∞.


Exercises 1. 4

In Exercises 1 - 5, use transformations of y =1

xand y =

1

x2to sketch the graph of the given

function. Label any asymptotes. Find and plot any intercepts. Determine the domain andrange and state these in interval notation.

1. f(x) =1

x+ 1 2. g(x) =

1

x+ 1− 2 3. h(x) = −1

x− 3

4. F (x) =1

(x+ 2)25. G(x) =

2x− 1

x− 1[Hint: Use long division �rst.]

In Exercises 6 - 11, �nd the vertical and horizontal asymptotes of the graph of the givenfunction. Determine any intercepts.

6. f(x) =x+ 2

x− 47. g(x) =

x− 3

x2 − 3x+ 28. h(x) =

x2

x+ 1

9. F (x) =2x2 + 1

x2 − 410. G(x) =

3x+ 6

x2 + x11. H(x) =

8x3 + 1

x2 + 4

12. Match the functions with their graphs in the �gure. Hint: Use intercepts and asymptotesto make your decisions.

(a) y =1

x2 − 1(b) y =

x2

x2 + 1(c) y =

1

x2 + 1(d) y =

x

x2 − 1

I II III IV

13. Sketch the graph of a rational function f that satis�es the following conditions:

(a) The graph of f has a horizontal asymptote y = 2.

(b) The graph of f has a vertical asymptote x = 1.

(c) As x → 1−, f(x) → ∞.

(d) As x → 1+, f(x) → −∞.

1. 5 Composition of Functions 35

1. 5 Composition of Functions

In Section 1.1, we noted that x in a function f(x) can be viewed as a placeholder and thatany value or expression can be inputted in its place. When the input is itself a function, thisprocess is called composition of functions.

De�nition: Let f and g be functions of x. Then the function, (f ◦ g)(x), read "fcomposed with g of x", is de�ned as

(f ◦ g)(x) = f(g(x)).

In order to �nd the formula for (f ◦ g)(x), replace x by g(x) in the formula for f . SeeExamples 1− 3. In order to evaluate f(g(x)) at a speci�c value of x = a, we must �rst �ndg(a) and then �nd f(g(a)). See Example 4.

x g−−−→ g(x) f

−−−→ f(g(x))

For (f ◦ g)(x) to be de�ned, both g(x) and f(g(x)) must be de�ned. Therefore, x must bein the domain of g and g(x) must be in the domain of f . So the domain of f ◦ g consists ofall x in the domain of g such that g(x) is in the domain of f .

Example 1 Let f(x) = x2 + 1 and g(x) = 2x − 1. Find the following functions anddetermine their domains.

(a) (f ◦ g)(x) (b) (g ◦ f)(x) (c) (f ◦ f)(x) (d) (g ◦ g)(x)

Solution:

(a) (f ◦ g)(x) = f(g(x)) = f(2x− 1) = (2x− 1)2 + 1 = 4x2 − 4x+ 1 + 1 = 4x2 − 4x+ 2.

The functions f and g are de�ned for all real numbers. So for any real number x, x isin the domain of g and g(x) is in the domain of f . Therefore, the domain of f ◦ g is theset of all real numbers, that is, Df◦g = (−∞, ∞).

(b) (g ◦ f)(x) = g(f(x)) = g(x2 + 1) = 2(x2 + 1) − 1 = 2x2 + 2 − 1 = 2x2 + 1 andDg◦f = (−∞, ∞).

Note that the order of composition is important. In general, f ◦ g ̸= g ◦ f .

(c) (f ◦ f)(x) = f(f(x)) = f(x2 + 1) = (x2 + 1)2 + 1 = x4 + 2x2 + 1+ 1 = x4 + 2x2 + 2 andDf◦f = (−∞, ∞).

(d) (g ◦ g)(x) = g(g(x)) = g(2x − 1) = 2(2x − 1) − 1 = 4x − 2 − 1 = 4x − 3 and Dg◦g =(−∞, ∞).

Another way to �nd the domain of f ◦ g is to �nd the domain of the expression obtainedfrom the composition before simpli�cation.


Example 2 Find f ◦ g and g ◦ f and their domains for the following pairs of functions.

(a) f(x) = 2x2 − 4, g(x) =√x+ 2 b) f(x) =

x− 1

x+ 1, g(x) =

1

x

Solution:

(a) (f ◦ g)(x) = f(g(x)) = f(√

x+ 2)= 2

(√x+ 2

)2 − 4 = 2(x+ 2)− 4 = 2x.

The domain of unsimpli�ed expression 2(√

x+ 2)2 − 4 is x ≥ −2. Hence, the domain

of f ◦ g is x ≥ −2.

(g ◦ f)(x) = g(f(x)) = g (2x2 − 4) =√(2x2 − 4) + 2 =

√2x2 − 2.

To �nd the domain of√(2x2 − 4) + 2 , we must solve the quadratic inequality 2x2−2 ≥ 0

to get that Dg◦f = (−∞,−1] ∪ [1,∞).

(b) (f ◦ g)(x) = f(g(x)) = f

(1

x

)=

1

x− 1

1

x+ 1

=

1− x

x1 + x

x

=1− x

1 + x

From the �rst expression, we see that x ̸= 0 and that1

x+ 1 ̸= 0. But

1

x+ 1 ̸= 0 if and

only if1

x̸= −1. By cross-multiplying and solving for x, we see that x ̸= −1. Hence,

the domain of f ◦ g is all real numbers except x = −1 and x = 0.

Example 3 Composition of Three Functions

Let f(x) = x2 + 1, g(x) =1

x, and h(x) =

√x− 1. Find g ◦ h ◦ f .

Solution:

(g ◦ h ◦ f) = g (h(f(x)) = g(h(x2 + 1)) = g(√

(x2 + 1)− 1)= g(

√x2) = g(|x|) = 1

|x|.

Example 4 Evaluating Compositions of Functions

Let f(x) =√2x− 5 and g(x) = x3 − 4. Find (f ◦ g)(2), (g ◦ f)(7) and (g ◦ g)(1).

Solution:

(f ◦ g)(2) = f(g(2)). g(2) = 23 − 4 = 4 and f(g(2)) =√2(4)− 5 =

√3.

(g◦f)(7) = g(f(7)). f(7) =√2 · 7− 5 =

√9 = 3 and g(f(7)) = g(3) = 33−4 = 27−4 = 23.

(g ◦ g)(1) = g(g(1)). g(1) = 13 − 4 = −3 and g(g(1)) = (−3)3 − 4 = −27− 4 = −31.

Example 5 Recognizing the Composition of Two Functions

For each of the following, �nd two functions f(x) and g(x) such that h(x) = (f ◦ g)(x) andf(x) is a power of x.

(a) h(x) = (2x2 + 100)3

(b) h(x) =√x+ 2 (c) h(x) =

1

(x3 − 2)2

Solution: (a) f(x) = x3 and g(x) = 2x2 + 100; (b) f(x) = x1/2 and g(x) = x + 2; (c)f(x) = x−2 and g(x) = x3 − 2.


Exercises 1. 5

For each pair of functions f and g, �nd f ◦ g, g ◦ f , and their domains.

1. f(x) = 6x− 1, g(x) = −2x+ 5 2. f(x) = 5x2 − 2x+ 1, g(x) = x− 2

3. f(x) =4

x, g(x) =

1

x− 54. f(x) =

2x

x+ 3, g(x) =

5

x

5. f(x) =√x, g(x) = 2x+ 1 6. f(x) = x2 + 1, g(x) =

√x− 5

7. f(x) = | x− 5| , g(x) = 2x+ 5 8. f(x) = 3x− 1, g(x) =x+ 1

3

9. f(x) =1

x− 1, g(x) = 1 +

1

x

10. Given f(x) = x2 + x and g(x) = x − 3, use the de�nition of the composition of twofunctions to �nd the following values.

(a) (f ◦ g)(0) (b) (g ◦ f)(0) (c) (f ◦ f)(0) (d)(g ◦ g)(0) (e) [(fg) ◦ (f + g)] (0)

11. Let f and g be functions some of whose values are given in the table below. Find (a)-(e).

x −3 −2 −1 0 1 2 3

f(x) 0 −3 −1 2 3 1 −2

g(x) 3 1 5 2 −3 −4 0

(a) (g ◦ f)(−2) (b) (f ◦ g)(3) (c) (g ◦ g)(0) (d) (f ◦ f)(3) (e) (f ◦ f ◦ g)(3)

12. Let f(x) = x2 − 1, g(x) =√x+ 1, and h(x) =

1

x+ 2. Find the following.

(a) f ◦ f (b) f ◦ g (c) f ◦ h (d) g ◦ f (e) g ◦ g (f) g ◦ h

(g) h ◦ f (h) h ◦ g (i) h ◦ h (j) f ◦ g ◦ h (k) g ◦ f ◦ h (l) h ◦ g ◦ f

13. For each of the following, �nd two functions f(x) and g(x) such that h(x) = (f ◦ g)(x)and f(x) is either a power of x or the absolute value of x.

(a) h(x) = (x− 5)4 (b) h(x) =√2x2 + 5 (c) h(x) =

1

3x− 1

(d) h(x) =3√

x3 − 1 (e) h(x) = |5x− 1| (f) h(x) =1

(x2 + 1)2

1. 6 Inverse Functions 38

1. 6 Inverse Functions

De�nition of Inverse Functions

Let f(x) = x3 and g(x) = 3√x. Then (f ◦ g)(x) = ( 3

√x)

3= x and (g ◦ f)(x) = 3

√x3 = x for

all real x. So in e�ect g reverses the action of f and f reverses the action of g.

De�nition: Let f(x) be a function with domain D and range R and g(x) be a functionwith domain R and range D. We say that g is the inverse function of fif for any x in D, g(f(x)) = x and for any x in R, f(g(x)) = x.

If f has an inverse, we say that f is invertible and we write f−1(x) for the inverse functionof f(x). So

(f ◦ f−1)(x) = f(f−1(x)) = x and (f−1 ◦ f)(x) = f−1(f(x)) = x.

Suppose that f has an inverse and y = f(x). Then f−1(y) = f−1(f(x)) = x. So if y = f(x),then x = f−1(y). Now if we interchange x and y in the equation x = f−1(y), we obtainy = f−1(x). This leads to the following procedure for �nding the inverse function.

Step 1: Set y = f(x).

Step 2: Solve for x in terms of y.

Step 3: Interchange x and y.

Example 1 Let f(x) = 3x− 2. Find f−1(x) and verify the answer.

Step 1: Set y = 3x− 2.

Step 2: y = 3x− 2 implies 3x = y + 2 and x =y + 2

3.

Step 3: y =x+ 2

3

Therefore, f−1(x) =x+ 2

3. Check:(f ◦ f−1)(x) = 3

(x+ 2

3

)− 2 = x + 2 − 2 = x and

(f−1 ◦ f)(x) = (3x− 2) + 2

3=

3x

3= x.

If (x, y) lies on the graph of f(x), then the point (y, x) lies on the graph of f−1(x). Forexample, (2, 8) lies on the graph of f(x) = x3 and so (8, 2) lies on the graph of f−1(x) = 3

√x.

Therefore, Df = Rf−1 and Rf = Df−1 .

Example 2 Let f(x) = 2 +1

x. Find f−1(x). Determine the domains and ranges of each

function, stating the answers in interval notation.


Step 1: Set y = 2 +1

x.

Step 2: y − 2 =1

xx(y − 2) = 1 x =

1

y − 2

Step 3: y =1

x− 2

Hence, f−1(x) =1

x− 2. Df = (−∞, 0)∪ (0,∞) = Rf−1 and Df−1 = (−∞, 2)∪ (2,∞) = Rf .

One-to-One Functions

Consider f(x) = x2. Solving y = x2 for x yields two solutions x =√y and x = −√

y.Interchanging x and y, we get y = ±

√x. But this equation does not represent a function

since there are two values of y for every x ̸= 0. Therefore, f(x) = x2 does not have an inverse.In general, if more than one value of x corresponds to any y in the equation y = f(x), thenf is not invertible.

De�nition: A function f is called one-to-one if for every y in the range of f , there isexactly one x in the domain of f such that y = f(x).

A function f is invertible if and only if f is one-to-one.

De�nition: A graph of a function represents a one-to-one function if and only if anyhorizontal line intersects the graph in at most one point.

Example 3 Determine which of the following graphs represent a one-to-one function.

(a) (b)

Solution:

(a) This function is not one-to-one since its graph does not pass the horizontal line test.

(b) This function is one-to-one since it passes the horizontal line test: every horizontal lineintersects the graph in either one point or none at all.


Inverse Functions and Their Graphs

Suppose f(x) is invertible. Then if (x, y) lies on the graph of f(x), the point (y, x) lies on thegraph of f−1(x). These two points are symmetric with respect to the line y = x. Therefore,the graphs of f(x) and f−1(x) are symmetric with respect to the line y = x.

Example 4 Let f(x) = x3 and f−1(x) = 3√x. Sketch the graphs of f and f−1, along with

the line y = x, on the same coordinate system. Find the domain and range of each function,stating the answers in interval notation.

Solution:

f-1

fy = x

-5 5

-5

5

(x, y) becomes (y, x)

Do this mentally when graphing.

f(x) = x3 f−1(x) = 3√x

(−2,−8) −→ (−8,−2)

(−1,−1) −→ (−1,−1)

(0, 0) −→ (0, 0)

(1, 1) −→ (1, 1)

(2, 8) −→ (8, 2)

Df = (−∞,∞) = Rf−1 and Df−1 = (−∞,∞) = Rf .

Example 5 The graph of f is given below. Sketch the graphs of f and f−1, along with theline y = x, on the same coordinate system. State the domain and range of each function.

f

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

Solution: The graph of f consists of straight line segments and by symmetry, so does thegraph of f−1(x). All we have to do is interchange the x- and y-coordinates of the endpoints of


each of the segments in the graph of f and then connect the dots with straight line segments.See the �gure below.

y = x

f

f-1

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

Df = [−3, 3] = Rf−1 and Rf = [−2, 3] = Df−1 .

Restricting Domains

We know that the graph of f(x) = x2 − 1 is a parabola and therefore f(x) = x2 − 1 is nota one-to-one function. However, if we restrict the domain of f to x ≥ 0 or to x ≤ 0, theresulting functions become one-to-one and hence are invertible.

For example, g(x) = x2 − 1, x ≥ 0 has inverse g−1(x) =√x+ 1. (Why?) Dg−1 = Rg =

[−1,∞) while Rg−1 = Dg = [ 0,∞).

For h(x) = x2 − 1, x ≤ 0, h−1(x) = −√x+ 1. (Why?) Dh−1 = Rh = [−1,∞) while

Rh−1 = Dh = (−∞, 0 ].

g-1HxL = x + 1

h-1HxL = - x+1


Example 6 The function f(x) =√x − 1 is one-to-one. Find f−1 and verify the answer.

Sketch the graphs of f and f−1, along with the line y = x, on the same coordinate system.Find the domain and range of each function, stating the answers in interval notation. Also,�nd all intercepts.

Solution: To �nd f−1, we follow the steps outlined previously. However, there is a domainissue with f−1. The range of f is [−1,∞) and therefore the domain of f−1 must also be[−1,∞). If the domain is not restricted, the resulting function would not be one-to-one.

Step 1: Set y =√x− 1, y ≥ −1

Step 2: y + 1 =√x implies x = (y + 1)2, y ≥ −1

Step 3: y = (x+ 1)2 , x ≥ −1.

Check: (f ◦ f−1)(x) =√

(x+ 1)2 − 1 = x + 1 − 1 = x for x ≥ −1 and (f−1 ◦ f)(x) =

((√x− 1) + 1)

2= x.

Remark:√(x+ 1)2 = x + 1 only if x + 1 ≥ 0, that is, for x ≥ −1. In general,√(x+ 1)2 = |x+ 1| =

{−(x+ 1) if x < −1

x+ 1 if x ≥ −1.

f -1

f

y = x

-2 2 4

-4

-2

2

4

Df = [ 0,∞) = Rf−1 and Rf = [−1,∞) = Df−1

To �nd the y-intercept of f(x), compute f(0) =√0 − 1 = −1. Hence, the y-intercept for

f(x) is (0,−1). Similarly, f−1(0) = (0 + 1)2 = 1 and so the y-intercept of f−1(x) is (0, 1).

To obtain the x-intercepts of the two functions, one can solve the equations f(x) = 0 andf−1(x) = 0. However, there is an easier way to obtain these intercepts. Since any point (x, y)on the graph of a function will become (y, x) on the graph of its inverse, the x-intercept forf(x) is (1, 0) and the x-intercept for f−1(x) is (−1, 0).


Exercises 1. 6

1. Determine which of the following graphs represent a one-to-one function.

(a) (b) (c) (d)

2. Use the graphs of the functions below to sketch the graph of their inverses. State thedomain and range of each function.

-3 -2 -1 1 2 3

-3

-2

1

2

3

H1,0L

(a) (b)

3. (a) Suppose f(2) = 5. Find f−1(5).

(b) Suppose g−1(3) = −2. Find g(−2).

(c) Find h(h−1(1)) and h−1(h(7)).

In Exercises 4 - 10, the function f is one-to-one. Find f−1 and verify your answer by showingthat (f ◦ f−1)(x) = x and (f−1 ◦ f)(x) = x. Find the domain and range of each function,stating the answers in interval notation. Do not forget to restrict the domain of f−1 whenneeded.

4. f(x) = 3− 2x 5. f(x) = (x+ 1)3 − 2 6. f(x) = 3√2x+ 1

7. f(x) =√x+ 1 8. f(x) = x2 + 2, x ≥ 0 9. f(x) =

1

x+ 1

10. f(x) =1− x

3x+ 2

In Exercises 11-14, the function f is one-to-one. Find f−1. Sketch the graphs of f and f−1,along with the line y = x, on the same coordinate system. Find the domain and range ofeach function, stating the answers in interval notation. Find all intercepts.

11. f(x) = 2x+ 1 12. f(x) = x3 + 1

13. f(x) =√x− 2 + 1 14. f(x) = (x− 1)2 − 2, x ≥ 1

Chapter 1 Additional Practice Problems 44

Chapter 1 Additional Practice Problems

1. Evaluate the following functions at the indicated values. Simplify where possible.

a) f(x) = −2x2 + x− 1 b) g(x) =x

2x+ 1

f(−1), f(x− 1), f(2x) g(1/2), g(2− x), g

(1

x

)2. Find

f(x+ h)− f(x)

h, h ̸= 0 for the following functions. Simplify fully.

a) f(x) =1

x+ 3b) f(x) = −3x2 + 4x

3. Find the domains of the following functions. State your answers in interval notation.

a) f(x) = x2 + 2 b) g(x) =x− 1

x3 − xc) h(x) =

√3x− 12

d)F (x) =1√

4− x2

4. Let f(x) =x+ 1

xand g(x) =

1

x. Find f + g, f − g, fg, f/g, g/f , and their domains.

5. For each of the functions whose graphs are given below, determine

a) the domain e) all x for which f(x) = −3

b) the range f) the intervals on which f(x) > 0, if any

c) the intercepts, if any g) the intervals on which f(x) < 0, if any

d) f(3)

I II

-6 -4 -2 2 4 6 8

-2

-1

1

2

3

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

4


6. For each function, determine

a) the domain d) all x for which f(x) = −2

b) the intercepts, if any e) the intervals on which f(x) > 0, if any

c) f(−2) f) the intervals on which f(x) < 0, if any

i) f(x) = −3x2 + 5x ii. f(x) =x+ 2

x− 3

7. For each of the functions whose graphs are given below, determine (a) the intervals onwhich the function is increasing, (b) the intervals on which the function is decreasing, (c) thevalues of x where f has a local maximum, (d) the values of x where f has a local minimum,(e) if f is even, odd, or neither.

I II

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

-2 -1 1 2

-2

-1

1

2

8. Determine if the following functions are even, odd, or neither.

a) f(x) = 3x2 + 5 b) g(x) =x

x2 − 1c) h(x) =

x2

x2 + 9

9. Find the points of intersection of the graphs of the two functions given.

a) y = 3x− 5 ; y = 4− 2x b) y = x2 − 1 ; y = 2x+ 2

10. Graph f(x + 1) − 2 and f(x − 2) + 1 for the (a) f(x) = |x| and (b) f(x) =1

x. Label

any intercepts and asymptotes.

11. Graph g(x) =

1

x+ 1, if x < −1

√x+ 1− 1, if x ≥ −1

, labeling any intercepts and asymptotes.

Find g(−5), g(−1) and g(5).


12. Find the domains and intercepts of the following functions. Determine the vertical andhorizontal asymptotes.

a) f(x) =x2 + 1

x4 − 1b) g(x) =

2x2 − 1

3x2 − 5x− 2

13. For each pair of functions f and g, �nd f ◦g, g ◦f , and their domains. Find (f ◦g)(−4)and (g ◦ f)(2) for each pair of functions.

a) f(x) = x+ 1, g(x) = x2 + 4 b) f(x) =3

x− 1, g(x) =

2

x

c) f(x) =√x, g(x) = 2x2 + 3

14. For each one-to-one function below, �nd f−1 and verify your answer by showing that(f ◦ f−1)(x) = x and (f−1 ◦ f)(x) = x. Determine the domain and range of f and f−1 andthe intercepts of both functions. Graph f and f−1 on the same grid, as well as the liney = x, for parts (a) and (b).

a) f(x) = x3 − 1 b) f(x) =√x+ 1 c) f(x) =

2x− 3

x+ 4

2. 1 Exponential Functions and Their Graphs 47

2. 1 Exponential Functions and Their Graphs

Review of Exponential Notation

In this section, we will de�ne f(x) = ax for a > 0 and x a real number. For a positive integern,

an = a · a · · · a · a︸︷︷︸n times

.

For example, a2 = a · a, a3 = a · a · a, a4 = a · a · a · a, and so on.

For positive integers n and m,

a1/n = n√a and am/n = n

√am =

(n√a)m

.

This de�nes ax for all positive rational numbers x.

a0 = 1, by convention. As for negative rational powers, we de�ne a−x =1

ax. So for positive

integers n and m,

a−n =1

anand a−m/n =

1n√am

.

If x is irrational, x can be approximated by a sequence of rational numbers, x1, x2, x3, . . .The values ax1 , ax2 , ax3 , . . . then approximate the value of ax. Such a process of approxi-mation is called a limit. For example, the irrational number

√2 = 1.414213562 . . . Then

the rational numbers x1 = 1, x2 = 1.4, x3 = 1.41, . . . , x10 = 1.414213562, and so on, ap-proximate the exact value of

√2 with an increasing degree of precision. Hence, the values

a1, a1.4, . . . , a1.414213562, and so on, give progressively better approximations for the exactvalue of a

√2.

We have now de�ned ax, a > 0 for all real numbers x. Observe that for a > 0,

ax > 0 for all x,

since products, quotients and roots of positive numbers are always positive.

However, we can not do the same if a ≤ 0. a−n =1

anis not de�ned for a = 0 and a1/n is not

de�ned when n is even for a < 0. Therefore, we only consider functions f(x) = ax whena > 0.

Furthermore, for a = 1, 1x = 1 for all x. No matter how many times you multiply or divide1 by itself or take roots of 1, the result is still 1. So the function f(x) = 1x is simply theconstant function f(x) = 1. Therefore this function is not considered to be an exponentialfunction.


Laws of Exponents

You have learned the laws of exponents for rational exponents in algebra. It can be shownthat the same laws apply to all real exponents. You must know these laws and be ableto use them.

If x, y are real numbers, m,n are positive integers, and a, b > 0, then:

ax · ay = ax+y ax

ay= ax−y (ax)y = axy a0 = 1 a1 = a

a−x =1

ax=

(1

a

)x

(ab)x = ax · bx(ab

)x=

ax

bx1x = 1

a1/n = n√a and am/n = n

√am = ( n

√a)

m

Example 1 Use the laws of exponents to simplify the following expressions: (a) 2 x−4·2 4−2x,

(b)2 3x+1

4x−1, (c)

√3x, (d)

(3√x · 5

√x)√x

.

Solution

(a) 2 x−4 · 2 4−2x = 2x−4+4−2x = 2−x

(b)2 3x+1

4x−1=

2 3x+1

(22)x−1=

2 3x+1

22x−2= 23x+1−(2x−2) = 2x+3

(c)√3x = (3x)

12 = 3x·

12 = 3x/2

(d)(3√x · 5

√x)√x

= (3√x)

√x · (5

√x)

√x = 3x · 5x = (3 · 5)x = 15x

Exponential Functions and Their Graphs

De�nition: An exponential function is a function of the form f(x) = ax,where a > 0, a ̸= 1. In the expression ax, a is called the baseand x is called the exponent.

We will consider two di�erent cases, a > 1 and 0 < a < 1.

a > 1 :

∗ If we raise a, a > 1, to two di�erent exponents, the bigger exponent will result in abigger value of ax. For example, 23 = 8 < 24 = 16. Therefore, f(x) = ax is an increasingfunction when a > 1.

∗ As x → ∞, ax → ∞. In fact, f(x) = ax grows very rapidly. For example, 210 = 1024and 2100 ≈ 1.26765 × 1030. This type of exponential function grows much faster thanany power of x.


∗ As x → −∞, ax → 0. For x < 0, x = −|x| and ax = a−|x| =1

a|x|. As x → −∞, |x|

becomes large and positive and so does a|x|. Hence,1

a|x|is a very small positive number.

For example, 2−10 =1

210=

1

1024= 0.000976563 and 2−100 =

1

2100≈ 7.88861× 10−31.

∗ Since ax → 0 as x → −∞, f(x) = ax has a (one-sided) horizontal asymptote, y = 0, onthe left.

∗ Finally, a0 = 1. Therefore, for a > 1, the graph of f(x) = ax has the following shape.

H0,1L

f HxL= ax, a > 1

Properties of the graph of f(x) = ax, a > 1

a. Df = (−∞,∞) and Rf = (0,∞)

b. Increasing and hence one-to-one

c. No x-intercept; y-intercept (0, 1)

d. Horizontal asymptote y = 0 on the left;

no vertical asymptote

0 < a < 1 : Let b = 1/a. Then b > 1 and b−x =1

bx=

(1

b

)x

= ax. Therefore, the

graph of f(x) = ax is the same as the graph of b−x, b > 1. So to obtain this graph, we cansimply take the graph of g(x) = bx and re�ect the graph over the y-axis. This implies thatf(x) = ax, 0 < a < 1 is a decreasing function.

As x → ∞, ax → 0. Hence, f(x) = ax has a (one-sided) horizontal asymptote, y = 0, on theright. And as x → −∞, ax → ∞. Finally, a0 = 1. Therefore, for 0 < a < 1, the graph off(x) = ax has the following shape.

H0,1L

f HxL= ax, 0 < a < 1

Properties of the graph of f(x) = ax, 0 < a < 1

a. Df = (−∞,∞) and Rf = (0,∞)

b. Decreasing and hence one-to-one

c. No x-intercept; y-intercept (0, 1)

d. Horizontal asymptote y = 0 on the right;

no vertical asymptote


Example 2 Sketch the graphs of f(x) = 2x and g(x) =

(1

2

)x

on the same coordinate

system.

Solution Since 2 and 1/2 are rational numbers, we can easily compile a table of values for

both f(x) = 2 x and g(x) =

(1

2

)x

. We will see that the graph of g(x) =

(1

2

)x

= 2−x is

simply the graph of f(x) = 2 x re�ected across the y-axis.

f(x) = 2 x

x y

−3 1/8

−2 1/4

−1 1/2

0 1

1 2

2 4

3 8

g(x) = 2−x

x y

−3 8

−2 4

−1 2

0 1

1 1/2

2 1/4

3 1/8H0,1L

f HxL= 2xgHxL= 2-x

-3 -2 -1 1 2 3

2

4

6

8

Example 3 Use transformations to sketch the graphs of (a) f(x) = 3x+1 and (b) g(x) =3−x − 1. Identify any asymptotes and intercepts. State the domain and range of eachfunction.

Solution

(a) The graph of f(x) = 3x+1 is the graph of y = 3x shifted to the left by 1 unit. Since thereis no vertical shift, the horizontal asymptote, y = 0 remains the same, and there is stillno x-intercept. However, the y-intercept has changed to (0, 3) since f(0) = 30+1 = 3.Df = (−∞,∞) and Rf = (0,∞). See the �gure below on the left.

(b) The graph of g(x) = 3−x−1 is the graph of y = 3−x shifted down by 1 unit. This verticalshift down by 1 unit changes the horizontal asymptote to y = −1 and the y-interceptto (0, 0). Since an exponential function is a one-to-one function, (0, 0) is also the onlyx-intercept. Dg = (−∞,∞) and Rg = (−1,∞). See the �gure below on the right.


f HxL = 3x +1

-2 -1 1 2

3

6

9

12

15

18

21

24

27

gHxL= 3-x- 1

-2 -1 1 2

-2

2

4

6

8

The Number e

For reasons that will become clear when you take calculus, the most important exponentialfunction is f(x) = ex. The number e = 2.71828... can be obtained as the limit of the

sequence of rational numbers,

(1 +

1

n

)n

, n = 1, 2, 3, . . . Below is a small table of values of

the elements of this sequence.

n 1 +1

n

(1 +

1

n

)n

1 2 2

10 1.1 2.59374246

100 1.01 2.704813829

1000 1.001 2.716923932

1,000,000 1.000001 2.718280469

f(x) = ex is arguably the most important function in mathematics. In fact, any otherexponential function can be written in terms of ex and hence in higher mathematics, otherexponential functions are rarely used.

The number e is irrational, so it is impossible to plot points when graphing y = ex preciselyexcept for the y-intercept. Therefore, when graphing y = ex or a transformation of y = ex

by hand, you should concentrate on the basic shape of the graph, the horizontal asymptote,and any intercepts.


Example 4 Let f(x) = x2ex − 3xex. (a) Compute f(0), f(1), and f(−1). (b) Find thevalues of x where f(x) = 0. (c) Find the intervals on which f(x) > 0 and the intervals onwhich f(x) < 0.

Solution

(a) f(0) = 0, f(1) = e1 − 3 · e1 = −2e, and f(−1) = e−1 + 3e−1 = 4e−1.

(b) x2ex − 3xex = 0 implies ex(x2 − 3x) = 0. Since ex > 0 for all x, we have x2 − 3x = 0and so x(x− 3) = 0. Hence x = 0 or x = 3.

(c) Since ex > 0, the sign of ex(x2 − 3x) is the same as the sign of x2 − 3x. Using a signdiagram, we see that f(x) > 0 on (−∞, 0) ∪ (3, ∞) and f(x) < 0 on (0, 3).

+ + ++ + + - - -0 3

Some Exponential Equations

In Section 2.3, we will discuss a systematic procedure for solving exponential equations afterwe introduce the logarithmic functions in Section 2.2. In this section, we will only discussequations that can solved due to the following observation:

For a > 0, a ̸= 1, au = av if and only if u = v.

This is true because f(x) = ax is a one-to-one function which implies that every value of ax

comes from precisely one value of x.

Example 5 Solve the following equations: (a) 5x+2 = 25−2 · 125x , (b) (ex)2 · e5 = 3√ex,

(c)

(2 x2)2

4x= 16.

Solution In order to solve these equations using the observation above, we must transformeach equation to the form au = av. Notice that there is only one exponential expression oneach side and that the bases are the same. Otherwise, this method CAN NOT be applied.

(a) 5x+2 = 25−2 · 125x

5x+2 = (52)−2 · (53)x

5x+2 = 5−4 · 53x

5x+2 = 53x−4

x+ 2 = 3x− 4

x = 3

(b) (ex)2 · e5 = 3√ex

e2x+5 = ex/3

2x+ 5 =1

3x

5

3x = −5

x = −3

(c)

(2 x2)2

4x= 16

2 2x2

2 2x= 2 4

2 2x2−2x = 2 4

2x2 − 2x = 4

2(x− 2)(x+ 1) = 0

x = −1, x = 2


Exercises 2. 1

In Exercises 1-7, simplify the expressions.

1. 2 x+1 · 2 x−1 2.2 x+1

2 x−13.(3√2)−√

2

4. (ex)3 · e1−2x 5.3√ex2 6.

8x+1

2 3x

7. ex · e−x

In Exercises 8-10, write the expression in the form cax where c is a constant and a > 0.

8. (3x)1/2 9. 22x · 3x 10.3x+1

4x

11. Match each graph below to one of the following functions.

(a) y = ex (b) y = e−x (c) y = ex + 1 (d) y = ex − 1

(e) y = e−x + 1 (f) y = e−x − 1 (g) y = ex+1 (h) y = ex−1

(I) (II) (III)

-2 -1 1 2

-2

2

4

6

-2 -1 1 2

-2

2

4

6

-2 -1 1 2

-2

2

4

6

(IV) (V) (VI)

-2 -1 1 2

-2

2

4

6

-2 -1 1 2

-2

2

4

6

-1 1 2 3

-2

2

4

6

In Exercises 12-15, use transformations to sketch the graphs of the following functions.Identify any asymptotes and intercepts. Identify the domain and range of the functions.

12. f(x) = 2x − 2 13. g(x) = 2−x+1 = 2−(x−1)

14. h(x) = e−x − 1 15. F (x) = ex−2 + 1


In Exercises 16-19, �nd the exact solutions to each equation.

16. 2x = 8 17. 27 = 9−x 18. 3 x+2 · 9 x−1 =1

319.

ex2

e5x= e6

20. Suppose 3x = 2. Find (a) 3x+1 (b) 3−x (c) 32x (d) 3x/2 (e) 27x.

For the functions in Exercises 21-25, �nd

(a) f(0), f(1), and f(−1) (b) all values of x where f(x) = 0

(c) the intervals on which f(x) > 0 (d) the intervals on which f(x) < 0

21. f(x) = xe−x 22. f(x) = (3− x)2x 23. f(x) = xex + x

24. f(x) = (6− x− x2)e−x 25. f(x) = x2ex + 2xex


26. (ex + e−x)2 − (ex − e−x)

227. ex(e−x + 1) + e−x(ex + 1)

28. ex − 1− ex(1− e−x) 29. (ex − 1)(ex + 1)

2. 2 Logarithmic Functions and Their Graphs 55

2. 2 Logarithmic Functions and Their Graphs

Logarithmic Functions

Recall that for f(x) = ax, a > 0 and a ̸= 1, f(x) = ax is a one-to-one function. Therefore,it has an inverse. This inverse is called the logarithm to the base a of x, denoted byy = loga x and is de�ned by

y = loga x if and only if x = ay.

These are called the logarithmic form and the exponential form, respectively. Noticethat loga x is the EXPONENT to which one must raise a in order to obtain x.

By properties of inverse functions, f(f−1(x)) = x and f−1(f(x)) = x. So for f(x) = ax withf−1(x) = loga x,

aloga x = x and loga ax = x. (∗)

Notation: It is common practice to write log10 x as simply log x and loge x as lnx. lnx iscalled the natural log of x and is very widely used in calculus. Finally, x is called theargument of loga x.

Example 1 Find (a) log2 8, (b) log 100, 000, (c) ln√e, (d) log3

1

9, (e) loga 1, (f) e

ln 2, (g)

10log√5.

Solution

(a) Since 8 = 23, log2 8 = 3. (b) 100, 000 = 105. Therefore log 100, 000 = 5.

(c)√e = e1/2 and hence ln

√e = 1/2 (d)

1

9= 3−2, so log3

1

9= −2.

(e) Since 1 = a0, loga 1 = 0. (f) eln 2 = 2 by properties of inverses.

(g) 10log√5 =

√5 by properties of inverses.

Example 2 (a) Convert the exponential form 25 = 7x to logarithmic form. (b) Convertthe logarithmic form lnx = 2 to exponential form.

Solution

(a) The statement 25 = 7x is equivalent to the statement x = log7 25.

(b) The statement lnx = 2 is equivalent to the statement x = e2.

Example 3 Suppose that f(x) = ex + 1 and g(x) = ln(x− 2). Find f ◦ g and g ◦ f .

Solution(f ◦ g) = f(g(x)) = eln(x−2) + 1 = x− 2 + 1 = x− 1 since elnM = M .

(g ◦ f) = g(f(x)) = ln(ex + 1 − 2) = ln(ex − 1). This expression can not be simpli�edany further. The properties * do not apply to this expression.


Graphs of Logarithmic Functions

Example 4 Sketch the graphs of f(x) = 2x and g(x) = log2 x on the same coordinatesystem.

Solution As in the previous section, we can compile a table of values for f(x) = 2 x.By simply reversing the x− and y−values, we obtain the corresponding table of values forg(x) = log2 x. (Why?) Observe that the graph of g(x) = log2 x is simply the graph off(x) = 2 x re�ected across the line y = x.

f(x) = 2 x

x y

−3 1/8

−2 1/4

−1 1/2

0 1

1 2

2 4

3 8

g(x) = log2 x

x y

1/8 −3

1/4 −2

1/2 −1

1 0

2 1

4 2

8 3

f HxL= 2x

gHxL= Log 2x

y = x

-3 -1 1 3 5 7

-3

-1

1

3

5

7

From the graphs, we see that D2x = (−∞,∞) = Rlog2 x and R2x = (0,∞) = Dlog2 x. Thisagrees with what we know about inverses.

More generally, the domain of f(x) = ax is (−∞,∞) and hence the range of f−1(x) = loga xis also (−∞,∞). The range of f(x) = ax is (0,∞) and so the domain of f−1(x) = loga x is(0,∞). This means that loga x is only de�ned for x > 0. You must always remember thatthe log of zero or the log of a negative number is UNDEFINED.

We will only consider loga x for a > 1 since the functions loga x for 0 < a < 1 are virtuallynever used in mathematics. For a > 1, f(x) = ax is increasing and so is f−1(x) = loga x(compare the two graphs in Example 3 for a = 2). Furthermore, the graph of y = loga x isobtained from the graph of y = ax by interchanging x− and y−coordinates (compare thetwo tables in Example 3 for a = 2). This implies that the horizontal asymptote y = 0 forthe graph of y = ax becomes the vertical asymptote x = 0 for the graph of y = loga x andthe y−intercept (0, 1) for y = ax becomes the x−intercept (1, 0) for y = loga x.


H1,0L

f HxL= log ax, a > 1Properties of the graph of

f(x) = loga x, a > 1

a. Df = (0,∞) and Rf = (−∞,∞)

b. Increasing and hence one-to-one

c. No y-intercept; x-intercept (1, 0)

d. Vertical asymptote x = 0;

no horizontal asymptote

Example 5 Find the domains of the following functions: (a) f(x) = log2(x + 2), (b)g(x) = 1− lnx, (c) h(x) = ln(1− x2).

Solution Notice in Part (a), the argument of the logarithm is x + 2, as clearly indicatedby the parentheses. In Part (b), where there are no parentheses, the argument is simply x,and in Part(c), the argument is 1−x2. Since the logarithm is only de�ned for positive valuesof the argument,

to �nd the domain of a logarthmic function, we must determine when theargument is greater than zero.

Solution

(a) f(x) = log2(x+ 2) is de�ned when x+ 2 > 0, that is, x > −2. So Df = (−2,∞).

(b) g(x) = 1− lnx is de�ned when x > 0. So Dg = (0,∞)

(c) h(x) = ln(1−x2) is de�ned when 1−x2 > 0, that is, when (1−x)(1+x) > 0. By usingsign diagrams, we would see that Dh = (−1, 1).

Example 6 Use transformations to sketch the graphs of (a) f(x) = log2(x + 2) and (b)g(x) = ln x−2. Identify any asymptotes and intercepts. State the domain and range of eachfunction.

Solution

(a) The graph of f(x) = log2(x+ 2) is the graph of y = log2 x shifted to the left by 2 units.Since there is a horizontal shift, the vertical asymptote is now x = −2. This can also beseen by �nding the domain of f(x). In Example 3a, we saw that Df = (−2,∞).

To �nd the y-intercept, we must compute f(0). f(0) = log2(0 + 2) = log2 2 = 1, and sothe y-intercept is (0, 1). (We have a y−intercept here because of the shift to the left.)


To �nd the x-intercept, we must solve the equation log2(x + 2) = 0. This logarithmicexpression is equivalent to the exponential expression x+2 = 2 0. But 2 0 = 1 and hencex = 1− 2 = −1, and so the x-intercept is (−1, 0).

Finally, Df = (−2,∞) and Rf = (−∞,∞). See the �gure below to the left.

(b) The graph of g(x) = ln x − 2 is the graph of y = ln x shifted down by 2 units. Thisvertical shift down does not a�ect the vertical asymptote, and so it remains x = 0.Moreover, there is no y-intercept since lnx is not de�ned for x = 0. Setting g(x) = 0yields lnx− 2 = 0 and by converting to exponential form, we obtain x = e2 ≈ 7.39. Sothe x-intercept is (e2, 0). Furthermore, Df = (0,∞) and Rf = (−∞,∞). See the �gureto the right.

f HxL= log 2Hx+2L

-2 -1 1 2

-3

-2

-1

1

2

gHxL= ln x - 2

He2,0L

5 10 15 20 25 30

-3

-2

-1

1

2

Example 7 Let f(x) = x lnx − x. Find (a) the domain of f , (b) the values of x wheref(x) = 0, (c) the intervals on which f(x) > 0, and (d) the intervals on which f(x) < 0.

Solution

(a) lnx is only de�ned for x > 0. So Df = (0,∞).

(b) x lnx− x = 0 implies x(lnx− 1) = 0 and hence x = 0 or lnx = 1. But x = 0 is not inthe domain of f and must be rejected. Converting the logarithmic statement lnx = 1to the exponential form, we obtain x = e1 = e. Hence, f(x) = 0 when x = e.

(c) In Df , x > 0. Therefore, f(x) = x(lnx − 1) > 0 when lnx − 1 > 0. This occurs whenx > e. (Why?) So f(x) > 0 on (e,∞).

(d) f(x) = x(lnx− 1) < 0 when lnx− 1 < 0. This occurs when 0 < x < e. So f(x) < 0 on(0, e).


Example 8 By the Richter scale, the formula

M =2

3log

E

104.40

gives the magnitude M of an earthquake measured in terms of the energy released by theearthquake, E, in joules. Use this formula to answer the following questions. (a) What isthe magnitude of an earthquake which releases 5.96× 1016 joules? (b) How much energy isreleased by an earthquake of magnitude 7?

Solution

(a) When E = 5.96× 1016, M =2

3log

5.96× 1016

104.40= 8.25.

(b) When M = 7, 7 =2

3log

E

104.40. Then log

E

104.40=

21

2= 10.50. Converting this

expression to the exponential form, we obtainE

104.40= 1010.50 and E = 1010.50+4.40 =

1014.90 = 7.94× 1014 joules.

Recall that y = loga x if and only if x = ay. To use this fact, we isolated the logarithmicexpression above before converting the equation to its exponential form.


Exercises 2. 2

In Exercises 1-10, �nd the exact value of the expression.

1. log3 9 2. log21

643. log

√103

4. log 0.0001 5. log4 2 6. ln1

e7. 5log5 7 8. elnπ 9. ln 3

√e

10. log7 1

In Exercises 11-12, convert the exponential form to logarithmic form.

11. 23 = 8 12. 5x = 7

In Exercises 13-14, convert the logarithmic form to exponential form.

13. ln√e =

1

214. log3 x = −2

In Exercises 15-16, given f(x) and g(x), �nd f ◦ g and g ◦ f .

15. f(x) = ex − 3 ; g(x) = ln(x+ 2) 16. f(x) = ln x+ 2 ; g(x) = ex+1

In Exercises 17-21, �nd the domain of the function.

17. f(x) = ln(x− 1) 18. g(x) = ln(−x) 19. h(x) = ln

(x+ 1

x

)20. F (x) = ln(4− x2) 21.G(x) = ln |x|

In Exercises 22-24, use transformations to sketch the graph of the function. State the domainand range of each function. Identify any intercepts and asymptotes.

22. f(x) = log2(x− 1)− 2 23. g(x) = ln(x+ 3) 24. h(x) = ln(x+ 1) + 2

In Exercises 25-27, Find (a) the domain of f , (b) the values of x where f(x) = 0, (c) theintervals on which f(x) > 0, and (d) the intervals on which f(x) < 0.

25. f(x) = x lnx+ 2x 26. f(x) = x lnx− lnx 27. f(x) = x2 log(x+ 2)

28. By the Richter scale, the formula M =2

3log

E

104.40gives the magnitude M of an earth-

quake measured in terms of the energy released by the earthquake, E, in joules. Usethis formula to answer the following questions. (a) What is the magnitude of an earth-quake which releases 1013.4 joules? (b) Approximately how much energy is released byan earthquake of magnitude 7.25?

29. pH is a measure of the acidity or basicity of a solution. It can be approximated by theformula pH = − log[H+], where [H+] is the concentration of hydrogen ions in moles perliter. The pH level is essential to a pool's water balance. A pH reading of 7.0-7.6 isrequired, with 7.2 being the ideal level. (a) What is the pH of a pool's water for which[H+] is 10−7? (b) Determine the hydrogen ion concentration of the pH of the water ina pool with the ideal pH level of 7.2.

2. 3 Properties of Logarithms 61

2. 3 Properties of Logarithms

Properties of Logarithms

Any property of the exponential function has a corresponding property of the logarithmicfunction.

Suppose M = ak and N = aℓ, then MN = ak+ℓ. The equivalent logarithmic statementis: if loga M = k and loga N = ℓ, then loga(MN) = k + ℓ. Therefore, loga(MN) =loga M + loga N .

Similarly,M

N=

ak

aℓ= ak−ℓ implies loga

(M

N

)= k − ℓ and so loga

(M

N

)= loga M −

loga N .

Furthermore, sinceM = ak, thenM b =(ak)b

= akb for any real number b. Then loga(M b)=

bk and hence loga

(M b

)= b loga M .

We summarize the Properties of Logarithms as follows: for M > 0, N > 0, a > 0, a ̸= 1,

loga(MN) = loga M + loga N

loga

(M

N

)= loga M − loga N

loga Mb = b loga M

Example 1 Simplify the following expressions: (a) e2 ln 5, (b) log 4+ log 25, (c) 3log3 4−log3 8.

Solution

(a) e2 ln 5 = eln 52 = eln 25 = 25.

(b) log 4 + log 25 = log(4 · 25) = log 100 = 2.

(c) 3log3 4−log3 8 = 3log3(4/8) = 4/8 = 1/2.

Example 2 Given that ln 2 ≈ 0.69 and ln 3 ≈ 1.10, use the properties of logs to �nd the

approximate values of (a) ln 6, (b) ln 12, (c) ln(1/3), and (d) ln33√2.

Solution

(a) ln 6 = ln(2 · 3) = ln 2 + ln 3 ≈ 0.69 + 1.10 = 1.79

(b) ln 12 = ln(4 · 3) = ln(2 2 · 3) = ln 22 + ln 3 = 2 ln 2 + ln 3 ≈ 2(.69) + 1.10 = 2.48

(c) ln(1/3) = ln 1− ln 3 = 0− ln 3 ≈ −1.10. Equivalently, ln(1/3) = ln 3−1 = − ln 3 ≈ −1.10

(d) ln33√2= ln 3− ln 3

√2 = ln 3− ln 21/3 = ln 3− 1

3ln 2 ≈ 1.10− 1

3(0.69) = 0.87


Example 3 Suppose that loga u = 2 and loga v = 3, �nd (a) loga(uv), (b) loga(u3v4), (c)

logau

a2, and (d) loga

√a

uv.

Solution(a) loga(uv) = loga u+ loga v = 2 + 3 = 5

(b) loga(u3v4) = 3 loga u+ 4 loga v = 3(2) + 4(3) = 18

(c) logau

a2= loga u− loga a

2 = 2− 2 = 0

(d) loga

√a

uv= loga a

1/2 − loga u− loga v =1

2− 2− 3 = −9/2

Example 4 Write the following as a sum/di�erence of logarithms. Express all powers

as factors: (a) log((x + 1)2(2x + 1)3), x > −1/2, (b) ln

(x2ex

3√x2 + 1

), x > 0, and (c)

loga

√x+ 2

x− 3, x > 3.

Solution(a) log((x+ 1)2(2x+ 1)3) = log(x+ 1)2 + log(2x+ 1)3 = 2 log(x+ 1) + 3 log(2x+ 1)

(b) ln

(x2ex

3√x2 + 1

)= ln x2 + ln ex − ln(x2 + 1)1/3 = 2 ln x+ x− 1

3ln(x2 + 1)

(c) loga

√x+ 2

x− 3= loga

(x+ 2

x− 3

)1/2

=1

2loga

(x+ 2

x− 3

)=

1

2(loga(x+ 2)− loga(x− 3)) =

1

2loga(x+ 2)− 1

2loga(x− 3)

Example 5 Express 2 ln(x− 3) + ln x− 3 ln(x+ 2) as a single logarithm.

Solution

2 ln(x− 3) + ln x− 3 ln(x+ 2) = ln(x− 3)2 + ln x− ln(x+ 2)3

= ln(x(x− 3)2)− ln(x+ 2)3

= ln

(x(x− 3)2

(x+ 2)3

)In the �rst step, we brought the constant factors inside the log by making them expo-nents. In the next step, we made the sum of logs into the log of a product and �nally,we made the di�erence of logs into the log of a quotient. But notice that the middle stepcan be omitted by observing that if a term in the right-hand side of the �rst line has a+ sign, the argument of the log in that term will appear as a factor in the numeratorof the combined log. If a term has a − sign, the argument of the log for that term willappear as a factor in the denominator of the combined log.


Some Logarithmic Equations

In Section 2. 1, we saw that for a > 0, a ̸= 1, au = av if and only if u = v. Likewise, forM, N > 0,

logaM = loga N if and only if M = N .

This follows from the fact that f(x) = loga x is a one-to-one function. This property can beused to solve certain logarithmic equations. However, in order to use this property properlyto solve an equation, we may �rst need to use properties of logarithms to rewrite the equationin the form loga M = loga N .

Warning: Logarithmic functions have restricted domains. The argument of any log mustalways be positive. Therefore, after �nding a possible solution to a logarithmic equation,one must check that the possible solution does not give the log of zero or thelog of a negative number when substituted into the ORIGINAL EQUATION. Ifit does, the solution is invalid and must be rejected. This issue only arises if one must usethe properties of logarithms in order to convert an equation to a form that one can solve.

Example 6 Solve the following equations. (a) log2(x−5) = log2 3, (b) 2 ln x = ln(x−1)+ln(x+ 3), and (c) ln 3x = ln(x+ 2)− lnx.

Solution

(a) log2(x− 5) = log2 3 implies x− 5 = 3 and so x = 8.

Since 8− 5 > 0, the solution is valid.

(b) 2 ln x = ln(x− 1) + ln(x+ 3)

lnx2 = ln((x− 1)(x+ 3)) Use properties of logarithms.

x2 = (x− 1)(x+ 3) logaM = logaN if and only if M = N .

x2 = x2 + 2x− 3

0 = 2x− 3 implies x = 3/2.

For x = 3/2, x > 0, x− 1 > 0, and x+ 3 > 0. So x = 3/2 is a valid solution.

(c) ln 3x = ln(x+ 2)− lnx

ln 3x = ln

(x+ 2

x

)Use properties of logarithms.

3x =x+ 2

xlogaM = logaN if and only if M = N .

3x2 = x+ 2 Cross multiply.

x = −2/3 or x = 1.

For x = −2/3, 3x < 0 and so ln 3x is not de�ned at x = −2/3. So reject x = −2/3.

For x = 1, 3x > 0, x+ 2 > 0, and x > 0. So x = 1 is the only solution.


Exercises 2. 3


1. log 5 + log 200 2. log3 105− log3 35 3. e2 ln 2

4. 53 log5 2−2 log5 3 5. 2 ln√e+ eln 3

In Exercises 6-10, given that ln 2 ≈ 0.69 and ln 5 ≈ 1.61, �nd the approximate values of thefollowing.

6. ln 10 7. ln5

28. ln

1

259. ln 3

√2 10. ln 100

In Exercises 11-14, assume that loga x = 3 and loga y = 4. Find the following.

11. loga(xy) 12. logay

x213. loga 4

√y 14. loga

(a2

xy

)

In Exercises 15 -18, write the expression as a sum/di�erence of logarithms. Express all powersas factors.

15. ln((x2 + 1)2

√x− 1

)16. log

((x− 1)3(x+ 2)2

3√x

)17. ln

(√x2 + 1

(x− 1)5

)18. ln

(xex

(x− 1)(x+ 2)

)In Exercises 19-23, express the following as a single logarithm. Simplify where possible.

19. log(x+ 1) + log(x− 1) 20. 2 ln x− 3 ln(2− x)

21. loga(x2 + x)− loga x+

1

2loga(x+ 1) 22. log2(x

2 − 1)− log2 x− log2(x− 1)

23. ln

(x+ 2

x− 1

)+ ln

(x− 2

x+ 1

)+ ln(x2 − 1)

In Exercises 24-27, solve the equation.

24. log(x+ 3) = log(2x+ 1) 25. ln(x+ 2)− lnx = ln 5

26. log2(x+ 3) + log2(x− 2) = log2 6 27. ln(x+ 1) + ln(2x− 2) = 2 lnx

In Exercises 28-29, given f(x) and g(x), �nd f ◦ g and g ◦ f .

28. f(x) = ln(x+ 1) ; g(x) = e2x 29. f(x) = ex−1 ; g(x) = 1− lnx

2. 4 Exponential and Logarithmic Equations 65

2. 4 Exponential and Logarithmic Equations

Basic Exponential and Logarithmic Equations

The simplest exponential equation is an equation of the form au = b, where u is someexpression involving x and a and b are positive real numbers, a ̸= 1. To solve this equation,convert the equation to its logarithmic form, u = loga b, and then solve for x.

Similarly, the simplest logarithmic equation is an equation of the form loga M = c, where Mis an expression involving x, a is a positive real number, a ̸= 1, and c is any real number.To solve this equation, convert it to its exponential form, M = ac, and then solve for x.

Example 1 Solve the following equations: (a) e−x = 5, (b) log x = 3, (c) 2e2x = 3, (d)

2x2−1 = 5, (e) ln(2x+ 7) = −1, and (f) log3

(x+ 1

x

)= 2

Solution

(a) e−x = 5 implies that −x = ln 5 and hence x = − ln 5.

(b) log x = 3 implies that x = 103 = 1000.

(c) 2e2x = 3 implies e2x = 3/2 and 2x = ln(3/2). So x =ln(3/2)

2. Notice that we isolated

the exponential expression before converting the equation to its logarithmic form.

(d) Converting 2 x2−1 = 5 to x2− 1 = log2 5 gives x2 = log2 5+1 and hence x =

√log2 5 + 1

or x = −√

log2 5 + 1.

(e) ln(2x+ 7) = −1 implies 2x+ 7 = e−1. Hence, 2x = e−1 − 7 and x =e−1 − 7

2. Since we

did not use properties of logarithms, this answer is automatically valid.

(f) Converting log3

(x+ 1

x

)= 2 to exponential form, we obtain

x+ 1

x= 32. Cross-

multiplying gives x+ 1 = 9x. Therefore, 8x = 1 and x = 1/8.

More Complicated Equations

Many exponential and logarithmic equations may require the use the laws of exponents,properties of logarithms, and factoring in order to reduce the equations to the simple formsdiscussed above.

Example 2 Solve the following equations: (a) (2x)3 · 22−x = 3, (b) e2x − 3ex = 0, (c)e2x − ex = 12, (d) log x+ log(x+ 3) = 1, and (e) x lnx+ ln x2 = 0.


Solution

(a) (2x)3 · 22−x = 3

23x · 22−x = 3

22x+2 = 3

2x+ 2 = log2 3

x =log2 3− 2

2

(b) e2x − 3ex = 0

ex(ex − 3) = 0

ex = 0 or ex = 3 ex = 0 has no solutions.

x = ln 3

(c) e2x − ex = 12

e2x − ex − 12 = 0

(ex − 4)(ex + 3) = 0

ex = 4 or ex = −3 ex = −3 has no solutions. Why?

x = ln 4

(d) log x+ log(x+ 3) = 1

log(x(x+ 3)) = 1

x2 + 3x = 10

x2 + 3x− 10 = 0

(x+ 5)(x− 2) = 0

x = −5 or x = 2 log x is unde�ned for x = −5. Reject x = −5.

x = 2

(e) x lnx+ ln x2 = 0

x lnx+ 2 ln x = 0

lnx(x+ 2) = 0 Factor out lnx.

lnx = 0 or x+ 2 = 0

x = e0 = 1 or x = −2 ln x is unde�ned for x = −2. Reject x = −2.

x = 1

Example 3 Let f(x) = ln(x + 2) − 1. Find f−1. Verify that f and f−1 are inverses.State the domain and range of both functions. Find any intercepts and asymptotes of bothfunctions. Plot f and f−1 on separate coordinate systems labeling any asymptotes andintercepts.

Solution To �nd f−1, use the steps outlined in Section 1. 6.

y = ln(x+ 2)− 1

y + 1 = ln(x+ 2)

ey+1 = x+ 2 Convert to exponential form.

x = ey+1 − 2

y = ex+1 − 2 Switch x's with y's.

f−1(x) = ex+1 − 2


To verify that two functions are inverses, we check that f(f−1)(x) = x and f−1(f(x)) = x.

f(f−1)(x) = ln(ex+1 − 2 + 2)− 1 = ln(ex+1)− 1 = x+ 1− 1 = x.

f−1(f(x)) = eln(x+2)−1+1 − 2 = eln(x+2) − 2 = x+ 2− 2 = x.

ln(x + 2) is only de�ned when x > −2. So Df = (−2,∞) = Rf−1 . Since an exponentialfunction is de�ned for all values of x, Df−1 = (−∞,∞) = Rf .

To �nd the y-intercept of f(x), we compute f(0) = ln 2− 1. Hence the y-intercept of f(x) is(0, ln 2− 1). Similarly, compute f−1(0) = e− 2 and so the y-intercept of f−1(x) is (0, e− 2).Now recall that any point (x, y) on the graph of a function will become (y, x) on the graphof its inverse. Hence the x-intercept for f(x) is (e − 2, 0) and the x-intercept for f−1(x) is(ln 2 − 1, 0). The x-intercepts can also be obtained by solving the equations f(x) = 0 andf−1(x) = 0.

The graph of f(x) = ln(x + 2)− 1 has a vertical asymptote, x = −2. (Why?) This impliesthat the graph of f−1(x) = ex+1 − 2 has a horizontal asymptote, y = −2. (Why?)

To sketch these functions on a grid with a given scale would require knowing that ln 2− 1 ≈−.31 and e − 2 ≈ .72. However, it is the shapes of these graphs that matter most in yoursketches.

f HxL= lnHx+2L - 1

H0, ln 2 - 1L He-2,0L

VA: x = -2

5

-3

-2

-1

1

2

f -1HxL= e x +1- 2

Hln2-1, 0LH0,e-2L

HA: y = -2

-2 -1 1 2

-2

2

4

6

8

Exponential Equations Involving More Than One Base

For equations of the form au = bv involving more than one base, conversion to a log form maybe di�cult and, in some cases, may even be impossible. So we will use a di�erent approachto solve this type of equation.

Since y = lnx is a one-to-one function, we can take the natural log of both sides of theequation au = bv to obtain an equivalent equation, ln au = ln b v. Then we use properties oflogarithms to solve for x.


Example 4 Solve 2 x−3 = π x.

Solution

* Taking the natural log of both sides of this equation yields ln 2 x−3 = ln π x.

* We bring down the exponents and then distribute to obtain (x − 3) ln 2 = x ln π andx ln 2− 3 ln 2 = x ln π.

* Collecting all terms with x to one side yields x ln 2− x lnπ = 3 ln 2.

* Factoring out x gives x(ln 2− ln π) = 3 ln 2.

* Therefore, x =3 ln 2

ln 2− lnπ.


Exercises 2. 4

In Exercises 1-16, solve the equations.

1. 3x2+1 = 10 2. ln(3x− 5) = 0 3. log

(x− 1

x+ 1

)= −1

4. 2(ex)3 = 5 5.ex · ex+1

ex−1= 8 6. 2 log5 x = 3 log5 4

7. log(2x) + log(x+ 5) = 2 8. log9(x− 5) + log9(x+ 3) = 1

9. ln(x+ 1)− lnx = 2 10. e2x − 2ex = 0 11. xex + ex = 0

12. x lnx− x = 0 13. e2x + ex − 2 = 0 14. e2x − 5ex + 6 = 0

15. x log x− log x3 = 0 16. 3x−1 = 4x+2

In Exercises 17-18, a function f is given. Find f−1. Verify that f and f−1 are inverses.State the domain and range of both functions. Find any intercepts and asymptotes of bothfunctions. Plot f and f−1 on separate coordinate systems.

17. f(x) = ex−2 − 1 18. f(x) = ln(x− 1) + 2

2. 5 Applications 70

2. 5 Applications

Compound Interest

Suppose that you deposit $100 into a saving account earning 5% interest on your originalbalance per year. What will your balance be at the end of one, two, and three years? Ineach year, you will earn $100 × 0.05 = $5 in interest. So at the end of the Year 1, yourbalance will be $105. At the end of Year 2, your balance will be $105 + $5 = $110. At theend of Year 3, your balance will be $115. This is an example of an account earning simpleinterest, where interest is computed solely on the principal amount originally deposited.

Most �nancial institutions today use compound interest, where not only is interest earnedon the principal amount, but it is also earned on the interest already accumulated. Forexample, you deposit $100 at 5% per year, compounded annually. At the end of the �rstyear, your balance will still be $105. However, at the end of the second year, you willhave $105 + (105 × 0.05) = $110.25. At the end of the third year, your balance will be$110.25 + (110.25 × 0.05) = $115.76, rounded to the nearest cent. So you will earn moremoney with compounded interest, albeit a small di�erence in this example.

Suppose a principal amount, P, is deposited into a saving account earning r% interestcompounded annually. We wish to compute the accumulated balance, A(t), at the end oft years.

At the end of Year 1, A(1) = P + (P × r) = P (1 + r).

At the end of Year 2, A(2) = A(1) + (A(1)× r) = A(1)(1 + r) = P (1 + r)2.

At the end of Year 3, A(3) = A(2) + (A(2)× r) = A(2)(1 + r) = P (1 + r)3.

Continuing this process, we see that A(t) = P (1+r)t. To use this formula, one must convertr from a percent to a decimal.

Now suppose that the interest is compounded n times a year. The amount earned in interest

every compounding period is A × r

n, where A is the accumulated balance at the beginning

of the compounding period. Therefore the new accumulated balance at the end of the

compounding period is A + A × r

n= A

(1 +

r

n

). In the course of one year, the interest is

compounded n times. Therefore, the balance after one year is A(1) = P(1 +

r

n

)n. After t

years, the interest is compounded nt times, and so the accumulated balance is given by

A(t) = P

(1 +

r

n

)nt

.


Example 1 Suppose $1000 is invested in an account that pays 4% interest. How muchmoney will be in the account at the end of 3 years if the interest is compounded (a) annually,(b) monthly, and (c) daily? Round the answers to the nearest cent.

Solution

(a) A(3) = $1000 (1 + .04)3 ≈ $1124.86

(b) A(3) = $1000

(1 +

.04

12

)12·3

≈ $1127.27

(c) A(3) = $1000

(1 +

.04

365

)365· 3

≈ $1127.49. As can be seen, the ending balance is greater

the more times interest is compounded per year.

Recall from Section 2. 1, that as n → ∞,

(1 +

1

n

)n

→ e. We can replace n by any real

number x, that is, as x → ∞,

(1 +

1

x

)x

→ e. Setting x = n/r, we see that as n → ∞,(1 +

r

n

)n/r→ e. Therefore,

(1 +

r

n

)n=

((1 +

r

n

)n/r)r

→ er, and A(t) = P(1 +

r

n

)nt→

Pert. This limit gives us the formula for continuous compounding. If interest is com-

pounded continuously, then the accumulated balance after t years is given by

A(t) = Pert.

Example 2 (a) Suppose $1000 is invested in an account that pays 4% interest compoundedcontinuously. How much money will be in the account at the end of 3 years? Round theanswer to the nearest cent. (b) How long will it take for the amount in the account to reach$10,000 if the interest is compounded continuously? Round the answer to one decimal place.

Solution

(a) A(3) = $1000 e0.04×3 ≈ $1127.50.

(b) 10000 = 1000 e0.04 t

10 = e0.04 t

0.04 t = ln 10

t =ln 10

0.04≈ 57.6 years


Exponential Growth and Decay

Any quantity that grows or decays at a rate proportional to the amount present can bedescribed by the exponential model

Q(t) = Q0ekt

where Q0 is the quantity present at time t = 0, and k is the rate of growth if k > 0 or therate of decay if k < 0.

Example 3 The population of the United States in 2008 was 307 million people. Thepopulation growth rate is 0.98%. Assuming that the growth rate remains constant, (a) whatwill the population of the United States be in the year 2018. Round the answer to onedecimal place. (b) How long will it take for the population to reach 400 million people?Round the answer to nearest year. (c) In what year will the population reach 400 million?

Solution

(a) The population of the U.S. can be estimated using the formula P (t) = 307e0.0098 t, whereP (t) is in millions of people and t is the number of years since 2008. For 2018, t = 10.Hence, the population of the U.S. in 2018 will be P (10) = 307e0.0098·10 ≈ 338.6 millionpeople.

(b) Setting P (t) = 400, we obtain 400 = 307e0.0098 t which implies that t =ln(400/307)

0.0098≈ 27

years.

(c) The population will reach 400 million people in the year 2035. (Why?)

Example 4 Polonium-210 is a rare radioactive element that occurs naturally in the earth'scrust. The amount of Polonium-210 at time t is given by Q(t) = Q0e

−.005022 t, where Q0 isthe initial amount and t is measured in days. (a) What percentage of an initial amount ofpolonium-210 is left after 30 days? Round the answer to the nearest percent. (b) How longwill it take until only 10% of the original amount of polonium-210 remains? Round to thenearest day.

Solution

(a) Q(30) = Q0e−.005022·30 ≈ 0.86Q0. Hence, 86% of the original amount remains after 30

days.

(b) Setting Q(t) = .1Q0 yields 0.1Q0 = Q0e−0.005022 t. This implies that 0.1 = e−0.005022 t and

so t =ln(0.1)

−0.005022≈ 458 days.


Exercises 2. 5

1. Suppose $1000 is invested in an account that pays 3% interest. How much money willbe in the account at the end of 5 years if the interest is compounded (a) annually, (b)quarterly, (c) monthly, and (d) continuously? Round the answers to the nearest cent.

2. How long will it take for an investment to double if the interest is compounded contin-uously at a rate of 5%? Round the answer to one decimal place.

3. Suppose an amount invested triples in 20 years with continuous compounding. What isthe interest rate earned? State your �nal answer as a percentage rounded to one decimalplace.

4. How much money should be invested now in an account earning 6% interest so that20 years from now, the balance in the account will be $100, 000 if the interest is (a)compounded annually (b) compounded continuously? Round to the nearest cent.

5. The population of a city is given by P (t) = 200e0.012 t, where P is in thousands of peopleand t is in years with t = 0 corresponding to the year 2005. (a) What will the populationof the city be in the year 2013? Round the answer to one decimal place. (b) In whatyear will the population of the city reach 301 thousand people?

6. The number of bacteria in a culture at time t is given by N(t) = N0e0.07 t, where t

is measured in hours. (a) How long does it take for the population of the bacteria todouble? (b) to triple? (c) to quadruple? Round the answers to one decimal place.

7. The amount of radioactive C14 present in an object is given by Q(t) = Q0e−0.000121 t,

where t is measured in years. (a) If 10 gram of C14 are present at time t = 0, howmany grams will be present in the object 1000 years from now. Round the answer totwo decimal places. (b) What percentage of an initial amount of C14 is left after 10,000years? Round the answer to the nearest percent. (c) How long will it take until only 15%of the amount of C14 in the object remains? Round the answer to the nearest hundredsof years.




1. a) (e2x)2 b) 5 2−x · 5x c)3 2x

3 x+2d)

2 x

√2 x

e)

(3√2

ex

)−3

f)4x+1

2 2x

2. a) log3 1 b) ln e c) log5 25 d) ln√e e) log8 2 f) log 0.01 g) log3

(1

27

)3. a) ln e−4 b) eln 8 c) 3 4 log3 2 d) e1/2 ln 4

e) eln 2 − 3 ln( 3√e) f) log8 3 + log8

(64

3

)g) 2log2 5−log2 3

4. Assume lnx = 2 and ln y = −3. Find a) ln(xy) b) lnx3 c) ln

(y√x

)d) ln(e3y).

5. Write the following expressions as a sum/di�erence of logarithms. Express any powers asfactors.

a) log

(x3√x+ 1

(x− 2)(x− 1)

), x > 2 b) ln

((x− 4)2

x2 − 1

)2/3

, x > 4 c) ln

(xex

2

(x2 + 1)3

), x > 0

6. Express the following as a single logarithm. Simplify when possible.

a) log2 x2 − 3 log2

√x b) log(x+ 2) + log(x− 2)− log(x2 − 4)

c) ln(x2 − 1) + ln(x− 1)− 3 ln(x+ 1)

7. Find the domains of the following functions.

a) f(x) = e2x2+5 − 1 b) g(x) = ln(x2 − 2x) c) h(x) = log2

(x

2− x

)8. Determine on which intervals the following functions are positive and on which intervalsthey are negative.

a) f(x) = (x+ 1)e2x b) g(x) = 2ex − x2ex c) h(x) = xex − x

d) F (x) =1

2x+ x lnx e) G(x) = x lnx+ 2 ln x

9. Solve the following equations:

a)

(1

5

)2−x

= 25 b) ex2= (e3x) · 1

e2

c) log3(x− 1) + log3(x+ 3) = log3 5 d) log(x+ 2)− log x = log 5

e) 32x = 4 f) 2ex+1 = 3 g) log4 x = 5 h) ln(x− 4) = 0

i)2x+5

4x−2= 5 j) e2x − 4ex = 0 k) e2x − 3ex − 10 = 0

l) lnx− ln(x+ 2) = −1 m) log3 x+ log3(x− 8) = 2 n) x lnx3 + 2x = 0

o) 2x+2 = 3x−3


10. Given f(x) and g(x), �nd f ◦ g and g ◦ f .

a) f(x) = ex − 1, g(x) = ln(x2 + 1) b) f(x) = ln(x− 3), g(x) = e2x + 3.

11. For each of the following functions, �nd f−1 . Verify that f and f−1 are inverses.State the domain and range of both functions. Find any intercepts and asymptotes of bothfunctions. Plot both f and f−1 on separate coordinate systems labeling any intercepts andasymptotes.

a) f(x) = ex+1 + 3 b) f(x) = ln(x+ 2)− 1

3. 1 Angles and Right Triangular Trigonometry 76

3. 1 Angles and Right Triangular Trigonometry

Angles

In geometry, a ray is a half-line that starts from a point, called the vertex of the ray, andcontinues inde�nitely in one direction. A geometric �gure formed by two rays with a commonvertex is called an angle. The two rays are called the initial side and the terminal sideof the angle.

The measure of an angle is the magnitude of the rotation from the initial side to theterminal side of the angle. If the rotation is in the counterclockwise direction, the measureof the angle is positive; if the rotation is in the clockwise direction, the measure of the angleis negative. An angle θ in the xy-plane is said to be in standard position if its initial sidelies on the positive x-axis and its vertex is at the origin.

In Euclidean geometry, angles are measured using degrees. The rotation by a full circle

equals 360◦. One degree, 1◦, is obtained by rotating the initial side by1

360th of a circle.

135°

-90°

If the terminal side of an angle θ in standard position lies on one of the axes, as in the angleon the right above, the angle is said to be a quadrantal angle. If the terminal side of θlies in a quadrant, as in the angle on the left above, then θ is said to lie in that quadrant.For example, the angle θ = 135◦ lies in the Quadrant II.

Two angles in standard position are said to be coterminal if they have the same terminalside. For example, the angle of 45◦ and the angle of 405◦ are coterminal angles.

45° 405°


In general, two angles are coterminal if one can be obtained from the other by adding orsubtracting multiples of 360◦.

Trigonometric Functions of Acute Angles

An angle that is less that 90◦ is called an acute angle. In a right triangle, one angle is a90◦ angle and the other two are acute angles that add up to 90◦. (Why?) Consider the tworight triangles below that have the same acute angle θ. These triangles are similar trianglessince they have the same angles, 90◦, θ, and 90◦− θ, but they have sides of di�erent lengths.

Θ

a

bc

Θ

a'

b'c'

The sides of similar triangles are proportional to each other, that is, the ratio of correspondingsides are equal. Therefore

b

c=

b′

c′a

c=

a′

c′b

a=

b′

a′a

b=

a′

b′c

a=

c′

a′c

b=

c′

b′

These ratios depend only on the angle θ and not on the lengths of the sides of a right triangleand are called the sine, cosine, tangent, cotangent, secant, and cosecant of the angleθ, respectively. They de�ne the six trigonometric functions of θ.

sin θ =b

c=

opposite

hypotenusecos θ =

a

c=

adjacent

hypotenusetan θ =

b

a=

opposite

adjacent

csc θ =c

b=

hypotenuse

oppositesec θ =

c

a=

hypotenuse

adjacentcot θ =

a

b=

adjacent

opposite

We write the opposite and the adjacent above to denote the length of the leg opposite to θand the length of the leg adjacent to θ, respectively.


Example 1 For the right triangle below, �nd the six trigonometric functions of θ.

Θ

2

3

Solution Before computing the values of the trigonometric functions of θ, we must �rst�nd the length of the missing side, in this case, the hypotenuse. The Pythagorean Theoremtells us that a2 + b2 = c2, where a and b are the legs of the right triangle and c is thehypotenuse. Hence, c2 = 22 + 32 = 13 and therefore c =

√13. So we have

a = adjacent = 2 b = opposite = 3 c = hypotenuse =√13

sin θ =opposite

hypotenuse=

3√13

cos θ =adjacent

hypotenuse=

2√13

tan θ =opposite

adjacent=

3

2

csc θ =hypotenuse

opposite=

√13

3sec θ =

hypotenuse

adjacent=

√13

2cot θ =

adjacent

opposite=

2

3

Notice that csc θ is the reciprocal of sin θ, sec θ is the reciprocal of cos θ, and cot θ is thereciprocal of tan θ.

Example 2 Solving a right triangle

Solve the right triangle if (a) c = 20 and β = 50◦, (b) a = 5 and α = 20◦. See the diagrambelow. Round the answers to one decimal place.

Β

a

bc

Α

Solution To solve a right triangle, one must �nd the measure of its angles and the lengthsof its sides.


(a) In this triangle, we need to �nd α, a, and b. The sum of α and β is 90◦. Therefore,α = 90◦ − 50◦ = 40◦. Since this is a right triangle, we can use trigonometric functionsto solve for the remaining sides.

sin β =b

cimplies b = c sin β = 20 sin 50◦ ≈ 15.3.

cos β =a

cimplies a = c cos β = 20 cos 50◦ ≈ 12.9.

(b) In this triangle, we need to �nd β, b, and c. β = 90◦ − 20◦ = 70◦.

sinα =a

cimplies c =

a

sinα=

5

sin 20◦≈ 14.6.

tanα =a

bimplies b =

a

tanα=

5

tan 20◦≈ 13.7.

Radian Measure

De�nition: An angle is called a central angle if its vertex lies at thecenter of a circle. The radian measure of a positive centralangle, θ, is the ratio of the intercepted arc length, s, to the

radius of the circle, r. In other words, θ =s

r.

Θ

r

s

The circumference of a circle of radius r is 2πr. So the radian measure of an angle formedby one complete counterclockwise rotation is 2π radians. Therefore, 2π radians = 360◦. Itis customary not to include the word "radians" when stating the measure of an angle inradians. Therefore, we can simply state that 2π = 360◦. It follows that π = 180◦. Hence,

1 degree =π

180radians 1 radian =

180

πdegrees.

Example 3 Convert each angle in degrees to radians. (a) 30◦, (b) 135◦, (c) −270◦.

Solution (a) 30◦ · π

180◦=

π

6, (b) 135◦ · π

180◦=

3π

4, (c) − 270◦ · π

180◦= −3π

2.


Example 4 Convert each angle in radians to degrees. (a)π

2, (b)

2π

3, (c)

9π

4.

Solution (a)π

2· 180

◦

π= 90◦, (b)

2π

3· 180

◦

π= 120◦, (c)

9π

4· 180

◦

π= 405◦.

Radian measure is in many cases easier to use than degree measure. Since 1 revolution = 2πradians, half a revolution corresponds to π radians. So you can think of the xy-plane as beingsplit in half with π radians in the upper half-plane and π radians in the lower half-plane.Then use geometry to �nd the location of an angle measured in radians.

Example 5 Sketch each angle in standard position: (a)π

6, (b)

π

2, (c)

2π

3, (d)

5π

4, (e) −π

2.

Solution Refer to the �gures below. (a) To sketch the angleπ

6in standard position, divide

π into 6 equal parts in the upper half-plane. Then starting at the positive x-axis, rotate inthe counterclockwise direction until you reach the �rst division line. Use an arrow to indicate

the direction of rotation. This angle lies in Quadrant I. (b)π

2is obtained by dividing the

upper half-plane into 2 equal pieces and using the �rst piece. Soπ

2is a quadrantal angle. (c)

To draw the angle2π

3, divide the upper half-plane into 3 equal parts and use two of them

in the counterclockwise direction.2π

3lies in Quadrant II. (d) For the angle

5π

4, dividing the

upper-half into 4 parts is not enough. We need �ve fourths and so we will also divide the

lower-half into 4 parts. Using 5 parts, we see that5π

4is in Quadrant III. (e) This angle is a

quadrantal angle formed by rotating in the clockwise direction byπ

2.

Π

6

Π

22 Π

3

5 Π

4

-

Π

2


Below are all the quadrantal angles that lie in the interval [0, 2π].

0Π

2Π

3 Π

2

2Π

All other quadrantal angles can be obtained from the above angles by adding or subtractingmultiples of 2π.

Trigonometric Functions of Special Angles

In this section, we derive the values of the trigonometric functions of three special angles:π/6 = 30◦, π/4 = 45◦, and π/3 = 60◦.

Π

4

Π

4

1

12

Π

3

Π

6

1

32

Consider the isosceles right triangle whose legs have length 1 in the �gure to the left above.Since this is an isosceles right triangle, both of its acute angle are equal and have measureπ/4. Furthermore, by the Pythagorean Theorem, the hypotenuse must have length

√2.

Therefore, the trigonometric functions of π/4 are as follows:


sinπ

4=

1√2=

√2

2cos

π

4=

1√2=

√2

2tan

π

4= 1

cscπ

4=

√2 sec

π

4=

√2 cot

π

4= 1

For π/6 and π/3, consider the equilateral triangle with side of length 2 in the �gure to theright above. All of its angles are equal and have each angle has measure π/3. Now drop aperpendicular from top vertex to the base of the triangle. This perpendicular bisects boththe top angle and the base of the triangle. As a result, we obtain a right triangle with anglesπ/3 and π/6, the hypotenuse of length 2 and a leg of length 1. By the Pythagorean Theorem,the remaining leg has length

√3 (Why?). So the values of the trigonometric functions of

π/3 and π/6 are as follows.

sinπ

3=

√3

2cos

π

3=

1

2tan

π

3=

√3

cscπ

3=

2√3

secπ

3= 2 cot

π

3=

1√3

sinπ

6=

1

2cos

π

6=

√3

2tan

π

6=

1√3

cscπ

6= 2 sec

π

6=

2√3

cotπ

6=

√3

You MUST KNOW these values.


Exercises 3. 1

In Exercises 1-3, consider the right triangle below. With the given values, �nd the sixtrigonometric functions of θ.

1. a = 5, b = 12

2. a = 3, c = 7

3. b = 1, c = 4 Θ

a

bc

In Exercises 4-6, consider the right triangle below. Find the missing angles and sides of theright triangle. Round the answer to one decimal place.

4. β = 25◦, c = 10

5. α = 75◦, b = 3

6. β = 50◦, b = 7 Β

a

bc

Α

In Exercises 7-10, convert each angle in degrees to radians.

7. 60◦ 8. 225◦ 9. −150◦ 10. 450◦

In Exercises 11-13, convert each angle in radians to degrees.

11.π

1212.

7π

613. 3π

In Exercises 14-19, sketch each angle in standard position.

14.π

315. −π 16.

3π

4

17. −5π

618.

5π

319.

5π

2


20. sec2(π4

)− 2 tan

π

421.

(tan

π

3

)(cot

π

3

)22. sin

π

6+ cos

π

3+ cot

π

6

In Exercises 23-24, consider the right triangle given in Exercises 1-3.

23. If θ =π

6and c = 4, �nd the exact values of a and b.

24. If θ =π

4and b = 3, �nd the exact values of a and c.

3. 2 Trigonometric Functions of General Angles; Basic Identities 84

3. 2 Trigonometric Functions of General Angles; Basic Identities

Trigonometric Functions of General Angles

Consider an acute angle θ in standard position and a circle of radius r centered at the origin.See the �gure below. Let (x, y) be the point where the terminal side of θ intersects the circle.If we drop a perpendicular from this point to the x-axis, we obtain a right triangle whose legshave lengths x and y and whose hypotenuse is r. We can restate the trigonometric functionsas follows.

Θ

Hx, yL

x

yr

sin θ =y

rcsc θ =

r

y

cos θ =x

rsec θ =

r

x

tan θ =y

xcot θ =

x

y

This de�nition can be extended to any angle; if θ is a central angle in standard positionwhose terminal side intersects a circle of radius r at the point (x, y), then we use the sixratios above to de�ne the trigonometric functions of θ. We can even dispense with the circlesince by the Pythagorean Theorem, r =

√x2 + y2. So one can obtain the six trigonometric

values simply from a point (x, y) on the terminal side of an angle in standard position. Forthe rest of this chapter, we assume that all angles are in standard position.

Example 1 The point (−2, 1) lies on the terminal side of an angle θ. Find the six trigono-metric functions of θ.

Θ

H-2, 1L

È-2È=2

1


Solution By the Pythagorean Theorem, r =√4 + 1 =

√5. So

sin θ =y

r=

1√5

csc θ =r

y=

√5

cos θ =x

r= − 2√

5sec θ =

r

x= −

√5

2

tan θ =y

x=

−1

2cot θ =

x

y= −2

Trigonometric Identities

An identity is an equality that holds for all values of the variable in its domain. A closer lookat the ratios that de�ne the trigonometric functions will reveal the following basic identities:

Reciprocal Identities

csc θ =1

sin θsec θ =

1

cos θcot θ =

1

tan θ

Quotient Identities

tan θ =sin θ

cos θcot θ =

cos θ

sin θ

Furthermore, the sum of the two acute angles in a right triangle isπ

2. So if θ is one of the

acute angles in a right triangle, then the other acute angle isπ

2− θ. It is easy to see that

the opposite leg of θ is the adjacent leg ofπ

2− θ and vice versa. Therefore,

Cofunction Identities

sin

(π

2− θ

)= cos θ tan

(π

2− θ

)= cot θ sec

(π

2− θ

)= csc θ

cos

(π

2− θ

)= sin θ cot

(π

2− θ

)= tan θ csc

(π

2− θ

)= sec θ

An easy way to remember these identities is to keep in mind that the cofunctions of com-plementary angles are equal.

However, the most fundamental identity is derived from the Pythagorean Theorem. This

theorem implies that r2 = y2+x2 and dividing by r2, we obtain 1 =y2

r2+x2

r2=(yr

)2+(xr

)2=

sin2 θ + cos2 θ.


Furthermore, by dividing the fundamental identity by cos2 θ, we see thatsin2 θ

cos2 θ+1 =

1

cos2 θ

and hence tan2 θ + 1 = sec2 θ. Similarly, by dividing by sin2 θ, we get 1 + cot2 θ = csc2 θ.

Pythagorean Identities

sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ cot2 θ + 1 = csc2 θ

Example 2 Suppose cos θ = 4/5. Use trigonometric identities to �nd (a) sec θ, (b) sin2 θ,

(c) tan2 θ, and (d) sin(π2− θ).

Solution

(a) sec θ =1

cos θ=

5

4.

(b) sin2 θ = 1− cos2 θ = 1−(4

5

)2

= 1− 16

25=

9

25.

(c) tan2 θ =sin2 θ

cos2 θ=

9/25

16/25=

9

16.

(d) sin(π2− θ)= cos θ =

4

5.

Quadrantal Angles and the Unit Circle

To �nd the trigonometric values of quadrantal angles, it is best to consider the unit circle.This is the circle of radius, r = 1, centered at the origin. If the terminal side of an angleintersects the unit circle at the point (x, y), then x = cos θ and y = sin θ. By examining thecircle below, the terminal side of the angle 0 intersects the unit circle at (1, 0) and hencecos 0 = 1, sin 0 = 0, sec 0 = 1, csc 0 is unde�ned, tan 0 = 0, and cot 0 is unde�ned. The

terminal side of the angleπ

2is (0, 1) and so cos π/2 = 0, sin π/2 = 1, sec π/2 is unde�ned,

cscπ/2 = 1, tanπ/2 is unde�ned, and cotπ/2 = 0. The remaining quadrantal angles are leftas an exercise.

1-1

1

-1


Signs of Trigonometric Functions

Notice in Quadrant I, x and y are both positive. In Quadrant II, x is negative and y ispositive. In Quadrant III, both x and y are negative. In Quadrant IV, x is positive and y isnegative.

H+,+LH-,+L

H-,-L H+,-L

All positivesin Θ positive

tan Θ positive cos Θ positive

csc Θ positive

cot Θ positive sec Θ positive

Since r is always positive, all trigonometric functions of an angle in Quadrant I are positive.For an angle θ in Quadrant II, only sin θ and csc θ are positive. Only tan θ and cot θ arepositive for an angle θ in Quadrant III. Finally, for an angle θ in Quadrant IV, only cos θ andsec θ are positive. You should keep this in mind when determining the values of trigonometricfunctions of an angle.

Perhaps the easiest way to �nd the trigonometric values is to use the reference angle.

De�nition: The reference angle α of a nonquadrantal angle θ is theacute angle that the terminal side of θ makes with either thepositive or the negative x-axis.

The reference angle, α, for any nonquadrantal angle θ will be one of the following anglesdepending on the quadrant in which θ lies. Notice that α is not in standard position.

Α Α

Α Α

If (x, y) is a point on the terminal side of θ, then sinα =| y|r

and cosα =|x|r. This implies

that sin θ = ± sinα and cos θ = ± cosα. The same applies to the other four trigonometricfunctions. Thus, to �nd the trig values of θ, we can �nd instead the trig values of α andthen adjust for the sign.


Example 3 Suppose that sin θ = −2/3 and θ is in Quadrant III. Find the values of the�ve remaining trigonometric functions of θ.

Solution Draw an angle θ in the third quadrant and consider its reference angle α. sinα =2/3. So α is an angle in a right triangle with hypotenuse 3 and opposite leg 2. See Figure Abelow. The adjacent leg of this triangle is

√9− 4 =

√5. After adjusting for the appropriate

signs in Quad III, we obtain

csc θ = −3

2cos θ = −

√5

3sec θ = − 3√

5tan θ =

2√5

cot θ =

√5

2.

Α

Θ

3

È-2È

Figure A

Α

Θ

4

È-3È

Figure B

Example 4 Suppose that tan θ = −3/4 and cos θ > 0. Find the values of the �ve remainingtrigonometric functions of θ.

Solution Since tan θ < 0 and cos θ > 0, θ lies in Quadrant IV. Draw an angle θ in thefourth quadrant and consider its reference angle α with tanα = 3/4. α is an angle in a righttriangle with opposite leg 3 and adjacent leg 4. See Figure B above. The hypotenuse is 5(why?). After adjusting for the appropriate signs in Quad IV, we obtain

cot θ = −4

3sin θ = −3

5csc θ = −5

3cos θ =

4

5sec θ =

5

4.

Special Angles Revisited

In Section 3. 1, we derived the exact values of the trigonometric functions of the anglesπ/6, π/4, and π/3. Here, we consider the angles whose reference angles are π/6, π/4, and

π/3. These are angles of the formmπ

6,mπ

4, or

mπ

3, m an integer. When these fractions

are written in reduced form, then the reference angles will simply be π/6, π/4, and π/3,respectively.


Example 5 Find the exact values of the trigonometric functions of (a)5π

6, (b)

4π

3, (c)

7π

4.

Solution

(a)5π

6is in Quadrant II and its reference angle is

π

6. (Draw the angle and its reference angle

to verify this.) So sin5π

6=

1

2, cos

5π

6= −

√3

2, tan

5π

6= − 1√

3, csc

5π

6= 2 , sec

5π

6=

− 2√3, cot

5π

6= −

√3 .

(b)4π

3is in Quadrant III and its reference angle is

π

3. So sin

4π

3= −

√3

2, cos

4π

3=

−1

2, tan

4π

3=

√3 , csc

4π

3= − 2√

3, sec

4π

3= −2 , cot

4π

3=

1√3.

(c)7π

4is in Quadrant IV and its reference angle is

π

4. So sin

7π

4= − 1√

2, cos

7π

4=

1√2, tan

7π

4= −1 , csc

7π

4= −

√2 , sec

7π

4=

√2 , cot

7π

4= −1 .

Coterminal Angles

Since coterminal angles have the same terminal side, they have the same trigonometric values.Coterminal angles are obtained from each other by adding or subtracting a multiple of 2π.This means that for any integer n,

sin(θ + 2πn) = sin θ cos(θ + 2πn) = cos θ tan(θ + 2πn) = tan θ

csc(θ + 2πn) = csc θ sec(θ + 2πn) = sec θ cot(θ + 2πn) = cot θ

Example 6 Find the sine and cosine of (a) −7π

6and (b) 3π.

Solution (a)−7π

6+2π =

5π

6and so these two angles are coterminal. Therefore, sin

(−7π

6

)=

sin5π

6=

1

2and cos

(−7π

6

)= cos

5π

6= −

√3

2. (See Example 4.) (b) 3π − 2π = π. Hence

3π and π are coterminal and so sin 3π = sin π = 0 and cos 3π = cos π = −1 (why?).


Even-Odd Properties of Trigonometric Functions

Another observation that one can make by considering the unit circle is that

sin(−θ) = − sin θ and cos(−θ) = cos θ.

This is apparent from the picture below. If the point where the terminal side of an angle θintersects the unit circle is (x, y), then the corresponding point on the terminal side of theangle −θ will be (x,−y).

1-1

1

-1

Θ

- Θ

Hx, yL

Hx, -yL

The even-odd properties of the other trigonometric functions can be derived from the prop-erties of sine and cosine. For example,

tan(−θ) =sin(−θ)

cos(−θ)=

− sin θ

cos θ= − tan θ.

Similarly,cot(−θ) = − cot θ sec(−θ) = sec θ csc(−θ) = − csc θ.

The proof of these last three properties will be left as an exercise.

Example 7 Suppose sin θ = 3/7. Find the following: (a) sin(θ − 2π) − sin(−θ) and (b)sin(−θ + 2π).

Solution (a) sin(θ− 2π)− sin(−θ) = sin θ+sin θ = 3/7+ 3/7 = 6/7. (b) sin(−θ+2π) =sin(−θ) = − sin θ = −3/7.

Example 8 Use the odd-even identities to �nd sin(−π

6

)and cos

(−π

6

).

Solution sin(−π

6

)= − sin

π

6= −1

2and cos

(−π

6

)= cos

π

6=

√3

2.

The last example illustrates that you do not need reference angles when computing thetrigonometric functions of a negative angle. Simply use the even-odd identities and thecorresponding values for the positive angle.


Exercises 3. 2

1. The point (2, 5) lies on the terminal side of an angle θ. Find the six trigonometricfunctions of θ.

2. The point (4,−3) lies on the terminal side of an angle θ. Find the six trigonometricfunctions of θ.

3. Suppose sin θ = 3/7. Use trigonometric identities to �nd the exact value of eachexpression. In each part, state the identity that you are using.

(a) cos2 θ (b) csc θ (c) sec(π2− θ)

(d) cot2 θ.

4. Suppose tan θ = 3. Use trigonometric identities to �nd the exact value of eachexpression. In each part, state the identity that you are using.

(a) sec2 θ (b) cos2 θ (c) sin2 θ (d) cot(π2− θ).

5. Suppose that cos θ = 3/4 and θ is in Quadrant IV. Find the values of the �ve remainingtrigonometric functions of θ.

6. Suppose that sec θ = 3 and θ is in Quadrant I. Find the values of the �ve remainingtrigonometric functions of θ.

7. Suppose that tan θ = 3/2 and sin θ < 0. Find the values of the �ve remaining trigono-metric functions of θ.

8. Suppose that csc θ = 3/2 and cos θ < 0. Find the values of the �ve remaining trigono-metric functions of θ.

In Exercises 9-20, �nd the exact value of each expression.

9. sinπ

210. cos π 11. cos

2π

312. tan

5π

6

13. csc π 14. sin7π

615. sin

3π

216. cos

5π

2

17. sec7π

418. csc

π

619. tan

(−π

4

)20. cos(−2π)

21. cot7π

322. tan

(−π

2

)

In Exercises 21-24, assume that cos θ = −1

2. Use trigonometric identities to �nd the

exact value of each expression. In each part, state the identities that you are using.

23. cos(−θ) + cos(θ + 2π) 24. sin(π2− θ)and sin

(θ − π

2

)25. sec2 θ 26. tan2 θ

3. 3 Graphs of Trigonometric Functions 92

3. 3 Graphs of Trigonometric Functions

Graphs of the Sine and Cosine Functions

In this section, we will present the graphs of the sine and cosine functions. Since it iscustomary to use x for the independent variable in functional notation, we will now writesinx and cos x instead of sin θ and cos θ.

To understand the behavior of y = sin x and y = cosx, we can consider the unit circle.If (a, b) is the point where the terminal side of an angle x intersects the unit circle, thenb = sinx and a = cos x. See the �gure below. When x = 0, sinx = 0. As x increases, thepoint (a, b) moves in the counterclockwise direction and b = sin x increases until it reachesits maximum value of 1 at x = π/2. Now b = sinx starts to decrease. It reaches the value of0 at x = π and continues decreasing until it attains its minimum value of −1 at x = 3π/2.From that point on, b starts to increase again until it reaches the value of 0 at x = 2π.

1-1

1

-1

x

Ha,bL

We have now gone through one full positive rotation of the circle. The pattern now repeatsitself as we continue in the positive direction. This happens because the angle x and theangle x+2π are coterminal and therefore sin(x+2π) = sinx. This leads us to the notion ofa periodic function.

De�nition: A function f is said to be periodic with a period c if for all x in thedomain of f , f(x + c) = f(x). The smallest such c is called the periodof f .

For a trigonometric function, the period is the length of one cycle in the graph of thefunction. Both sinx and cos x are periodic with period 2π.

Another property that helps us draw the graph of y = sin x is the fact that sin(−x) = − sinx.This means that sinx is an odd function and therefore the graph of y = sin x is symmetricwith respect to the origin.


We are now in the position to draw the graph of y = sinx. See the �gure below. The graphshows two full cycles of y = sinx from −2π to 2π. The ticks on the x-axis correspond to thequadrantal angles in this interval. The ticks on the y-axis correspond to the maximum andminimum values of y = sin x.

y = sinx

-2 Π -3 Π2

-Π -Π

2Π

2Π

3 Π2

2 Π

-1

1

Properties of the graph of f(x) = sinx

a. Df = (−∞,∞) and Rf = [−1, 1];

b. x-intercepts at x = πn, where n is an integer; y-intercept (0, 0);

c. Local maximum value of 1 at x =π

2+ 2πn;

d. Local minimum value of −1 at x =3π

2+ 2πn;

e. Periodic with period 2π;

f. Odd function.

When graphing the basic trigonometric functions, you must make sure that all the interceptsand any maximum and minimum points are in their correct positions. These points aresometimes called key points.

The graph y = sin x runs through one cycle as x runs from 0 to 2π. The key points for thiscycle correspond to x = 0, x = π/2, x = π, x = 3π/2, and x = 2π. When graphing one cycleof sinx, divide the interval [0, 2π] on the x-axis into four equal subintervals. The endpointsof these subintervals will be the x-coordinates of these �ve key points. You should alwaysplot key points when graphing the sine and cosine curves.

Using the same procedure for y = cos x, we see that cos 0 = 1 and as x increases, cosx isdecreasing, passing through 0 at x = π/2, and continuing to decrease until it reaches theminimum value of −1 at x = π. cosx then turns and starts increasing, passing through 0 atx = 3π/2, and reaching its maximum value of 1 at x = 2π. The pattern then repeats itselfsince cos(x + 2π) = cosx. Furthermore, since cos(−x) = cosx, cos x is an even functionand hence its graph is symmetric with respect to the y-axis. We now graph two cycles ofy = cosx below as we did for the graph of y = sin x. Notice that the key points for y = cosxoccur at the same values of x as for y = sin x.


y = cosx

-2 Π -3 Π2

-Π -Π

2Π

2Π

3 Π2

2 Π

-1

1

Properties of the graph of f(x) = cosx

a. Df = (−∞,∞) and Rf = [−1, 1];

b. x-intercepts at x =π

2+ πn, where n is an integer; y-intercept (0, 1);

c. Local maximum value of 1 at x = 2πn;

d. Local minimum value of −1 at x = π + 2πn;

e. Periodic with period 2π;

f. Even function.

Example 1 Using transformations, graph (a) y = cos x+1, −2π ≤ x ≤ 2π, (b) y = − sinx,−2π ≤ x ≤ 2π. Be sure that the key points are in their correct positions.

Solution

(a) The graph of y = cosx+ 1 is the graph of y = cos x shifted up by 1 unit.

-2 Π -3 Π2

-Π -Π

2Π

2Π

3 Π2

2 Π

2

(b) The graph of y = − sinx is the graph of y = sin x re�ected across the x-axis.

-2 Π -3 Π2

-Π -Π

2Π

2Π

3 Π2

2 Π

-1

1


Graphs of y = A sin(Bx) and y = A cos(Bx)

The functions y = A sin(Bx) and y = A cos(Bx) play an important role in physics andmathematics. For example, they describe harmonic motion. Many other periodic phenomenacan be modeled using these functions. We will only consider functions of these forms whenB > 0.

Example 2 Graph one cycle of y = 3 sinx. Be sure that the key points are in their correctpositions.

Solution

The graph of y = 3 sinx is obtained from the graph of y = sinx by multiplying every y-valueby 3. This does not change the period, x-intercepts, or the locations of local maxima andminima. But it does a�ect the range of the function. sinx takes on values between −1 and1, whereas 3 sin x takes on values between −3 and 3 as can be seen on the graph below.

Π

2Π

3 Π2

2 Π

-3

3

In general, −1 ≤ sin x ≤ 1 and this implies that −|A| ≤ A sinx ≤ |A|. |A| is called theamplitude of the function of A sin x. This is the maximum distance by which the graphof y = A sinx deviates from the x-axis. Furthermore, it is easy to see that the functiony = A cosx also has the amplitude |A|.

In the previous example, B was equal to 1. If B ̸= 1, then the graphs of y = sin(Bx)and y = cos(Bx) are horizontal compressions or expansions of the graphs of y = sinx andy = cos x, respectively, by the factor of 1/B. If 0 < B < 1, an expansion occurs. If B > 1,a compression occurs. In these cases, the period is no longer 2π. Instead, the

Period =2π

B.

The shape of the graphs remains the same, except now the length of one cycle is 2π/Binstead of 2π. The key points of y = sin x and y = cos x occur every quarter of a cycle.Therefore, to obtain the corresponding points for y = A sin(Bx) or y = A cos(Bx), simplytake one cycle of the function and divide it into four equal parts.


Example 3 Find the amplitude and period of y = 4 cos(2x). Then graph y = 4 cos(2x),0 ≤ x ≤ 2π. Be sure that the key points are in their correct positions.

Solution Since A = 4, the amplitude of y = 4 cos(2x) is 4. Hence, the range of thisfunction is [−4, 4]. As this is a cosine curve, the graph starts at (0, 4) instead of at (0, 0)

as with the sine curve. The period of this function is2π

B=

2π

2= π. Therefore, one cycle

has length π and the interval 0 ≤ x ≤ 2π contains 2 cycles of y = 4 cos(2x). Furthermore,to obtain the x-coordinates of the key points, simply take each period interval and divide itinto four equal parts.

Π

2Π

4Π

3 Π4

5 Π4

7 Π4

3 Π2

2 Π

-4

4

Example 4 Find the amplitude and period and then graph one cycle of y = − sin(π4x).

Be sure that the key points are in their correct positions.

Solution Here, A = −1 and B = π/4. So the amplitude = | − 1| = 1 and the period

=2π

π/4= 2π · 4

π= 8. The key points will occur at x = 0, x = 2, x = 4, x = 6, and x = 8.

As this is a sine curve, the graph will start at (0, 0). However, since A < 0, the graph willgo down instead of up from this point.

2 4 6 8

-1

1


Example 5 Below is the graph of a function of the form y = A sin(Bx) or y = A cos(Bx).Find the amplitude and period of this function. Use this information to determine itsequation.

Π 2 Π 3 Π 4 Π 5 Π 6 Π 7 Π 8 Π

-2

2

Solution The amplitude of this function is 2 (the maximum distance from the graph andthe x-axis). So |A| = 2 and hence, A = 2 or A = −2. The period is the length of one

cycle, in this case, it is 4π. This implies that2π

B= 4π and B =

1

2. Since the graph begins

at (0,−2), this is the graph of a cosine function and moreover, A is negative. Therefore,

A = −2, B = 1/2, and y = −2 cos(x2

).

Graphs of the Remaining Trigonometric Functions

In Section 3. 2, we saw that for nonquadrantal angles x, tanx = ± tanα, where α is thereference angle of x. Now, x and x + π have the same reference angle and so tanx andtan(x + π) are either tanα or − tanα. However, since the sign of tanx and tan(x + π) arethe same, tanx = tan(x+ π). So the period of y = tanx is π. That is, tanx = tan(x+ πn),for any integer n.

Recall that tanx =sinx

cosx. So sin 0 = 0 and cos 0 = 1 implies that tan 0 = 0. As x increases,

sinx increases and cosx decreases until x = π/2. Hence, tanx is increasing on [0, π/2). As

x → π

2from the left, sin x → 1 and cos x → 0. Therefore, tanx → ∞ as x → π

2−. This

means that the graph of y = tanx has a vertical asymptote at x =π

2. Since tanx is an odd

function, tanx → −∞ as x → −π

2+ and the graph of y = tanx has a vertical asymptote at

x = −π

2. As the period of tanx is π, the graph of y = tan x for −π

2< x <

π

2represents one

cycle of the tangent curve. See the graph below.


y = tanx

Π-3 Π2

-Π

2-Π Π

Π

23 Π2

This graph shows three cycles of y = tanx.

Properties of the graph of f(x) = tanx

a. Df : x ̸= π

2+ πn, where n is an integer, and Rf = (−∞, ∞);

b. x-intercepts at x = πn; y-intercept (0, 0);

c. No local maximum or local minimum values;

d. Periodic with period π;

e. Odd function.

A similar analysis leads to the graphs of y = cotx, y = csc x, and y = secx.

y = cotx

Π-Π

2-Π Π

Π

23 Π2

2 Π


y = cscx

Π-Π

2-Π Π

Π

23 Π2

2 Π

-1

1

y = secx

Π-3 Π2

-Π

2-Π Π

Π

23 Π2

-1

1


Exercises 3. 3

1. Sketch the graph of f(x) = sin x, 0 ≤ x ≤ 4π, making sure that the key points are intheir correct positions. Then answer the following questions.

(a) What is sinπ

2, sin

7π

2, and sin 3π?

(b) On which intervals is f(x) increasing? On which intervals is f(x) decreasing?

(c) For which values of x does sin x = 0 on the interval 0 ≤ x ≤ 4π?

(d) On which intervals is f(x) > 0? On which intervals is f(x) < 0?

(e) For which values of x does f have a local maximum? What is the maximum value?

(f) For which values of x does sin x = −1 on the interval 0 ≤ x ≤ 4π?

2. Sketch the graph of f(x) = cos x, 0 ≤ x ≤ 4π, making sure that the key points are intheir correct positions. Then answer the following questions.

(a) What is cos π, cos7π

2, and cos 2π?

(b) On which intervals is f(x) increasing? On which intervals is f(x) decreasing?

(c) For which values of x does cos x = 0 on the interval 0 ≤ x ≤ 4π?

(d) On which intervals is f(x) > 0? On which intervals is f(x) < 0?

(e) For which values of x does f have a local minimum? What is the minimum value?

(f) For which values of x does cos x = 1 on the interval 0 ≤ x ≤ 4π?

3. Use the fact that the range of both y = sin x and y = cos x is [−1, 1] to �nd a and bthat satisfy each given inequality.

(a) a ≤ sinx+ 1 ≤ b.

(b) a ≤ cosx− 2 ≤ b

(c) a ≤ sin2 x ≤ b

In Exercises 4-5, use transformations to graph the function in the given interval. Be surethat the key points are in their correct positions.

4. y = sin x− 1, −2π ≤ x ≤ 2π 5. y = − cosx, −π ≤ x ≤ π

In Exercises 6-8, �nd the amplitude and period and use them to graph one cycle of thefunction. Be sure that the key points are in their correct positions.

6. y = −2 sinx 7. y = cos πx 8. y = −4 cos(x4

)


In Exercises 9-12, the graph of a function of the form y = A sin(Bx) or y = A cos(Bx) isgiven. Find the amplitude and period of the function. Use this information to determine itsequation.

9. 10.

Π

2Π

-3

3

2 4 6 8

-1

1

11. 12.

3 Π3 Π2

9 Π2

6 Π

-12

12

1 2

-5

5

In Exercises 13-14, �nd the period of the function.

13. y = tan(2x) 14. y = cot(4πx)

3. 4 Inverse Trigonometric Functions 102

3. 4 Inverse Trigonometric Functions

Inverse Sine Function

Clearly, the graph of y = sin x does not pass the horizontal line test. Hence the functionf(x) = sin x is not a one-to-one function and therefore is not invertible. However, if we

restrict the domain of sin x to[−π

2,π

2

], then sinx becomes one-to-one. See the graph

below. This restricted sine function has an inverse.

-Π

2Π

2

-1

1

The inverse of f(x) = sin x, −π

2≤ x ≤ π

2is called the inverse sine function or the arcsine

function. This function is denoted by sin−1 x or arcsinx. The domain of sin−1 x is therange of sin x, which is [−1, 1]. The range of sin−1 x is the domain of the restricted sine

function, that is,[−π

2,π

2

]. More explicitly, for x in [−1, 1],

y = sin−1 x if and only if x = sin y and −π

2≤ y ≤

π

2.

Thus, sin−1 x is the angle that lies between −π

2and

π

2, inclusive, whose sine is x.

As the restricted sine function is an odd function, f(x) = sin−1 x is also an odd function.

Example 1 Find (a) sin−1 1

2, (b) sin−1(−1), (c) sin−1 1√

2, (d) sin−1

(−√3

2

).

Solution

(a) sin−1 1

2=

π

6since sin

π

6=

1

2and −π

2≤ π

6≤ π

2.

(b) sin−1(−1) = −π

2. See the restricted sine curve.

(c) sin−1 1√2=

π

4.

(d) sin−1

(−√3

2

)= − sin−1

√3

2= −π

3.


By properties of inverse functions, f(f−1(x)) = x for any x in Df−1 and f−1(f(x)) = x forany x in Df . This means that for every x in [−1, 1], sin(sin−1 x) = x and for every x in[−π

2,π

2

], sin−1(sinx) = x. However, sin−1(sinx) ̸= x in general. This is only true when x

is between −π

2and

π

2, inclusive.

sin(sin−1 x) = x for all x in [−1, 1]

sin−1(sinx) = x for x in

[−π

2,π

2

]only

Example 2 Find (a) sin(sin−1(2/3)), (b) sin−1(sin 1), (c) sin−1(sin(2π)).

Solution

(a) −1 ≤ 2/3 ≤ 1 and hence sin(sin−1(2/3)) = 2/3.

(b) Since −π

2≤ 1 ≤ π

2, sin−1(sin 1) = 1.

(c) sin−1(sin(2π)) = sin−1 0 = 0. Notice that in this case, sin−1(sin(2π)) ̸= 2π.

Inverse Tangent Function

tanx is not a one-to-one function. However, tanx restricted to(−π

2,π

2

)is one-to-one. See

the graph below.

-Π

2Π

2


The inverse of this restricted tangent function is called the inverse tangent function or thearctangent function. This function is denoted by tan−1 x or arctanx. The domain oftan−1 x is the range of the tangent function, which is (−∞, ∞). The range of tan−1 x is the

domain of the restricted tangent function, that is,(−π

2,π

2

). More explicitly, for all x,

y = tan−1 x if and only if x = tan y and −π

2< y <

π

2.

Thus, tan−1 x is the angle that lies between −π

2and

π

2whose tangent is x. Furthermore,

tan(tan−1 x) = x for all x

tan−1(tanx) = x for x in

(−π

2,π

2

)only

Since the restricted tangent function is an odd function, f(x) = tan−1 x is also an oddfunction.

Example 3 Find (a) tan−1 0, (b) tan−1(−√3), (c) tan(tan−1 2), (d) tan−1

(tan

5π

4

).

Solution

(a) tan−1 0 = 0 since tan 0 = 0 and −π

2< 0 <

π

2.

(b) tan−1(−√3) = − tan−1

√3 = −π

3.

(c) tan(tan−1 2) = 2 .

(d) tan−1

(tan

5π

4

)= tan−1 1 =

π

4.

Inverse Cosine Function

cosx is not a one-to-one function. However, cosx restricted to [ 0, π ] is one-to-one. See thegraph below.

Π

2Π

-1

1


The inverse of this restricted cosine function is called the inverse cosine function or thearccosine function. This function is denoted by cos−1 x or arccosx. Its domain is [−1, 1]and its range is [ 0, π ]. For x in [−1, 1],

y = cos−1 x if and only if x = cos y and 0 ≤ y ≤ π .

Thus, cos−1 x is the angle that lies between 0 and π, inclusive, whose cosine is x. Also,

cos(cos−1 x) = x for all x in [−1, 1]

cos−1(cosx) = x for x in [ 0, π ] only

Notice that the restricted cosine function is NOT an even function, and hence f(x) = cos−1 xis NOT an even function. So �nding the values of cos−1 x when x < 0 is a bit tricky. However,it becomes easy once one realizes that cos−1(−x) = π − cos−1 x.

Example 4 Find (a) cos−1 0, (b) cos−1

√2

2, (c) cos−1

(−1

2

), (d) cos−1(−1) + cos−1 1, (e)

cos−1

(cos

7π

6

).

Solution

(a) cos−1 0 = π/2 since cos(π/2) = 0 and 0 ≤ π/2 ≤ π .

(b) cos−1

√2

2=

π

4.

(c) cos−1

(−1

2

)= π − cos−1 1

2= π − π

3=

2π

3.

(d) cos−1(−1) + cos−1 1 = π + 0 = π since cos π = −1 and cos 0 = 1 . See the restrictedcosine curve above.

(e) cos−1

(cos

7π

6

)= cos−1

(−√3

2

)=

5π

6.

Example 5 Find (a) cos(sin−1 0), (b) tan(sin−1(−1/2)), (c) sin(cos−1(−1/√2)),

(d) cos−1(sin(π/3)).

Solution

(a) cos(sin−1 0) = cos 0 = 1 .

(b) tan(sin−1(−1/2)) = tan(−π/6) = −1/√3 .

(c) sin(cos−1(−1/√2)) = sin(3π/4) = 1/

√2 .

(d) cos−1(sin(π/3)) = cos−1(√3/2) = π/6 .


Example 6 Find the values of the six trigonometric functions of θ = tan−1 2.

Solution We cannot �nd the exact value of tan−1 2 since 2 is not the tangent of any anglewith which we are familiar. However, we can �nd the exact values of the six trigonometricfunctions of θ.

Since 2 > 0, angle θ lies in the �rst quadrant. Moreover, tan θ = 2. So θ is an angle inthe right triangle with opposite leg 2 and adjacent leg 1. The hypotenuse of this righttriangle is

√5.

1

25

Θ

We see that sin θ = 2/√5, csc θ =

√5/2, cos θ = 1/

√5, sec θ =

√5, tan θ = 2, and

cot θ = 1/2.

The Other Inverse Trigonometric Functions

The three remaining inverse trigonometric functions are y = cot−1 x, y = sec−1 x and y =csc−1 x. While we will not concentrate on these functions here, you may see these in calculus.You can use the following identities to evaluate these functions.

sec−1 x = cos−1(1/x), csc−1 x = sin−1(1/x) cot−1 x = π/2− tan−1 x .


Exercises 3. 4

1. Fill in the following table.

x −1 −√3

2− 1√

2−1

20

1

2

1√2

√3

21

sin−1 x

cos−1 x

2. Fill in the following table.

x −√3 −1 − 1√

30

1√3

1√3

tan−1 x

In Exercises 3-11, �nd the exact value of each expression.

3. sin(sin−1 0.3) 4. cos−1(cos 1.2) 5. tan−1

(tan

5π

4

)6. tan(cos−1(−1)) 7. tan−1(cos π) 8. cos

(sin−1

(−√3

2

))9. sec

(tan−1 1√

3

)10. csc

(sin−1

(− 1√

2

))11. cot

(cos−1

(− 1√

2

))

In Exercises 12-13, �nd the values of the six trigonometric functions of the given angle.

12. θ = tan−1 3

213. θ = cos−1

(−2

3

)

3. 5 Trigonometric Identities 108

3. 5 Trigonometric Identities

Basic Identities Revisited

Below is the list of basic trigonometric identities that we have derived earlier. These canbe used to simplify trigonometric expressions and to establish other trigonometric identities.Here, we use x instead of θ to denote the angle as it is more appropriate for functionalnotation.

Reciprocal Identities

csc x =1

sin xsecx =

1

cosxcotx =

1

tanx

Quotient Identities

tanx =sinx

cosxcotx =

cos x

sin x

Pythagorean Identities

sin2 x+ cos2 x = 1 tan2 x+ 1 = sec2 x cot2 x+ 1 = csc2 x

Even-Odd Identities

sin(−x) = − sin x cos(−x) = cos x tan(−x) = − tanx.

Periodic Identities

sin(x+ 2πn) = sinx cos(x+ 2πn) = cosx tan(x+ πn) = tan x

csc(x+ 2πn) = csc x sec(x+ 2πn) = sec x cot(x+ πn) = cot x

Cofunction Identities

sin(π2− x)= cos x tan

(π2− x)= cot x sec

(π2− x)= csc x

cos(π2− x)= sin x cot

(π2− x)= tan x csc

(π2− x)= sec x

Example 1 Simplify the following trigonometric expressions. (a) cotx sec x sin x, (b)tan2 x

sec2 x, (c) cosx(tanx+ cotx), (d) (sinx+ 2 cos x)2 + (2 sin x− cos x)2.

Solution To simplify the expressions below, we rewrite each expression in terms of sineand cosine. This method can be used to simplify many expressions; if you are not sure howto approach a problem, try rewriting all of the trigonometric expressions in terms of sineand cosine.

(a) cotx sec x sin x =cos x

sin x· 1

cosx· sin x = 1


(b)tan2 x

sec2 x=

sin2 x

cos2 x

1

cos2 x

=sin2 x

cos2 x· cos

2 x

1= sin2 x

(c) cosx(tanx+ cotx) = cos x

(sinx

cos x+

cosx

sinx

)= cos x

(sin2 x+ cos2 x

cos x sin x

)= cos x

(1

cos x sinx

)= cscx

(d) (sinx+ 2 cos x)2 + (2 sin x− cosx)2

= sin2 x+ 4 sin x cos x+ 4 cos2 x+ 4 sin2 x− 4 sin x cos x+ cos2 x

= 5 sin2 x+ 5 cos2 x

= 5(sin2 x+ cos2 x)

= 5

Example 2 Simplify the following trigonometric expressions. (a)tan2 x

sec x− 1, (b)

sinx

1 + cos x+

sin(−x)

1− cos(−x), (c) (cotx− csc x)(cot x+ csc x).

Solution In Parts (a) and (c), we will not convert to sine and cosine. We will usePythagorean identities instead.

(a)tan2 x

sec x− 1=

sec2 x− 1

sec x− 1=

(secx+ 1)(sec x− 1)

secx− 1= sec x+ 1

(b)sin x

1 + cos x+

sin(−x)

1− cos(−x)=

sin x

1 + cos x− sinx

1− cos x

=sin x(1− cosx)− sin x(1 + cos x)

1− cos2 x

=sin x− sin x cos x− sin x− sin x cos x

sin2 x

=−2 sin x cosx

sin2 x

=−2 cos x

sinx= −2 cotx

(c) (cotx− cscx)(cot x+ csc x) = cot2 x− csc2 x = cot2 x− (cot2 x+ 1) = −1

Notice that good algebra skills are essential when simplifying trigonometric expressions.


Example 3 Verify the following identities. (a) sec2 x+csc2 x = sec2 x csc2 x, (b) tan x(2+cotx) = 2 tanx+ 1.

Solution When verifying an identity, you should start with the more complicated side ofthe identity and manipulate it using algebra and basic trigonometric identities until youarrive at the other side.

(a) sec2 x+ csc2 x =1

cos2 x+

1

sin2 x

=sin2 x+ cos2 x

cos2 x sin2 x

=1

cos2 x sin2 x

=1

cos2 x

1

sin2 x

= sec2 x csc2 x

(b) tanx(2 + cot x) = 2 tanx+ tanx cotx = 2 tan x+ 1

Sum and Di�erence Identities

The identities below allow us to compute the sine and cosine of the sum or di�erence oftwo angles. We will not give a proof of these identities here. Instead, we refer an interestedstudent to http://mathforum.org/library/drmath/view/54051.html.

Sum and Di�erence Identities

sin(x+ y) = sinx cos y + cos x sin y sin(x− y) = sin x cos y − cos x sin y

cos(x+ y) = cos x cos y − sinx sin y cos(x− y) = cos x cos y + sinx sin y

In this course, we will only make a limited use of these identities. However, these identitiesare important in mathematics. They are very useful to know.

Example 4 Use the sum and di�erence identities to simplify the following expressions. (a)

sin(π2− x), (b) cos(π − x), (c)

√2 cos

(π4+ x)

Solution

(a) sin(π2− x)= sin

(π2

)cos x− cos

(π2

)sinx = 1 · cos x− 0 · sinx = cos x

(b) cos(π − x) = cos π cos x+ sinπ sinx = − cos x+ 0 = − cosx


(c)√2 cos

(π4+ x)=

√2(cos

π

4cosx− sin

π

4sin x

)=

√2

(1√2· cos x− 1√

2· sin x

)= cos x− sin x

Double Angle Identities

We will use the sum and di�erence identities to derive the formulas for the sine and cosineof a double angle.

sin(2x) = sin(x+ x) = sinx cosx+ cos x sin x = 2 sin x cos x

cos(2x) = cos(x+ x) = cos x cos x− sin x sin x = cos2 x− sin2 x

We can use the latter identity together with the fundamental identity, sin2 x+cos2 x = 1, toderive two more identities which are important for calculus.

cos(2x) = cos2 x− sin2 x = (1− sin2 x)− sin2 x = 1− 2 sin2 x

cos(2x) = cos2 x− sin2 x = cos2 x− (1− cos2 x) = 2 cos2 x− 1

In summary, we haveDouble Angle Identities

sin(2x) = 2 sin x cosx

cos(2x) = cos2 x− sin2 x cos(2x) = 1− 2 sin2 x cos(2x) = 2 cos2 x− 1

Example 5 Verify the following identities. (a) (sinx+ cos x)2 = 1 + sin(2x) ,

(b)1− cos(2x)

1 + cos(2x)= tan2 x, (c) tanx− cotx = −2 cot(2x).

Solution

(a) (sinx+ cos x)2 = sin2 x+ 2 sin x cosx+ cos2 x = 1 + 2 sinx cosx = 1 + sin(2x)

(b)1− cos(2x)

1 + cos(2x)=

1− (1− 2 sin2 x)

1 + (2 cos2 x− 1)=

2 sin2 x

2 cos2 x= tan2 x

(c) tanx− cotx =sin x

cos x− cos x

sin x

=sin2 x− cos2 x

cos x sin x

=− cos(2x)12sin(2x)

= −2 cot(2x)


Example 6 Let θ be an angle in Quadrant II such that sin θ =2

5. Find sin(2θ) and cos(2θ).

Determine in which quadrant the angle 2θ lies.

Solution

The reference angle α of θ has sinα = 2/5. Hence α is an angle in a right trianglewith opposite leg 2 and hypotenuse 5. By the Pythagorean Theorem, the adjacent legis√25− 4 =

√21. Hence, cosα =

√21/5 and therefore cos θ = −

√21/5 (why?).

sin(2θ) = 2 sin θ cos θ = 2 · 25· −

√21

5= −4

√21

25

cos(2θ) = cos2 θ − sin2 θ =21

25− 4

25=

17

25

Since sin(2θ) < 0 and cos(2θ) > 0, θ lies in Quadrant IV.


Exercises 3. 5

In Exercises 1-12, simplify the trigonometric expression.

1. tanx cscx cosx 2.sin(−x)

cos(−x)3.

tanx cotx

csc x

4.cot2 x

csc2 x5. sin

(π2− x)cos x+ cos

(π2− x)sin x

6. cos x(secx− cos x) 7. sin x(tanx+ cotx) 8. (1 + tan x)2 − sec2 x

9.sin(2x)

sin2 x10.

1− tanx

1− cotx11. cotx(tan x+ cot x)

In Exercises 13-14, rewrite the given expression as a single logarithm and simplify.

12. ln | cos x| − ln | sin x| 13. ln | secx− 1|+ ln | secx+ 1| − 2 ln | tanx|

In Exercises 14-18, verify the identity.

14. cos x+ sin x tanx = sec x 15. sec4 x− tan4 x = sec2 x+ tan2 x

16. (cosx+ sin x)2 − (cosx− sinx)2 = 2 sin(2x)

17. cos4 x− sin4 x = cos(2x) 18.sin2(2x)

1 + cos(2x)= 2 sin2 x

In Exercises 19-20, use the sum and di�erence identities to simplify the expression.

19. sin

(3π

2+ x

)20. cos

(π2− x)

In Exercises 21-22, �nd sin(2θ) and cos(2θ). Determine in which quadrant the angle 2θ lies.

21. cos θ =3

4and θ lies in Quadrant IV 22. cot θ = 2 and sin θ < 0

3. 6 Trigonometric Equations 114

3. 6 Trigonometric Equations

Basic Trigonometric Equations

The simplest trigonometric equation is an equation of the form f(x) = c, where f(x) is oneof the six trigonometric functions.

An equation of the form sinx = c or cos x = c has no solutions if c < −1 or c > 1.

For −1 ≤ c ≤ 1, we can �rst consider just one cycle of the curves y = sinx and y = cosx,0 ≤ x < 2π. If c = 1 or c = −1, then the equations sinx = c and cosx = c have onlyone solution in the interval [ 0, 2π). Otherwise, these equations have two solutions in thisinterval. This is illustrated in the �gure below.

Π

2Π

3 Π2

2 Π

-1

c

1

Π

2Π

3 Π2

2 Π

-1

c

1

However, these are not the only solutions. An equation of this form has in�nitely manysolutions because of periodicity. To obtain the general solution, we must add 2πn to thesolutions in the interval [ 0, 2π), where n is an integer.

Example 1 Find all solutions in the interval [ 0, 2π) and the general solution for

(a) sin x =1

2, (b) cosx = −

√2

2, (c) sin x = 0, (d) sin(2x) = 1.

Solution

(a) sinx =1

2has two solutions in [ 0, 2π), x =

π

6in Quadrant I and x =

5π

6in Quadrant

II. The general solutions to this equation are x =π

6+ 2 πn and x =

5π

6+ 2 πn, where n

is an integer.

(b) The solutions of cos x = −√2

2in [ 0, 2π) are x =

3π

4in Quadrant II and x =

5π

4in

Quadrant III. The general solutions are x =3π

4+ 2 πn and x =

5π

4+ 2 πn.

(c) The solutions to the equation sinx = 0 in [ 0, 2π) are x = 0 and x = π. If we addmultiples of 2π to these solutions, we simply get all the multiples of π. So the generalsolution is x = πn.


(d) We will use substitution to solve sin(2x) = 1.

Letting u = 2x, the equation becomes sin(u) = 1. The general solution to this equation

is u =π

2+ 2πn.

So 2x =π

2+2πn and hence x =

π

4+πn. In the interval [ 0, 2π), the solutions are x =

π

4

and x =π

4+ π =

5π

4.

The equation tanx = c has in�nitely many solutions for any value of c. In the interval [ 0, π),this equation has one solution. Since the period of tanx is π, the general equation is thenobtained by adding πn to this solution, n an integer.

Example 2 Find all solutions in the interval [ 0, 2π) and the general solution for tanx =−√3.

Solution The solution to tanx = −√3 is x =

2π

3in the interval [ 0, π). However, in the

interval [ 0, 2π), there is another solution, namely x =2π

3+ π =

5π

3. The general solution

to tanx = −√3 is given by x =

2π

3+ πn.

More Complicated Equations

Many trigonometric equations can be solved by reducing them to basic trigonometric equa-tions. This is accomplished by using trigonometric identities and algebra.

Example 3 Find all solutions in the interval [ 0, 2π) for (a) cos2 x = 1, (b)√3 tan(x)+2 =

1, (c) sin(2x) = cosx.

Solution

(a) By taking square roots, we see that cos2 x = 1 implies cosx = 1 or cosx = −1. In[ 0, 2π), cosx = 1 for x = 0 and cos x = −1 for x = π. So the solutions are x = 0 andx = π.

(b)√3 tan x+ 2 = 1 implies that tanx = − 1√

3. So the two solutions in [ 0, 2π) are x =

5π

6

and x =11π

6.


(c) The sine function on the left of this equation has a di�erent argument than the cosinefunction on the right of this equation. So we can not solve this equation in the given form.Instead, we use the double angle identity, sin(2x) = 2 sin x cos x, to make the argumenton both sides the same. Using this identity, sin(2x) = cosx becomes 2 sin x cos x = cos x.Collecting all terms to one side and factoring, we obtain

2 sin x cosx− cosx = 0

cos x(2 sin x− 1) = 0

cosx = 0 or sinx = 1/2

In [ 0, 2π), cosx = 0 for x =π

2or x =

3π

2and sin x = 1/2 for x =

π

6or x =

5π

6. So the

solutions are x =π

6, x =

π

2, x =

5π

6, and x =

3π

2.

Example 4 Find all solutions in the interval [ 0, 2π) for (a) sin2 x − sinx − 2 = 0, (b)cos(2x) = − cosx.

Solution

(a) sin2 x − sinx − 2 = 0 implies that (sinx − 2)(sinx + 1) = 0 and hence sinx = 2 orsinx = −1.

But sin x = 2 has no solutions. Hence, in [ 0, 2π), the only solution is x = 3π/2.

(b) cos(2x) = 2 cos2 x − 1. We need to replace cos(2x) with an appropriate identity. Here,we will use cos(2x) = 2 cos2 x− 1 in order to obtain a quadratic form in terms of cos x.

cos(2x) = − cosx

2 cos2 x− 1 = − cosx

2 cos2 x+ cos x− 1 = 0

(2 cos x− 1)(cosx+ 1) = 0

2 cos x− 1 = 0 or cosx+ 1

cos x =1

2or cosx = −1

In [ 0, 2π), cosx =1

2for x =

π

3or x =

5π

3and cosx = −1 for x = π. So the solutions

in [ 0, 2π) are x =π

3, x = π, and x =

5π

3.


Solutions to Basic Trigonometric Equations - the General Case

For the values of c other than those corresponding to special angles or quadrantal angles,the equations sinx = c, cosx = c, and tanx = c can be solved using inverse trigonometricfunctions:

sin x = c x = sin−1 c+ 2πn or x = π − sin−1 c+ 2πn

cos x = c x = ± cos−1 c+ 2πn

tanx = c x = tan−1 c+ πn

These formulas can also be used for the special angles and the quadrantal angles. However,be aware that inverse trigonometric functions do not always have a value in [ 0, 2π). So thesolution may need to be adjusted by adding a period of 2 π (or π in the case of tanx) if oneis looking for solutions in the interval [ 0, 2π).

Example 5 Find the general solutions and all the solutions in the interval [ 0, 2π) for

(a) cos x =2

3, (b) tanx = −1.

Solution

(a) The general solutions to cosx =2

3are x = cos−1 2

3+ 2πn and x = − cos−1 2

3+ 2πn.

cos−1 2

3lies in the interval [ 0, 2π). So x = cos−1 2

3is one of the solutions in this interval.

However, − cos−1 2

3lies in [−π, 0] and so we must add 2π to it to have a solution in

[ 0, 2π). Hence, x = − cos−1 2

3+ 2π is the second solution in [ 0, 2π).

(b) The general solution to tanx = −1 is x = tan−1(−1) + πn = −π

4+ πn. −π

4is not in

[ 0, 2π) and hence must be adjusted. So the two solutions to tanx = −1 in [ 0, 2π) are

x = −π

4+ π =

3π

4and x = −π

4+ 2π =

7π

4.

Here, we used inverse trigonometric functions to solve the equation. If we had used

the other method, we would have obtained x =3π

4+ πn for our general solution. But

this actually is the same set of values as the set given by x = −π

4+ πn. If you plug

in di�erent values of n into both formulas, you will get the same values of x, but in adi�erent order.


Exercises 3. 6

In Exercises 1-6, �nd the general solution to the equation.

1. sin x = −1 2. 2 sin x =√3 3. tanx− 1 = 0

4.√3 cotx = −1 5. cos(2x) = −1 6. sinx− cosx = 0

In Exercises 7-14, �nd all solutions to the equation in the interval [ 0, 2π).

7. 4 sin2 x− 3 = 0 8. tan2 x =1

39. 2 sin2 x = 1

10.√2 cos2 x− cos x = 0 11. 2 sin2 x− sinx− 1 = 0 12. cos2 x+ 3 cos x− 4 = 0

13. cos(2x) = sin x 14. 2 sin2 x+ sin(2x) = 0

In Exercises 15-16, �nd the general solutions and all the solutions in the interval [ 0, 2π) tothe equation.

15. tanx = 2 16. sinx = −1

3

3. 7 Polar Coordinates 119

3. 7 Polar Coordinates

Consider the point (x, y) ̸= (0, 0) in the xy-plane. The position of this point can be describedin terms of rectangular coordinates x and y. But this point can also be described by r andθ, where r is the distance from the the point (x, y) to the origin and θ is the angle formedby the positive x-axis and the ray from the origin through the point (x, y). See the picturebelow.

Θ

Hx, yL

x

yr

Θ

Hr, ΘL

x

yr

The pair (r, θ) is called the polar coordinates of the point. The conversion from polarcoordinates to rectangular coordinates is straightforward.

x = r cos θ y = r sin θ

However, the conversion from rectangular coordinates to polar coordinates is more compli-

cated. r is easy to �nd: r =√x2 + y2. As for θ, tan θ =

y

xwhen x ̸= 0. We can solve this

equation to �nd θ making sure that our solution lies in the correct quadrant.

r =√

x2 + y2 tan θ =y

x, x ̸= 0

If x = 0 and θ is in ( 0, 2π), then a quick sketch will show you that θ =π

2if y > 0 and

θ =3π

2if y < 0.


Example 1 Convert from polar coordinates to rectangular coordinates. (a) (2, π/4), (b)(1, 7π/6), (c) (4, 3π/2).

Solution

(a) x = 2 cos(π/4) = 2 ·√2

2=

√2 and y = 2 sin(π/4) = 2 ·

√2

2=

√2. So the rectangular

coordinates are (√2,√2).

(b) x = 1 · cos(7π/6) = −√3

2and y = 1 · sin(7π/6) = −1

2. The rectangular coordinates are

(−√3/2,−1/2).

(c) x = 4 cos(3π/2) = 0 and y = 4 sin(3π/2) = 4 · (−1) = −4. The rectangular coordinatesare (0,−4).

Example 2 Convert from rectangular coordinates to polar coordinates with r > 0 and θin [ 0, 2π). (a) (1,−1), (b) (

√3, 1), (c) (0, 2), (d) (−2, 3).

Solution

(a) r =√12 + (−1)2 =

√2 and tan θ = −1. Since (1,−1) lies in Quadrant IV, θ =

7π

4. So

the polar coordinates of the point are (√2, 7π/4).

(b) r =√(√3)2 + 12 = 2 and tan θ =

1√3. As (

√3, 1) lies in Quadrant I, θ =

π

6. The

polar coordinates of the point are (2, π/6).

(c) r = 2. This point lies on the positive y-axis, so θ =π

2. The polar coordinates of the

point are (2, π/2).

(d) r =√

(−2)2 + 32 =√13 and tan θ = −3

2. Since (−2, 3) lies in Quadrant II, θ =

tan−1(−3/2) + π. So the polar coordinates of the point are (√13 , tan−1(−3/2) + π).

In the examples above, we assumed that r is positive and that θ lies between 0 and 2π,including 0. This made the polar coordinates of each point unique. However, in general, oneneed not assume that r > 0 and that θ is in [ 0, 2π). So there are actually in�nitely manyways to represent a point by polar coordinates.


Exercises 3. 7

In Exercises 1-4, convert from polar coordinates to rectangular coordinates.

1.

(1

2,π

6

)2.

(6,

3π

4

)3. (e, π)

4.(100,

π

2

)In Exercises 5-8, convert from rectangular coordinates to polar coordinates with r > 0 andθ in [ 0, 2π).

5. (0,−3) 6. (−2, 2) 7. (−4,−4√3)

8. (3, 4)

3. 8 Supplementary Exercises Involving Trigonometric Functions 122

3. 8 Supplementary Exercises Involving Trigonometric Functions

1. Suppose f(x) = sin x, g(x) = ex + 1, and h(x) =1

(x− 1)2. Find

a) f ◦ g b) g ◦ f c) f ◦ h ◦ g d) f ◦ f .

2. Determine if the following functions are even, odd, or neither.

a) f(x) = x sinx b) g(x) = x2 cosx c) h(x) = x4 tanx

3. Let f(x) = ecosx. Find

a) the y−intercept, if any b) the x−intercept, if any c) the intervals on whichf(x) > 0, if any. d) the intervals on which f(x) < 0, if any.

4. Repeat Exercise 3 for f(x) = esinx − 1, 0 ≤ x ≤ 2π.

The example below will help you with exercises 5 and 6.

Example: Suppose x = 3 sin θ and −π

2< θ <

π

2. Rewrite

x√9− x2

in terms of θ. Simplify

your answer.

Given that cos θ > 0 for −π

2< θ <

π

2,

x√9− x2

=3 sin θ√

9− 9 sin2 θ=

3 sin θ

3√1− sin2 θ

=3 sin θ

3√

cos2 θ=

3 sin θ

3 cos θ= tan θ.

5. Suppose x = 2 sin θ and −π

2< θ <

π

2. Rewrite the following expressions in terms of θ.

Simplify your answer.

a)x√

4− x2b) x

√4− x2

6. Suppose x = 3 tan θ and −π

2< θ <

π

2. Rewrite the following expressions in terms of θ.

Simplify your answer.

a)√9 + x2 b)

x2

9 + x2

3. 8 Supplementary Exercises Involving Trigonometric Functions 123

The example below will help you with exercises 7 and 8.

Example: Suppose sin θ =x

3and 0 < θ <

π

2. Write cot θ in terms of x. Simplify your

answer.

Θ

x3

Θ

9- x2

x3

The angle θ in the triangle on the left has sin θ =x

3. By using the Pythagorean

Theorem, we �nd the missing leg of this triangle to obtain the triangle on the right. Wethen see that

cot θ =

√9− x2

x.

7. Suppose sin θ =x

2and 0 < θ <

π

2. Write the following expressions in terms of x.

Simplify your answers.

a) cos θ b) tan θ c) θ [Hint: Use an inverse trigonometric function.]

8. Suppose tan θ =x

3and −π

2< θ < 0. Write the following expressions in terms of x.

Simplify your answers. (Quad IV angle)

a) sin θ b) cos θ c) θ.



In Exercises 1 and 2, �nd the six trigonometric functions of θ and the following: sin(π2− θ),

cot(π2− θ), cos(4π + θ), tan(θ − π), sin(−θ), cos(−θ), tan(−θ), sin(2θ), and cos(2θ).

1. The point (−2,−3) lies on the terminal side of an angle θ.

2. Suppose cot θ = −6

5and that cos θ > 0.

3. Suppose sec θ = −10. Use trigonometric identities to �nd the exact value of a) cos θ b)

sec(−θ) c) tan2 θ d) sin2 θ e) csc(π2− θ)

f) csc(θ − π

2

). In each part, state

the identity that you are using.

4. Find the exact values of each of the following expressions.

a) cos 0 − cosπ b) sin 0 + sin π c) csc3π

4d) tan

11π

6e) − cos

(−π

3

)f) sin

7π

3

g) tan2π

3· cot π

3h) sin2

(9π

4

)+ cos2

(9π

4

)i) 2 sin

π

3+ 4 cos

π

6j) csc2

(7π

6

)−

cot2(7π

6

)k) tan

π

4+ sec2

(π4

)l)

(3 sin

π

2− 2 cos

3π

2

)·(3 sin

3π

2− 2 cos

π

2

)5. a) Use transformations to graph y = − cos x − 1. Be sure that the key points are in

their correct positions.

b) Find the amplitude and period and use them to graph one cycle of the followingfunctions. Be sure that the key points are in their correct positions. I) y = 3 sin 4x

II) y = −2 cos(πx

6

).

6. Find the exact value of each expression.

a) tan−1 1 b) tan−1(−1) c) sin−1 1 d) sin−1(−1) e) cos−1 1 f) cos−1(−1)

g) sin−1

(−√3

2

)h) tan−1

(1√3

)i) cos−1

(− 1√

2

)j) sin−1

(sin

π

12

)k) cos−1(cos 3π) l) cot

(tan−1 3

2

)m) tan−1

(sin

π

2

)n) sin

(sin−1−3

5

)o) cos

(sin−1−2

3

)7. Simplify the following trigonometric expressions.

a) 1 + sin(−x) cos(−x) tan x b)1

1− cos x+

1

1 + cos xc)

cosx

1 + sin x+ tanx

d) (1 + tanx)(1− tanx) + sec2 x e)(cot x+ 1)2 − csc2 x

cos xf)

secx+ csc x

tanx+ cotx


8. Verify the following identities.

a) (tanx+ cotx)2 − (secx+ csc x)2 = −2 sec x cscx

b) (sinx)(tanx cosx− cotx cos x) = − cos(2x)

c) cotx− tanx = 2 cot(2x) d) cos4 x− sin4 x = cos(2x)

9. Use the sum and di�erence identities to simplify the following expressions.

a) cos(π + x) b) sin

(5π

2− x

)c)

√2 cos

(x+

π

4

)10. Find the general solution to the following equations.

a) sinx cosx = 0 b) cosx− 1 = 0 c) tanx = −1 d) 2 sin x = 1

11. Find all solutions to the following equations in the interval [0, 2π) .

a) cos2 x =1

2b) 4 sin2 x = 1 c) 2 sin2 x−

√3 sin x = 0 d) 2 cos2 x− 5 cos x− 3 = 0

e)√3 sin x− cos x = 0 f) ex sinx+ ex cosx = 0

12. Convert from polar coordinates to rectangular coordinates.

a)

(4,

2π

3

)b)(2√2,

π

4

)13. Convert from rectangular coordinates to polar coordinates.

a) (−5, 0) b) (2,−2) c) (−2,−1)

Chapter 1: Answer Key 126

Section 1.1

1. (a) and (d) 2. 3, 7−2√2, 2x2+2x+3, 2x 3.

1

3, −5,

1

7,1− 2a

2a+ 3,

a

4− a4. −7, 4, 0, 5, 5

5. 2, 2, −3 6. x2 + 4x + 3, x2 + 2 7. 8x3 +1

2x, 2x3 +

2

x8. 3 9. −2x − h 10.

− 1

x(x+ h)11. − 2

(2x+ 1)(2x+ 2h+ 1)12. 4x+2h−1 13. (−∞,∞) 14. (−∞,−3)∪

(−3,∞) 15. (−∞,−2) ∪ (−2, 2) ∪ (2,∞) 16. [10,∞) 17. (−∞,∞) 18. (−∞,∞)

19.(−∞, 3

2

]20. (−∞, 2) 21.

[0, 1

2

)∪(12,∞)

22. (−∞,−2 ] ∪ [2,∞) 23. [ 1,∞)

24. [ 0, 9)∪ ( 9,∞) 25. (−∞,−1 )∪ [2,∞) 26. [ 0, 2 ] 27. (f + g)(x) = x2+x− 2, (f −g)(x) = x2 − x− 6, (fg)(x) = (2+ x) (x2 − 4) , (f/g)(x) = x− 2, (g/f)(x) =

1

x− 2, Df+g =

Df−g = Dfg = (−∞,∞), Df/g = (−∞,−2) ∪ (−2,∞), Dg/f = (−∞,−2) ∪ (−2, 2) ∪

(2,∞) 28. (f + g)(x) =2x2

x2 − 1, (f − g)(x) =

2x

x2 − 1, (fg)(x) =

x2

x2 − 1, (f/g)(x) =

x+ 1

x− 1, (g/f)(x) =

x− 1

x+ 1, Df+g = Df−g = Dfg = (−∞,−1) ∪ (−1, 1) ∪ (1,∞), Df/g =

Dg/f = (−∞,−1) ∪ (−1, 0) ∪ (0, 1) ∪ (1,∞) 29. (f + g)(x) =√x+ 4 +

√2− x, (f −

g)(x) =√x+ 4 −

√2− x+, (fg)(x) =

√x+ 4

√2− x, (f/g)(x) =

√x+ 4√2− x

, (g/f)(x) =√2− x√x+ 4

, Df+g = Df−g = Dfg = [−4, 2 ], Df/g = [−4, 2 ), Dg/f = (−4, 2 ] 30. 5, −1, 6,

2/3, 3/2 31. A(l) =l 2

232. P (l) = 3l 33. A(r) = πr2 34. r(A) =

√A

π35.

d(h) =√10000 + h2 36. V (x) = x(10− 2x)(12− 2x) = 4x3 − 44x2 + 120x

Section 1.2

1. (a) Df = [−3, 4], Rf = [−3, 1 ] (b) x = 1, x = 4; y = −1 (c) f(1) = 0, f(−2) = −3(d) f(x) = 1 at x = 2; f(x) = −3 at x = −2 (e) (1, 4) (f) [−3, 1) (g) neither 2. (a)Df = (−∞,∞)Rf = (−∞, 1 ] (b) x = −1, x = 1; y = 1 (c) f(1) = 0, f(−2) = −3(d) f(x) = 1 at x = 0; f(x) = −3 at x = −2,x = 2 (e) (−1, 1) (f) (−∞,−1)and (1,∞) (g) even 3. (a) Df = [−2, 2 ], Rf = [−2, 2 ] (b) x = −2, x = 0, x =2; y = 0 (c) f(1) = −2, f(−2) = 0 (d) f(x) = 1 at x ≈ −1.7,x ≈ −.3; f(x) = −3nowhere (e) (−2, 0) (f) (0, 2) (g) odd 4. (a) Df = (−∞,∞) (b) x = −1/3, x =

0; y = 0 (c) f(5) = 80 (d) x =−1−

√37

6,x =

−1 +√37

6(e) (−∞,−1/3) and

(0,∞) (f) (−1/3, 0) 5. (a) Df = (−∞,−1) ∪ (−1, 1) ∪ (1,∞) (b) x = −3; y = 3(c) f(5) = −1/3 (d) f(x) = 3 at x = 0, x = −1/3 (e) (−∞,−3) and (−1, 1) (f)(−3,−1) and (1,∞) 6. (a) Df = (−∞,−3] ∪ [ 3,∞) (b) x = −3, x = 3; no y-intercept (c) f(5) = 4 (d) f(x) = 3 at x = −3

√2;x = 3

√2 (e) (−∞,−3) and

( 3,∞) (f) None 7. (a) Df = (−∞,∞); Rf = (−∞,∞) (b) (−2, 2) (c) (−∞,−2)


and (5,∞) (d) (2, 5) 8. (a) Df = (−∞,∞); Rf = (−∞,∞) (b) (−∞,−1) and (1,∞)(c) (−1, 1) (d) None 9. (a) increasing on (−2, 2); decreasing on (−∞,−2) and (2,∞)(b) x = 2 (c) x = −2 (d) odd 10. (a) increasing on (−3, 0) and (3, 4); decreasing on(−4,−3) and (0, 3) (b) x = 0 (c) x = −3, x = 3 (d) even 11. (a) increasing on(−∞,−2) and (0,∞); decreasing on (−2, 0) (b) x = −2 (c) x = 0 (d) neither 12.

neither 13. even 14. odd 15. neither 16.(a)-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

(b)-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

17.

(a)-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

(b)-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

18. (5, 14) 19. (−2, 11), (1, 5) 20.

(a) neither, f(0) = g(0) (b) f(2) (c) (0, 4) (d) (−∞, 0) ∪ (4,∞) (e) x = 0, x = 4

21. One possibility:-2 -1 1 2 3 4 5

-3

-2

-1

1

2

3

4

Section 1.3

1. (a)-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

(b)-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

(c)-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

(d)-4 -3 -2 -1 1 2 3 4 5

-4

-3

-2

-1

1

2

3

4

(e)-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

2. (a) y =√x+ 5 (b) y = |x+ 1| − 1 (c) y = 3

√x− 2

(d) y = − (x+ 2)2 + 2 (e) y = (x− 2)3 + 1 (f) y = −√x+ 2


3.

1

1 Df = (−∞,∞); Rf = (−∞, 3 ]; (−3, 0), (3, 0), (0, 3)

4.1

1 Dg = (−∞,∞); Rg = [−2,∞);(2−√2, 0), (2 +

√2, 0), (0, 2)

5.1

1 Dh = (−∞,∞) = Rh; (−2, 0), (0, 8)

6.1

1

DF = (−∞,∞) = RF ; (−1, 0), (0, 1)

7.1

1

DG = [−2,∞); RG = (−∞, 0 ]; (−2, 0), (0,−√2)

8.

1

1

DH = [ 1,∞); RH = [ 4,∞); no intercepts

9.

1

1Df = (−∞,∞); Rf = [ 0,∞); (−3/2, 0), (0, 3)

10.1

1 Dg = (−∞,∞); Rg = (−∞, 4 ]; (−2, 0), (2, 0), (0, 4)


11.1

1

Df = (−∞,∞); Rf = (−1,∞)

12.1

1 Df = (−∞,∞); Rf = [−2,∞)

13. 1

1

Df = (−∞,∞); Rf = (−∞, 3]

14.1

2

Df = (−∞,∞); Rf = {−1, 0, 1}

Section 1.4

In 1-5, asymptotes are not labeled as required.

1.

-1 1-1

1VA: x = 0, HA: y = 1; (−1, 0), no y-intercept;Df = (−∞, 0) ∪ (0,∞); Rf = (−∞, 1) ∪ (1,∞)

2.-1 1-1

1

VA: x = −1, HA: y = −2; (−1/2, 0), (0,−1);Dg = (−∞,−1) ∪ (−1,∞); Rg = (−∞,−2) ∪ (−2,∞)


3.

-1 1-1

1

VA: x = 0, HA: y = −3; (−1/3, 0), no y-intercept;Dh = (−∞, 0) ∪ (0,∞); Rh = (−∞,−3) ∪ (−3,∞)

4.

-2

VA: x = −2, HA: y = 0; no x-intercept, (0, 1/4);DF = (−∞,−2) ∪ (−2,∞); RF = (0,∞)

5.-1 2

-1

2 VA: x = 1, HA: y = 2; (1/2, 0), (0, 1);DG = (−∞, 1) ∪ (1,∞); RG = (−∞, 2) ∪ (2,∞)

6. VA: x = 4, HA: y = 1; (−2, 0), (0,−1/2) 7. VA: x = 1, x = 2, HA: y = 0; (3, 0),(0,−3/2) 8. VA: x = −1, HA: None; (0, 0) 9. VA: x = −2, x = 2, HA: y = 2; nox-intercept, (0,−1/4) 10. VA: x = −1, x = 0, HA: y = 0; (−2, 0), no y-intercept 11.VA: None, HA: None; (−1/2, 0), (0, 1/4) 12. (a) III (b) II (c) IV (d) I 13. One

possibility:

-1 1-1

1

Section 1.5

R stands for "all real numbers"1. −12x + 29, R; −12x + 7, R 2. 5x2 − 22x + 25, R; 5x2 − 2x − 1, R 3. 4x − 20,

x ̸= 5;x

4− 5x, x ̸= 0, 4/5 4.

10

5 + 3x, x ̸= −5/3, 0;

5x+ 15

2x, x ̸= −3, 0 5.

√2x+ 1,

x ≥ −1/2; 1 + 2√x, x ≥ 0 6. x − 4, x ≥ 5;

√x2 − 4, x ≤ −2 and x ≥ 2 7. 2|x|,

R; 2|x − 5| + 5, R 8. x, R; x, R; 9. x, x ̸= 0; x, x ̸= 1; 10. (a) 6 (b) −3

(c) 0 (d) −6 (e) −36 11. (a) 3 (b) 2 (c) −4 (d) −3 (e) 1 12. (a) x4 − 2x2

(b) x (c)1

(x+ 2)2− 1 (d)

√x2 = |x| (e)

√√x+ 1 + 1 (f)

√x+ 3

x+ 2(g)

1

x2 + 1

(h)1√

x+ 1 + 2(i)

x+ 2

2x+ 5(j)

1

x+ 2(k)

1

|x+ 2|(l)

1

|x|+ 213. (a) f(x) = x4,

g(x) = x− 5 (b) f(x) =√x = x1/2, g(x) = 2x2 + 5 (c) f(x) =

1

x= x−1, g(x) = 3x− 1

(d) f(x) = 3√x = x1/3, g(x) = x3 − 1 (e) f(x) = |x|, g(x) = 5x − 1 (f) f(x) = x−2,

g(x) = x2 + 1


Section 1.6

1. b and d 2. (a)-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

Df = [−2, 3 ] = Rf−1 and Rf = [−3, 3 ] = Df−1

(b)H0,1L

Df = ( 0,∞) = Rf−1 and Rf = (−∞,∞) = Df−1 3. (a) 2 (b)

3 (c) 1, 7 4. f−1(x) =3− x

2; Df = (−∞,∞) = Rf−1 ; Rf = (−∞,∞) = Df−1 5.

f−1(x) = 3√x+ 2 − 1; Df = (−∞,∞) = Rf−1 ; Rf = (−∞,∞) = Df−1 6. f−1(x) =

1

2x3 − 1

2; Df = (−∞,∞) = Rf−1 ; Rf = (−∞,∞) = Df−1 7. f−1(x) = x2 − 1, x ≥ 0;

Df = [−1,∞) = Rf−1 ; Rf = [ 0,∞) = Df−1 8. f−1(x) =√x− 2; Df = [ 0,∞) =

Rf−1 ; Rf = [ 2,∞) = Df−1 9. f−1(x) =1

x− 1; Df = (−∞, 0) ∪ ( 0,∞) = Rf−1 ; Rf =

(−∞, 1) ∪ ( 1,∞) = Df−1 10. f−1(x) =1− 2x

3x+ 1; Df = (−∞,−2/3) ∪ (−2/3,∞) = Rf−1 ;

Rf = (−∞,−1/3) ∪ (−1/3,∞) = Df−1

11.

y = x

ff -1

-4 -2 2 4

-4

-2

2

4

f−1(x) =1

2x− 1

2

Df = (−∞,∞) = Rf−1 ; Rf = (−∞,∞) = Df−1

x-intercept of f is (−1/2, 0); y-intercept of f is (0, 1)

x-intercept of f−1 is (1, 0); y-intercept of f−1 is (0,−1/2)

12.

y = xf

f -1

-5 5

-5

5

f−1(x) = 3√x− 1

Df = (−∞,∞) = Rf−1 ; Rf = (−∞,∞) = Df−1

x-intercept of f is (−1, 0); y-intercept of f is (0, 1)

x-intercept of f−1 is (1, 0); y-intercept of f−1 is (0,−1)


13.

y = x

f

f -1

2 4 6

2

4

6

f−1(x) = (x− 1)2 + 2, x ≥ 1

Df = [ 2,∞) = Rf−1 ; Rf = [ 1,∞) = Df−1

Neither function has any intercepts.

14.

y = x

f

f -1

-2 2 4

-2

2

4

f−1(x) =√x+ 2 + 1

Df = [ 1,∞) = Rf−1 ; Rf = [−2,∞) = Df−1

x-intercept of f is (√2 + 1, 0); f has no y-intercept

f−1 has no x-intercept; y-intercept of f−1 is (0,√2 + 1)


1. (a)−4, −2x2+5x−4, −8x2+2x−1 (b) 1/4,x− 2

2x− 5,

1

x+ 22. (a)− 1

(x+ h+ 3)(x+ 3)

(b) −6x−3h+4 3. (a) (−∞,∞) (b) (−∞,−1)∪ (−1, 0)∪ (0, 1)∪ (1,∞) (c) [4,∞)

(d) (−2, 2) 4.x+ 2

x, 1,

x+ 1

x2, x + 1,

1

x+ 1, Df+g = Df−g = Dfg = Df/g : x ̸=

0;Dg/f : x ̸= 0, 1 5I. (a) (−∞,∞) (b) (−∞,∞) (c) x = −4, x = 0, x = 8; y = 0

(d) f(3) = 3 (e) None (f) (−∞,−4) and (0, 8) (g) (−4, 0) and (8,∞) 5II. (a)[−4, 4] (b) [−3, 4] (c) x = −3, x = 0, x = 2; y = 0 (d) f(3) = 2 (e) x = −2 (f)

[−4,−3) and (2, 4 ] (g) (−3, 0) and (0, 2) 6i. (a) (−∞,∞) (b) x = 0, x =5

3; y = 0

(c) f(−2) = −22 (d) x = −1

3, x = 2 (e) (0, 5/3) (f) (−∞, 0) and (5/3,∞) 6ii. (a)

(−∞, 3)∪(3,∞) (b) x = −2; y = −2/3 (c) f(−2) = 0 (d) x = 4/3 (e) (−∞,−2) and

(3,∞) (f) (−2, 3) 7I. (a) (−2, 2) (b) (−3,−2) and (2, 4) (c) x = 2 (d) x = −2

(e) neither 7II. (a) (−2,−1) and (1, 2) (b) (−1, 1) (c) x = −1 (d) x = 1 (e)

odd 8. (a) f(−x) = 3x2 + 5, even (b) g(−x) = − x

x2 − 1, odd (c) h(−x) =

x2

x2 + 9,

even 9. (a)

(9

5,2

5

)(b) (−1, 0) and (3, 8)

10a. f(x+ 1)− 2 = |x+ 1| − 2 and f(x− 2) + 1 = |x− 2|+ 1. The graphs are respectively:(Intercepts are not labeled as required.)


-4 -3 -2 -1 1 2

-2

-1

1

-1 1 2 3 4 5

-1

1

2

3

4

f(x+ 1)− 2 = |x+ 1| − 2 f(x− 2) + 1 = |x− 2|+ 1

x−intercept(s) (−3, 0), (1, 0) None

y−intercept (0,−1) (0, 3)

10b. f(x+ 1)− 2 =1

x+ 1− 2 and f(x− 2) + 1 =

1

x− 2+ 1.

The graphs are respectively:(Intercepts and asymptotes are not labeled as required.)

-4 -3 -2 -1 1 2

-4

-5

-3

-2

-1

1

-1 1 2 3 4

-3

-1

1

3

5

f(x+ 1)− 2 =1

x+ 1− 2 f(x− 2) + 1 =

1

x− 2+ 1

x−intercept(s) (−1/2, 0) (1, 0)

y−intercept (0,−1) (0, 1/2)

Vertical Asymptote(s) x = −1 x = 2

Horizontal Asymptote y = −2 y = 1

11. The Intercepts and asymptotes are not labeled as required.

1

1

x−intercept(s) (0, 0)

y−intercept (0, 0)

Vertical Asymptote

(one-sided) x = −1

Horizontal Asymptote

(one-sided) y = 0


12. (a) Df = (−∞,−1) ∪ (−1, 1) ∪ (1,∞), no x−intercepts, y−intercept is y = −1, VA:

x = −1, x = 1, HA: y = 0 (b) Df =

(−∞,−1

3

)∪(−1

3, 2

)∪ (2,∞), x−intercepts are

x = − 1√2, x =

1√2, y−intercept is y =

1

2, VA: x = −1

3, x = 2, HA: y =

2

3

13. (a) f(g(x)) = x2 + 5; g(f(x)) = x2 + 2x+ 5; Df◦g = Dg◦f = (−∞,∞); 21; 13

(b) f(g(x)) =3x

2− x; g(f(x)) =

2x− 2

3; Df◦g;x ̸= 0, 2;Dg◦f : x ̸= 1; −2; 2/3

(c) f(g(x)) =√2x2 + 3; g(f(x)) = 2x+ 3; Df◦g = (−∞,∞);Dg◦f = [0,∞);

√35; 7

14a. f−1(x) =3√x+ 1;Df = (−∞,∞) = Rf−1 ;Rf = (−∞,∞) = Df−1 . Graph to left

below.

f f−1

x−intercept(s) x = 1 x = −1

y−intercept y = −1 y = 1

y = xf

f -1

-5 5

-5

5

f -1

f

y = x

-2 2 4 6

-2

2

4

6

14b. f−1(x) = x2 − 1, x ≥ 0;Df = [−1,∞) = Rf−1 ;Rf = [ 0,∞) = Df−1 . Graph to rightabove.

f f−1

x−intercept(s) x = −1 x = 1

y−intercept y = 1 y = −1


14c. f−1(x) =4x+ 3

2− x;Df = (−∞, 4) ∪ (4,∞) = Rf−1 ;Rf = (−∞, 2) ∪ (2,∞) = Df−1

f f−1

x−intercept(s) x = 3/2 x = −3/4

y−intercept y = −3/4 y = 3/2


Section 2.1

1. 22x = 4x 2. 4 3. 19

4. ex+1 5. ex2/3 6. 8 7. 1 8. (

√3)x 9. 12x 10. 3

(34

)x11. I-c, II-b, III-g, IV-a; V-f; VI-h

12.

-2 -1 1 2 3

-2

2

4

6

Df = (−∞,∞); Rf = (−2, ∞); (1, 0); (0,−1); HA: y = −2

13.

-2 -1 1 2 3

-2

2

4

6

8

Dg = (−∞,∞); Rg = (0, ∞); no x-intercept, (0, 2) HA: y = 0

14.

-2 -1 1 2

-2

2

4

6

Dh = (−∞,∞); Rh = (−1, ∞); (0, 0); HA: y = −1

15.

-1 1 2 3

2

4

6

8

DF = (−∞,∞); RF = (1, ∞); no x-intercept; (0, e−2 + 1);

HA: y = 1 16. x = 3 17. x = −32

18. x = −13

19. x = −1, x = 6 20. (a) 6 (b)12

(c) 4 (d)√2 (e) 8 21. (a) f(0) = 0, f(1) = e−1, f(−1) = −e (b) x = 0 (c)

(0, ∞) (d) (−∞, 0) 22. (a) f(0) = 3, f(1) = 4, f(−1) = 2 (b) x = 3 (c) (−∞, 3)(d) (3, ∞) 23. (a) f(0) = 0, f(1) = e + 1, f(−1) = −1 − e−1 (b) x = 0 (c) (0, ∞)(d) (−∞, 0) 24. (a) f(0) = 6, f(1) = 4e−1, f(−1) = 6e (b) x = −3,x = 2 (c) (−3, 2)(d) (−∞,−3) ∪ (2, ∞) 25. (a) f(0) = 0, f(1) = 3e, f(−1) = −e−1 (b) x = −2,x = 0(c) (−∞,−2) ∪ (0, ∞) (d) (−2, 0) 26. 4 27. ex + e−x + 2 28. 0 29. e2x − 1


Section 2.2

1. 2 2. −6 3. 32

4. −4 5. 12

6. −1 7. 7 8. π 9. 13

10. 0 11. log2 8 = 312. x = log5 7 13. e1/2 =

√e 14. x = 3−2 15. (f ◦g)(x) = x−1 ; (g◦f)(x) = ln(ex−1)

16. (f ◦ g)(x) = x + 3 ; (g ◦ f)(x) = eln(x)+3 = e3x 17. (1,∞) 18. (−∞, 0) 19.(−∞,−1) ∪ (0, ∞) 20. (−2, 2) 21. (−∞, 0) ∪ (0, ∞)

22. 1 2 3 4 5 6 7

-2

2

Df = (1,∞); Rf = (−∞, ∞); (5, 0); no y-intercept; VA: x = 1

23.-3 -2 -1 1 2

-2

2

Dg = (−3,∞); Rg = (−∞, ∞); (−2, 0), (0, ln 3) VA: x = −3

24.

-2 -1 1 2 3

2

4

Dh = (−1,∞); Rh = (−∞, ∞); (e−2 − 1, 0); (0, 2); VA: x = −1

25. (a) Df = (0, ∞) (b) x = e−2 (c) (e−2, ∞) (d) (0, e−2) 26. (a) Df = (0, ∞)(b) x = 1 (c) (0, 1) ∪ (1, ∞) (d) None 27. (a) Df = (−2,∞) (b) x = −1,x = 0(c) (−1, 0) ∪ (0, ∞) (d) (−2,−1) 28. (a) 6 (b) 1.88 × 1015 joules 29. (a) 7 (b)6.3× 10−8 moles/liter

Section 2.3

1. 3 2. 1 3. 4 4. 89

5. 4 6. 2.30 7. 0.92 8. −3.22 9. 0.23 10. 4.6011. 7 12. −2 13. 1 14. −5 15. 2 ln(x2 + 1) + 1

2ln(x− 1)

16. 3 log(x− 1) + 2 log(x+ 2)− 13log x 17. 1

2ln(x2 + 1)− 5

2ln(x− 1)

18. lnx+ x − ln(x− 1)− ln(x+ 2) 19. log (x2 − 1) 20. ln

(x2

(2− x)3

)21. loga

((x+ 1)3/2

)22. log2

(x+ 1

x

)23. ln(x2 − 4) 24. x = 2 25. x =

1

226.

x = 3 27. x =√2 28. (f ◦g)(x) = ln(e2x+1) ; (g ◦f)(x) = (x+1)2 29. (f ◦g)(x) = 1

x;

(g ◦ f)(x) = 2− x


Section 2.4

1. x =√log3 10− 1, x = −

√log3 10− 1 2. x = 2 3. x = 11

94. x = 1

3ln(5

2) 5.

ln(8) − 2 6. x = 8 7. x = 5 8. x = 6 9. x = 1e2−1

10. x = ln 2 11. x = −1 12.

x = e 13. x = 0 14. x = ln 2, x = ln 3 15. x = 1, x = 3 16. x = ln 16+ln 3ln 3−ln 4

17. f−1(x) = ln(x+1)+ 2; Df = (−∞,∞) = Rf−1 ; Rf = (−1,∞) = Df−1 ; x-intercept of fis (2, 0); y-intercept of f is (0, e−2 − 1); x-intercept of f−1 is (e−2 − 1, 0); y-intercept of f−1

is (0, 2); f has an HA of y = −1 and no VA; f−1 has a VA of x = −1 and no HA;

graph of f : f

-1 1 2 3 4

-2

2

4

6

; graph of f−1:f -1

-2 2 4

-1

1

2

18. f−1(x) = ex−2 + 1; Df = (1,∞) = Rf−1 ; Rf = (−∞,∞) = Df−1 ; x-intercept of f is(e−2 + 1, 0); f has no y-intercept; f−1 has no x-intercept; y-intercept of f−1 is (0, e−2 + 1);f has a VA of x = 1 and no HA; f−1 has an HA of y = 1 and no VA;

graph of f :

f

2 4

-1

1

2

; graph of f−1:f -1

-1 1 2 3 4

-2

2

4

6

Section 2.5

1. (a) $1159.27 (b) $1161.18 (c) $1161.62 (d) $1161.83 2. 13.9 years 3. .055 =5.5% 4. (a) $31180.47 (b) $30119.42 5. (a) 220.2 thousand people (b) 2039 6.(a) 9.9 hours (b) 15.7 hours (c) 19.8 hours 7. (a) 8.86 grams (b) 30% (c) 15,700years


1. (a) e4x (b) 52 = 25 (c) 3x−2 (d) 2x/2 (e)e3x

2(f) 4 2. (a) 0 (b) 1 (c) 2

(d) 1/2 (e) 1/3 (f) −2 (g) −3 3. (a) −4 (b) 8 (c) 16 (d) 2 (e) 1 (f) 2

(g) 5/3 4. (a) −1 (b) 6 (c) −4 (d) 0 5. (a) 3 log x +1

2log(x + 1) − log(x −

2) − log(x − 1) (b)4

3ln(x − 4) − 2

3ln(x + 1) − 2

3ln(x − 1) (c) lnx + x2 − 3 ln(x2 + 1)

6. (a) log2(x1/2) (b) 0 (c) ln

((x− 1)2

(x+ 1)2

)7. (a) (−∞,∞) (b) (−∞, 0) ∪ (2,∞)

(c) (0, 2)


8.

Function Intervals where function >0 Intervals where function <0

f (−1,∞) (−∞,−1)

g(−√2,√2) (

−∞,−√2)∪(√

2,∞)

h (−∞, 0) ∪ (0,∞) None

F (e−1/2,∞) (0, e−1/2)

G (1,∞) (0, 1)

9. (a) x = 4 (b) x = 1, x = 2 (c) x = 2 (d) x = 1/2 (e) x =log3 4

2(f) x =

ln

(3

2

)− 1 (g) x = 45 = 1024 (h) x = 5 (i) x = 9− log2 5 (j) x = ln 4 (k) x = ln 5

(l) x =2e−1

1− e−1=

2

e− 1(m) x = 9 (n) x = e−2/3 (o) x =

3 ln 3 + 2 ln 2

ln 3− ln 2

10. (a)f(g(x)) = x2, g(f(x)) = ln(e2x − 2ex +2) (b)f(g(x)) = 2x, g(f(x)) = x2 − 6x+12

11a. f(x) = ex+1 + 3; f−1(x) = ln(x− 3)− 1;Df = (−∞,∞) = Rf−1 ;Rf = (3,∞) = Df−1 .

f f−1

x−intercept(s) None (e+ 3, 0)

y−intercept (0, e+ 3) None

VA None x = 3

HA y = 3 None

Asymptotes and intercepts are not labeled as required.

f

-3 -2 -1 1 2

1

3

5

7 f -1

1 3 5 7 9

-2

-1

1

2

11b. f(x) = ln(x+ 2)− 1; f−1(x) = ex+1 − 2;Df = (−2,∞) = Rf−1 ;Rf = (∞,∞) = Df−1 .

f f−1

x−intercept(s) (e− 2, 0) (ln 2− 1, 0)

y−intercept (0, ln 2− 1) (0, e− 2)

VA x = −2 None

HA None y = −2


Asymptotes and intercepts are not labeled as required.

f

-2 1 3 5 7 9

-2

-1

1

2

f -1

-3 -2 -1 1 2

-2

2

4


Section 3.1

1. sin θ =12

13, cos θ =

5

13, tan θ =

12

5, csc θ =

13

12, sec θ =

13

5, cot θ =

5

12

2. sin θ =2√10

7, cos θ =

3

7, tan θ =

2√10

3, csc θ =

7

2√10, sec θ =

7

3, cot θ =

3

2√10

3. sin θ =1

4, cos θ =

√15

4, tan θ =

1√15, csc θ = 4, sec θ =

4√15, cot θ =

√15

4. α = 65◦, a = 9.1, b = 4.2 5. β = 15◦, a = 11.2, c = 11.6 6. α = 40◦, a = 5.9, c = 9.1

7.π

38.

5π

49. −5π

610.

5π

211. 15◦ 12. 210◦ 13. 540◦ 14.

Π

3

15.

- Π

16.

3 Π

4

17.-

5 Π

6

18.5 Π

3

19.

5 Π

2

20. 0 21. 1 22. 1 +√3 23. a = 2

√3, b = 2 24. a = 3, c = 3

√2

Section 3.2

1. sin θ =5√29, cos θ =

2√29, tan θ =

5

2, csc θ =

√29

5, sec θ =

√29

2, cot θ =

2

5

2. sin θ = −3

5, cos θ =

4

5, tan θ = −3

4, csc θ = −5

3, sec θ =

5

4, cot θ = −4

3

3. (a)40

49(b)

7

3(c)

7

3(d)

40

94. (a) 10 (b)

1

10(c)

9

10(d) 3

5. sin θ = −√7

4, tan θ = −

√7

3, csc θ = − 4√

7, sec θ =

4

3, cot θ = − 3√

7


6. sin θ =2√2

3, cos θ =

1

3, tan θ = 2

√2, csc θ =

3

2√2, cot θ =

1

2√2

7. sin θ = − 3√13, cos θ = − 2√

13, csc θ = −

√13

3, sec θ = −

√13

2, cot θ =

2

3

8. sin θ =2

3, cos θ = −

√5

3, tan θ = − 2√

5, sec θ = − 3√

5, cot θ = −

√5

2

9. 1 10. −1 11. −1

212. − 1√

313. Unde�ned 14. −1

215. −1 16. 0 17.

√2

18. 2 19. −1 20. 1 21.1√3

22. Unde�ned 23. −1 24. −1

2;1

225. 4 26. 3

Section 3.3

1.Π

2Π

3 Π2

2 Π 5 Π2

3 Π 7 Π2

4 Π

-1

1

(a) 1,−1, 0 (b) increasing on (0, π/2) and (3π/2, 5π/2) and (7π/2, 4π); decreasing on(π/2, 3π/2) and (5π/2, 7π/2) (c) x = 0, π, 2π, 3π, 4π (d) f(x) > 0 on (0, π) and (2π, 3π);f(x) < 0 on (π, 2π) and (3π, 4π) (e) x = π/2, x = 5π/2; f(x) = 1 (f) x = 3π/2, x =7π/2

2.Π

2Π

3 Π2

2 Π 5 Π2

3 Π 7 Π2

4 Π

-1

1

(a) −1, 0, 1 (b) increasing on (π, 2π) and (3π, 4π); decreasing on (0, π) and (2π, 3π) (c)x = π/2, 3π/2, 5π/2, 7π/2 (d) f(x) > 0 on [0, π/2) and (3π/2, 5π/2) and (7π/2, 4π];f(x) < 0 on (π/2, 3π/2) and (5π/2, 7π/2) (e) x = π, x = 3π; f(x) = −1 (f) x = 0, x =2π, x = 4π

3. (a) a = 0, b = 2 (b) a = −3, b = −1 (c) a = 0, b = 1

4.-2 Π - 3 Π

2-Π -

Π

2Π

2Π

3 Π2

2 Π

-2

-1

1

5.-Π -

Π

2Π

2Π

-1

1


6.Π

2Π

3 Π2

2 Π

-2

2

Amplitude = 2; Period = 2π

7.12

1 32

2

-1

1

Amplitude = 1; Period = 2

8.6 Π2 Π 8 Π4 Π4 Π

-4

4

Amplitude = 4; Period = 8π

9. Amplitude = 3; Period = π; y = 3 cos(2x) 10. Amplitude = 1; Period = 8; y =

− sin(π4x)

11. Amplitude = 1/2; Period = 6π; y =1

2sin(x3

)12. Amplitude = 5;

Period = 1; y = −5 cos(2πx) 13. Period = π/2 14. Period = 1/4

Section 3.4

1.

x −1 −√3

2− 1√

2−1

20

1

2

1√2

√3

21

sin−1 x −π

2− π

3−π

4−π

60

π

6

π

4

π

3

π

2

cos−1 x π5π

6

3π

4

2π

3

π

2

π

3

π

4

π

60

2.

x −√3 −1 − 1√

30

1√3

1√3

tan−1 x −π

3−π

4− π

60

π

6

π

4

π

3


3. 0.3 4. 1.2 5.π

46. 0 7. −π

48.

1

29.

2√3

10. −√2 11. −1

12. sin θ =3√13, cos θ =

2√13, tan θ =

3

2, csc θ =

√13

3, sec θ =

√13

2, cot θ =

2

3

13. sin θ =

√5

3, cos θ = −2

3, tan θ = −

√5

2, csc θ =

3√5, sec θ = −3

2, cot θ = − 2√

5

Section 3.5

1. 1 2. − tanx 3. sin x 4. cos2 x 5. 1 6. sin2 x 7. sec x 8. 2 tan x 9. 2 cotx10. − tanx 11. csc2 x 12. ln | cot x| 13. 0 19. − cos x 20. sinx 21. sin(2x) =

−3√7

8; cos(2x) =

1

8; Quadrant IV 22. sin(2x) =

4

5; cos(2x) =

3

5; Quadrant I

Section 3.6

1. x = −π

2+ 2πn, or equivalently, x =

3π

2+ 2πn 2. x =

π

3+ 2πn, x =

2π

3+ 2πn

3. x =π

4+ πn 4. x =

2π

3+ πn, or equivalently, x = −π

3+ πn 5. x =

π

2+ πn, or

equivalently, x = −π

2+ πn 6. x =

π

4+ πn 7. x =

π

3, x =

2π

3, x =

4π

3, x =

5π

3

8. x =π

6, x =

5π

6, x =

7π

6, x =

11π

69. x =

π

4, x =

3π

4, x =

5π

4, x =

7π

410.

x =π

4, x =

π

2, x =

3π

2, x =

7π

411. x =

π

2, x =

7π

6, x =

11π

612. x = 0 13.

x =π

6, x =

5π

6, x =

3π

214. x = 0, x =

3π

4, x =

7π

4, x = π 15. x = tan−1 2 + πn;

x = tan−1 2 and x = tan−1 2 + π 16. x = sin−1

(−1

3

)+ 2πn, x = π − sin−1

(−1

3

)+ 2πn;

x = π − sin−1

(−1

3

)and x = sin−1

(−1

3

)+ 2π

Section 3.7

1.

(√3

4,1

4

)2.(−3

√2, 3

√2)

3. (−e, 0) 4. (0, 100) 5.

(3,

3π

2

)6.

(2√2,

3π

4

)7.

(8,

4π

3

)8.

(5, tan−1 4

3

)


Section 3.8

1. (a) sin(ex + 1) (b) esinx + 1 (c) sin(e−2x) (d) sin(sinx) 2. (a) even (b) even(c) odd 3. (a) y = e (b) none (c) (−∞,∞) (d) none 4. (a) y = 0 (b) x = 0,x = π, x = 2π (c) (0, π) (d) (π, 2π) 5. (a) tan θ (b) 2 sin(2θ) 6. (a) 3 sec θ

(b) sin2 θ 7. (a)

√4− x2

2(b)

x√4− x2

(c) sin−1(x2

)8. (a)

x√9 + x2

(b)

3√9 + x2

(c) tan−1(x3

)


1. θ is in Quad III: sin(θ) = − 3√13, cos(θ) = − 2√

13, tan(θ) =

3

2, csc(θ) = −

√13

3,

sec(θ) = −√13

2, cot(θ) =

2

3, sin

(π2− θ)= − 2√

13, cot

(π2− θ)=

3

2, cos(4π + θ) = − 2√

13,

tan(θ − π) =3

2, sin(−θ) =

3√13, cos(−θ) = − 2√

13, tan(−θ) = −3

2, sin(2θ) =

12

13, and

cos(2θ) = − 5

13. 2. θ is in Quad IV: sin(θ) = − 5√

61, cos(θ) =

6√61, tan(θ) = −5

6,

csc(θ) = −√61

5, sec(θ) =

√61

6, cot(θ) = −6

5, sin

(π2− θ)

=6√61, cot

(π2− θ)

= −5

6,

cos(4π + θ) =6√61, tan(θ − π) = −5

6, sin(−θ) =

5√61, cos(−θ) =

6√61, tan(−θ) =

5

6,

sin(2θ) = −60

61, and cos(2θ) =

11

61. 3. (a) − 1

10(b) −10 (c) 99 (d)

99

100(e) −10

(f) 10. The identities that you use must be stated for full credit. 4. (a) 2 (b)

0 (c)√2 (d) − 1√

3(e) −1

2(f)

√3

2(g) −1 (h) 1 (i) 3

√3 (j) 1 (k) 3 (l) −9

5. (a) y = − cos x− 1

Π

2Π

3 Π2

2 Π

-2

-1

1

(bI) y = 3 sin 4x (bII) y = −2 cos(πx

6

)


Π

8Π

43 Π8

Π

2

-3

-2

-1

1

2

3

3 6 9 12

-2

-1

1

2

6. (a)π

4(b) −π

4(c)

π

2(d) −π

2(e) 0 (f) π (g) −π

3(h)

π

6(i)

3π

4(j)

π

12

(k) π (l)2

3(m)

π

4(n) −3

5(o)

√5

37. (a) cos2 x (b) 2 csc2 x (c) secx (d)

2 (e) 2 csc x (f) sinx+cos x 9. (a) − cos x (b) cosx (c) cos x− sin x 10. (a)

x = πn, x =π

2+ πn (b) x = 2πn (c) x = −π

4+ πn or equivalently x =

3π

4+ πn

(d) x =π

6+ 2πn, x =

5π

6+ 2πn 11. (a) x =

π

4, x =

3π

4, x =

5π

4, x =

7π

4(b)

x =π

6, x =

5π

6, x =

7π

6, x =

11π

6(c) x = 0, x = π, x =

π

3, x =

2π

3(d) x =

2π

3, x =

4π

3

(e) x =π

6, x =

7π

6(f) x =

3π

4, x =

7π

412. (a) (−2, 2

√3) (b) (2, 2) 13. (a)

(5, π) (b)

(2√2,

7π

4

)(c)

(√5, tan−1

(1

2

)+ π

)

Math 1022 - Precalculus · Math 1022 - Precalculus: Table of Contents Chapter 1: unctionsF 1.1...

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Transcript of Math 1022 - Precalculus · Math 1022 - Precalculus: Table of Contents Chapter 1: unctionsF 1.1...