Stewart solution 6ed

1 FUNCTIONS AND MODELS1.1 Four Ways to Represent a Function

In exercises requiring estimations or approximations, your answers may vary slightly from the answers given here.

1. (a) The point (1,2) is on the graph of f , so f(1) = 2.(b) When x = 2, y is about 2.8, so f(2) 2.8.(c) f(x) = 2 is equivalent to y = 2. When y = 2, we have x = 3 and x = 1.(d) Reasonable estimates for x when y = 0 are x = 2.5 and x = 0.3.(e) The domain of f consists of all x-values on the graph of f . For this function, the domain is 3 x 3, or [3, 3].

The range of f consists of all y-values on the graph of f . For this function, the range is2 y 3, or [2, 3].(f ) As x increases from 1 to 3, y increases from 2 to 3. Thus, f is increasing on the interval [1, 3].

3. From Figure 1 in the text, the lowest point occurs at about (t, a) = (12,85). The highest point occurs at about (17, 115).Thus, the range of the vertical ground acceleration is85 a 115. Written in interval notation, we get [85, 115].

5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve failsthe Vertical Line Test.

7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [3, 2] and the rangeis [3,2) [1, 3].

9. The persons weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The persons weightdropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual

increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems.

11. The water will cool down almost to freezing as the ice

melts. Then, when the ice has melted, the water will

slowly warm up to room temperature.

13. Of course, this graph depends strongly on the

geographical location!

15. As the price increases, the amount sold decreases.

17.

9

10 CHAPTER 1 FUNCTIONS AND MODELS

19. (a) (b) From the graph, we estimate the number of

cell-phone subscribers worldwide to be about

92 million in 1995 and 485 million in 1999.

21. f(x) = 3x2 x+ 2.f(2) = 3(2)2 2 + 2 = 12 2 + 2 = 12.f(2) = 3(2)2 (2) + 2 = 12 + 2 + 2 = 16.f(a) = 3a2 a+ 2.f(a) = 3(a)2 (a) + 2 = 3a2 + a+ 2.f(a+ 1) = 3(a+ 1)2 (a+ 1) + 2 = 3(a2 + 2a+ 1) a 1 + 2 = 3a2 + 6a+ 3 a+ 1 = 3a2 + 5a+ 4.2f(a) = 2 f(a) = 2(3a2 a+ 2) = 6a2 2a+ 4.f(2a) = 3(2a)2 (2a) + 2 = 3(4a2) 2a+ 2 = 12a2 2a+ 2.f(a2) = 3(a2)2 (a2) + 2 = 3(a4) a2 + 2 = 3a4 a2 + 2.

[f(a)]2 = 3a2 a+ 2 2 = 3a2 a+ 2 3a2 a+ 2= 9a4 3a3 + 6a2 3a3 + a2 2a+ 6a2 2a+ 4 = 9a4 6a3 + 13a2 4a+ 4.

f(a+ h) = 3(a+ h)2 (a+ h) + 2 = 3(a2 + 2ah+ h2) a h+ 2 = 3a2 + 6ah+ 3h2 a h+ 2.

23. f(x) = 4 + 3x x2, so f(3 + h) = 4 + 3(3 + h) (3 + h)2 = 4 + 9 + 3h (9 + 6h+ h2) = 4 3h h2,

and f(3 + h) f(3)h

=(4 3h h2) 4

h=

h(3 h)h

= 3 h.

25. f(x) f(a)x a =

1

x 1a

x a =a xxa

x a =a x

xa(x a) =1(x a)xa(x a) =

1

ax

27. f(x) = x/(3x 1) is defined for all x except when 0 = 3x 1 x = 13

, so the domain

is x R | x 6= 13= , 1

3 1

3 , .

29. f(t) =t+ 3

t is defined when t 0. These values of t give real number results fort, whereas any value of t gives a real

number result for 3t. The domain is [0,).

31. h(x) = 1 4x2 5x is defined when x2 5x > 0 x(x 5) > 0. Note that x2 5x 6= 0 since that would result in

division by zero. The expression x(x 5) is positive if x < 0 or x > 5. (See Appendix A for methods for solvinginequalities.) Thus, the domain is (, 0) (5,).

SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION 11

33. f(x) = 5 is defined for all real numbers, so the domain is R, or (,).The graph of f is a horizontal line with y-intercept 5.

35. f(t) = t2 6t is defined for all real numbers, so the domain is R, or(,). The graph of f is a parabola opening upward since the coefficientof t2 is positive. To find the t-intercepts, let y = 0 and solve for t.

0 = t2 6t = t(t 6) t = 0 and t = 6. The t-coordinate of thevertex is halfway between the t-intercepts, that is, at t = 3. Since

f(3) = 32 6 3 = 9, the vertex is (3,9).

37. g(x) =x 5 is defined when x 5 0 or x 5, so the domain is [5,).

Since y =x 5 y2 = x 5 x = y2 +5, we see that g is the

top half of a parabola.

39. G(x) = 3x+ |x|x

. Since |x| =x if x 0x if x < 0 , we have

G(x) =

3x+ x

xif x > 0

3x xx

if x < 0=

4x

xif x > 0

2x

xif x < 0

=4 if x > 0

2 if x < 0

Note that G is not defined for x = 0. The domain is (, 0) (0,).

41. f(x) =x+ 2 if x < 0

1 x if x 0The domain is R.

43. f(x) =x+ 2 if x 1x2 if x > 1

Note that for x = 1, both x+ 2 and x2 are equal to 1.The domain is R.

45. Recall that the slope m of a line between the two points (x1, y1) and (x2, y2) is m =y2 y1x2 x1 and an equation of the line

connecting those two points is y y1 = m(x x1). The slope of this line segment is 7 (3)5 1 =

5

2, so an equation is

y (3) = 52(x 1). The function is f(x) = 5

2x 11

2, 1 x 5.


47. We need to solve the given equation for y. x+ (y 1)2 = 0 (y 1)2 = x y 1 = x y = 1x. The expression with the positive radical represents the top half of the parabola, and the one with the negativeradical represents the bottom half. Hence, we want f(x) = 1x. Note that the domain is x 0.

49. For 0 x 3, the graph is the line with slope 1 and y-intercept 3, that is, y = x+ 3. For 3 < x 5, the graph is the linewith slope 2 passing through (3, 0); that is, y 0 = 2(x 3), or y = 2x 6. So the function is

f(x) =x+ 3 if 0 x 32x 6 if 3 < x 5

51. Let the length and width of the rectangle be L and W . Then the perimeter is 2L+ 2W = 20 and the area is A = LW .

Solving the first equation for W in terms of L gives W = 20 2L2

= 10L. Thus, A(L) = L(10L) = 10LL2. Since

lengths are positive, the domain of A is 0 < L < 10. If we further restrict L to be larger than W , then 5 < L < 10 would be

the domain.

53. Let the length of a side of the equilateral triangle be x. Then by the Pythagorean Theorem, the height y of the triangle satisfies

y2 + 12x2= x2, so that y2 = x2 14x2 = 34x2 and y =

32 x. Using the formula for the area A of a triangle,

A = 12(base)(height), we obtain A(x) = 1

2(x)

32x =

34x2, with domain x > 0.

55. Let each side of the base of the box have length x, and let the height of the box be h. Since the volume is 2, we know that

2 = hx2, so that h = 2/x2, and the surface area is S = x2 + 4xh. Thus, S(x) = x2 + 4x(2/x2) = x2 + (8/x), with

domain x > 0.

57. The height of the box is x and the length and width are L = 20 2x, W = 12 2x. Then V = LWx and soV (x) = (20 2x)(12 2x)(x) = 4(10 x)(6 x)(x) = 4x(60 16x+ x2) = 4x3 64x2 + 240x.The sides L, W , and x must be positive. Thus, L > 0 20 2x > 0 x < 10;W > 0 12 2x > 0 x < 6; and x > 0. Combining these restrictions gives us the domain 0 < x < 6.

59. (a) (b) On $14,000, tax is assessed on $4000, and 10%($4000) = $400.

On $26,000, tax is assessed on $16,000, and

10%($10,000) + 15%($6000) = $1000 + $900 = $1900.

(c) As in part (b), there is $1000 tax assessed on $20,000 of income, so

the graph of T is a line segment from (10,000, 0) to (20,000, 1000).

The tax on $30,000 is $2500, so the graph of T for x > 20,000 is

the ray with initial point (20,000, 1000) that passes through

(30,000, 2500).

SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS 13

61. f is an odd function because its graph is symmetric about the origin. g is an even function because its graph is symmetric with

respect to the y-axis.

63. (a) Because an even function is symmetric with respect to the y-axis, and the point (5, 3) is on the graph of this even function,

the point (5, 3) must also be on its graph.

(b) Because an odd function is symmetric with respect to the origin, and the point (5, 3) is on the graph of this odd function,

the point (5,3) must also be on its graph.

65. f(x) = xx2 + 1

.

f(x) = x(x)2 + 1 =

xx2 + 1

= xx2 + 1

= f(x).

So f is an odd function.

67. f(x) = xx+ 1

, so f(x) = xx+ 1 =x

x 1 .

Since this is neither f(x) nor f(x), the function f isneither even nor odd.

69. f(x) = 1 + 3x2 x4.f(x) = 1 + 3(x)2 (x)4 = 1 + 3x2 x4 = f(x).So f is an even function.

1.2 Mathematical Models: A Catalog of Essential Functions

1. (a) f(x) = 5x is a root function with n = 5.

(b) g(x) =1 x2 is an algebraic function because it is a root of a polynomial.

(c) h(x) = x9 + x4 is a polynomial of degree 9.

(d) r(x) = x2 + 1

x3 + xis a rational function because it is a ratio of polynomials.

(e) s(x) = tan 2x is a trigonometric function.

(f ) t(x) = log10 x is a logarithmic function.

3. We notice from the figure that g and h are even functions (symmetric with respect to the y-axis) and that f is an odd function

(symmetric with respect to the origin). So (b) y = x5 must be f . Since g is flatter than h near the origin, we must have

(c) y = x8 matched with g and (a) y = x2 matched with h.


5. (a) An equation for the family of linear functions with slope 2

is y = f(x) = 2x+ b, where b is the y-intercept.

(b) f(2) = 1 means that the point (2, 1) is on the graph of f . We can use the

point-slope form of a line to obtain an equation for the family of linear

functions through the point (2, 1). y 1 = m(x 2), which is equivalentto y = mx+ (1 2m) in slope-intercept form.

(c) To belong to both families, an equation must have slope m = 2, so the equation in part (b), y = mx+ (1 2m),

becomes y = 2x 3. It is the only function that belongs to both families.

7. All members of the family of linear functions f(x) = c x have graphsthat are lines with slope 1. The y-intercept is c.

9. Since f(1) = f(0) = f(2) = 0, f has zeros of 1, 0, and 2, so an equation for f is f(x) = a[x (1)](x 0)(x 2),or f(x) = ax(x+ 1)(x 2). Because f(1) = 6, well substitute 1 for x and 6 for f(x).6 = a(1)(2)(1) 2a = 6 a = 3, so an equation for f is f(x) = 3x(x+ 1)(x 2).

11. (a) D = 200, so c = 0.0417D(a+ 1) = 0.0417(200)(a+ 1) = 8.34a+ 8.34. The slope is 8.34, which represents the

change in mg of the dosage for a child for each change of 1 year in age.

(b) For a newborn, a = 0, so c = 8.34 mg.

13. (a) (b) The slope of 95

means that F increases 95

degrees for each increase

of 1C. (Equivalently, F increases by 9 when C increases by 5

and F decreases by 9 when C decreases by 5.) The F -intercept of

32 is the Fahrenheit temperature corresponding to a Celsius

temperature of 0.

SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS 15

15. (a) Using N in place of x and T in place of y, we find the slope to be T2 T1N2 N1 =

80 70173 113 =

10

60=1

6. So a linear

equation is T 80 = 16(N 173) T 80 = 1

6N 173

6 T = 1

6N + 307

63076= 51.16 .

(b) The slope of 16

means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket

chirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1F.

(c) When N = 150, the temperature is given approximately by T = 16(150) + 307

6= 76.16 F 76 F.

17. (a) We are given change in pressure10 feet change in depth

=4.34

10= 0.434. Using P for pressure and d for depth with the point

(d, P ) = (0, 15), we have the slope-intercept form of the line, P = 0.434d+ 15.

(b) When P = 100, then 100 = 0.434d+ 15 0.434d = 85 d = 850.434 195.85 feet. Thus, the pressure is

100 lb/in2 at a depth of approximately 196 feet.

19. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form

f(x) = a cos(bx) + c seems appropriate.

(b) The data appear to be decreasing in a linear fashion. A model of the form f(x) = mx+ b seems appropriate.

Some values are given to many decimal places. These are the results given by several computer algebra systems rounding is left to the reader.

21. (a)

A linear model does seem appropriate.

(b) Using the points (4000, 14.1) and (60,000, 8.2), we obtain

y 14.1 = 8.2 14.160,000 4000 (x 4000) or, equivalently,

y 0.000105357x+ 14.521429.

(c) Using a computing device, we obtain the least squares regression line y = 0.0000997855x+ 13.950764.The following commands and screens illustrate how to find the least squares regression line on a TI-83 Plus.

Enter the data into list one (L1) and list two (L2). Press to enter the editor.

Find the regession line and store it in Y1. Press .


Note from the last figure that the regression line has been stored in Y1 and that Plot1 has been turned on (Plot1 is

highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing or by

pressing .

Now press to produce a graph of the data and the regression

line. Note that choice 9 of the ZOOM menu automatically selects a window

that displays all of the data.

(d) When x = 25,000, y 11.456; or about 11.5 per 100 population.

(e) When x = 80,000, y 5.968; or about a 6% chance.(f ) When x = 200,000, y is negative, so the model does not apply.

23. (a)

A linear model does seem appropriate.

(b)

Using a computing device, we obtain the least squaresregression line y = 0.089119747x 158.2403249,where x is the year and y is the height in feet.

(c) When x = 2000, the model gives y 20.00 ft. Note that the actual winning height for the 2000 Olympics is less than thewinning height for 1996so much for that prediction.

(d) When x = 2100, y 28.91 ft. This would be an increase of 9.49 ft from 1996 to 2100. Even though there was an increaseof 8.59 ft from 1900 to 1996, it is unlikely that a similar increase will occur over the next 100 years.

25. Using a computing device, we obtain the cubic

function y = ax3 + bx2 + cx+ d witha = 0.0012937, b = 7.06142, c = 12,823,and d = 7,743,770. When x = 1925,y 1914 (million).

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS 17

1.3 New Functions from Old Functions

1. (a) If the graph of f is shifted 3 units upward, its equation becomes y = f(x) + 3.

(b) If the graph of f is shifted 3 units downward, its equation becomes y = f(x) 3.

(c) If the graph of f is shifted 3 units to the right, its equation becomes y = f(x 3).

(d) If the graph of f is shifted 3 units to the left, its equation becomes y = f(x+ 3).

(e) If the graph of f is reflected about the x-axis, its equation becomes y = f(x).

(f ) If the graph of f is reflected about the y-axis, its equation becomes y = f(x).

(g) If the graph of f is stretched vertically by a factor of 3, its equation becomes y = 3f(x).

(h) If the graph of f is shrunk vertically by a factor of 3, its equation becomes y = 13f(x).

3. (a) (graph 3) The graph of f is shifted 4 units to the right and has equation y = f(x 4).

(b) (graph 1) The graph of f is shifted 3 units upward and has equation y = f(x) + 3.

(c) (graph 4) The graph of f is shrunk vertically by a factor of 3 and has equation y = 13f(x).

(d) (graph 5) The graph of f is shifted 4 units to the left and reflected about the x-axis. Its equation is y = f(x+ 4).

(e) (graph 2) The graph of f is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation is

y = 2f(x+ 6).

5. (a) To graph y = f(2x) we shrink the graph of fhorizontally by a factor of 2.

The point (4,1) on the graph of f corresponds to thepoint 1

2 4,1 = (2,1).

(b) To graph y = f 12x we stretch the graph of f

horizontally by a factor of 2.

The point (4,1) on the graph of f corresponds to thepoint (2 4,1) = (8,1).

(c) To graph y = f(x) we reflect the graph of f aboutthe y-axis.

The point (4,1) on the graph of f corresponds to thepoint (1 4,1) = (4,1).

(d) To graph y = f(x) we reflect the graph of f aboutthe y-axis, then about the x-axis.

The point (4,1) on the graph of f corresponds to thepoint (1 4,1 1) = (4, 1).


7. The graph of y = f(x) =3x x2 has been shifted 4 units to the left, reflected about the x-axis, and shifted downward

1 unit. Thus, a function describing the graph is

y = 1 reflect

about x-axis

f (x+ 4)

shift4 units left

1shift

1 unit left

This function can be written as

y = f(x+ 4) 1 = 3(x+ 4) (x+ 4)2 1 = 3x+ 12 (x2 + 8x+ 16) 1 = x2 5x 4 1

9. y = x3: Start with the graph of y = x3 and reflectabout the x-axis. Note: Reflecting about the y-axis

gives the same result since substitutingx for x givesus y = (x)3 = x3.

11. y = (x+ 1)2: Start with the graph of y = x2

and shift 1 unit to the left.

13. y = 1 + 2 cosx: Start with the graph of y = cosx, stretch vertically by a factor of 2, and then shift 1 unit upward.

15. y = sin(x/2): Start with the graph of y = sinx and stretch horizontally by a factor of 2.


17. y =x+ 3 : Start with the graph of

y =x and shift 3 units to the left.

19. y = 12 (x2 + 8x) = 12 (x

2 + 8x+ 16) 8 = 12 (x+ 4)2 8: Start with the graph of y = x2, compress vertically by a

factor of 2, shift 4 units to the left, and then shift 8 units downward.

0 0 0 0

21. y = 2/(x+ 1): Start with the graph of y = 1/x, shift 1 unit to the left, and then stretch vertically by a factor of 2.

23. y = |sinx|: Start with the graph of y = sinx and reflect all the parts of the graph below the x-axis about the x-axis.

25. This is just like the solution to Example 4 except the amplitude of the curve (the 30N curve in Figure 9 on June 21) is

14 12 = 2. So the function is L(t) = 12 + 2 sin 2365 (t 80) . March 31 is the 90th day of the year, so the model gives

L(90) 12.34 h. The daylight time (5:51 AM to 6:18 PM) is 12 hours and 27 minutes, or 12.45 h. The model value differs

from the actual value by 12.4512.3412.45 0.009, less than 1%.27. (a) To obtain y = f(|x|), the portion of the graph of y = f(x) to the right of the y-axis is reflected about the y-axis.

(b) y = sin |x| (c) y = |x|


29. f(x) = x3 + 2x2; g(x) = 3x2 1. D = R for both f and g.

(f + g)(x) = (x3 + 2x2) + (3x2 1) = x3 + 5x2 1, D = R.

(f g)(x) = (x3 + 2x2) (3x2 1) = x3 x2 + 1, D = R.

(fg)(x) = (x3 + 2x2)(3x2 1) = 3x5 + 6x4 x3 2x2, D = R.

f

g(x) =

x3 + 2x2

3x2 1 , D = x | x 6= 13

since 3x2 1 6= 0.

31. f(x) = x2 1, D = R; g(x) = 2x+ 1, D = R.(a) (f g)(x) = f(g(x)) = f(2x+ 1) = (2x+ 1)2 1 = (4x2 + 4x+ 1) 1 = 4x2 + 4x, D = R.(b) (g f)(x) = g(f(x)) = g(x2 1) = 2(x2 1) + 1 = (2x2 2) + 1 = 2x2 1, D = R.

(c) (f f)(x) = f(f(x)) = f(x2 1) = (x2 1)2 1 = (x4 2x2 + 1) 1 = x4 2x2, D = R.

(d) (g g)(x) = g(g(x)) = g(2x+ 1) = 2(2x+ 1) + 1 = (4x+ 2) + 1 = 4x+ 3, D = R.33. f(x) = 1 3x; g(x) = cosx. D = R for both f and g, and hence for their composites.

(a) (f g)(x) = f(g(x)) = f(cosx) = 1 3 cosx.(b) (g f)(x) = g(f(x)) = g(1 3x) = cos(1 3x).

(c) (f f)(x) = f(f(x)) = f(1 3x) = 1 3(1 3x) = 1 3 + 9x = 9x 2.

(d) (g g)(x) = g(g(x)) = g(cosx) = cos(cosx) [Note that this is not cosx cosx.]

35. f(x) = x+ 1x

, D = {x | x 6= 0}; g(x) = x+ 1x+ 2

, D = {x | x 6= 2}

(a) (f g)(x) = f(g(x)) = f x+ 1x+ 2

=x+ 1

x+ 2+

1x+ 1

x+ 2

=x+ 1

x+ 2+x+ 2

x+ 1

=(x+ 1)(x+ 1) + (x+ 2)(x+ 2)

(x+ 2)(x+ 1)=

x2 + 2x+ 1 + x2 + 4x+ 4

(x+ 2)(x+ 1)=

2x2 + 6x+ 5

(x+ 2)(x+ 1)

Since g(x) is not defined for x = 2 and f(g(x)) is not defined for x = 2 and x = 1,the domain of (f g)(x) is D = {x | x 6= 2,1}.

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

=

x+1

x+ 1

x+1

x+ 2

=

x2 + 1 + x

xx2 + 1 + 2x

x

=x2 + x+ 1

x2 + 2x+ 1=

x2 + x+ 1

(x+ 1)2

Since f(x) is not defined for x = 0 and g(f(x)) is not defined for x = 1,the domain of (g f)(x) is D = {x | x 6= 1, 0}.

(c) (f f)(x) = f(f(x)) = f x+ 1x

= x+1

x+

1

x+ 1x

= x+1

x+

1x2+1x

= x+1

x+

x

x2 + 1

=x(x) x2 + 1 + 1 x2 + 1 + x(x)

x(x2 + 1)=

x4 + x2 + x2 + 1 + x2

x(x2 + 1)

=x4 + 3x2 + 1

x(x2 + 1), D = {x | x 6= 0}


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

=

x+ 1

x+ 2+ 1

x+ 1

x+ 2+ 2

=

x+ 1 + 1(x+ 2)

x+ 2x+ 1 + 2(x+ 2)

x+ 2

=x+ 1 + x+ 2

x+ 1 + 2x+ 4=2x+ 3

3x+ 5

Since g(x) is not defined for x = 2 and g(g(x)) is not defined for x = 53 ,

the domain of (g g)(x) is D = x | x 6= 2, 53 .

37. (f g h)(x) = f(g(h(x))) = f(g(x 1)) = f(2(x 1)) = 2(x 1) + 1 = 2x 1

39. (f g h)(x) = f(g(h(x))) = f(g(x3 + 2)) = f [(x3 + 2)2]= f(x6 + 4x3 + 4) = (x6 + 4x3 + 4) 3 = x6 + 4x3 + 1

41. Let g(x) = x2 + 1 and f(x) = x10. Then (f g)(x) = f(g(x)) = f(x2 + 1) = (x2 + 1)10 = F (x).

43. Let g(x) = 3x and f(x) = x

1 + x. Then (f g)(x) = f(g(x)) = f( 3x ) =

3x

1 + 3x= F (x).

45. Let g(t) = cos t and f(t) =t. Then (f g)(t) = f(g(t)) = f(cos t) = cos t = u(t).

47. Let h(x) = x2, g(x) = 3x, and f(x) = 1 x. Then

(f g h)(x) = f(g(h(x))) = f(g(x2)) = f 3x2 = 1 3x2 = H(x).

49. Let h(x) =x, g(x) = secx, and f(x) = x4. Then

(f g h)(x) = f(g(h(x))) = f(g(x )) = f(secx ) = (secx )4 = sec4 (x ) = H(x).51. (a) g(2) = 5, because the point (2, 5) is on the graph of g. Thus, f(g(2)) = f(5) = 4, because the point (5, 4) is on the

graph of f .

(b) g(f(0)) = g(0) = 3

(c) (f g)(0) = f(g(0)) = f(3) = 0(d) (g f)(6) = g(f(6)) = g(6). This value is not defined, because there is no point on the graph of g that has

x-coordinate 6.

(e) (g g)(2) = g(g(2)) = g(1) = 4(f ) (f f)(4) = f(f(4)) = f(2) = 2

53. (a) Using the relationship distance = rate time with the radius r as the distance, we have r(t) = 60t.(b) A = r2 (A r)(t) = A(r(t)) = (60t)2 = 3600t2. This formula gives us the extent of the rippled area

(in cm2) at any time t.

55. (a) From the figure, we have a right triangle with legs 6 and d, and hypotenuse s.

By the Pythagorean Theorem, d2 + 62 = s2 s = f(d) = d 2 + 36.

(b) Using d = rt, we get d = (30 km/hr)(t hr) = 30t (in km). Thus,

d = g(t) = 30t.

(c) (f g)(t) = f(g(t)) = f(30t) = (30t)2 + 36 = 900t2 + 36. This function represents the distance between thelighthouse and the ship as a function of the time elapsed since noon.


57. (a)

H(t) =0 if t < 0

1 if t 0

(b)

V (t) =0 if t < 0

120 if t 0 so V (t) = 120H(t).

(c) Starting with the formula in part (b), we replace 120 with 240 to reflect the

different voltage. Also, because we are starting 5 units to the right of t = 0,

we replace t with t 5. Thus, the formula is V (t) = 240H(t 5).

59. If f(x) = m1x + b1 and g(x) = m2x + b2, then

(f g)(x) = f(g(x)) = f(m2x+ b2) = m1(m2x+ b2) + b1 = m1m2x+m1b2 + b1.So f g is a linear function with slope m1m2.

61. (a) By examining the variable terms in g and h, we deduce that we must square g to get the terms 4x2 and 4x in h. If we let

f(x) = x2 + c, then (f g)(x) = f(g(x)) = f(2x+ 1) = (2x+ 1)2 + c = 4x2 + 4x+ (1 + c). Since

h(x) = 4x2 + 4x+ 7, we must have 1 + c = 7. So c = 6 and f(x) = x2 + 6.

(b) We need a function g so that f(g(x)) = 3(g(x)) + 5 = h(x). But

h(x) = 3x2 + 3x+ 2 = 3(x2 + x) + 2 = 3(x2 + x 1) + 5, so we see that g(x) = x2 + x 1.

63. (a) If f and g are even functions, then f(x) = f(x) and g(x) = g(x).(i) (f + g)(x) = f(x) + g(x) = f(x) + g(x) = (f + g)(x), so f + g is an even function.

(ii) (fg)(x) = f(x) g(x) = f(x) g(x) = (fg)(x), so fg is an even function.

(b) If f and g are odd functions, then f(x) = f(x) and g(x) = g(x).(i) (f + g)(x) = f(x) + g(x) = f(x) + [g(x)] = [f(x) + g(x)] = (f + g)(x),

so f + g is an odd function.(ii) (fg)(x) = f(x) g(x) = f(x) [g(x)] = f(x) g(x) = (fg)(x), so fg is an even function.

65. We need to examine h(x).h(x) = (f g)(x) = f(g(x)) = f(g(x)) [because g is even] = h(x)

Because h(x) = h(x), h is an even function.

SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS 23

1.4 Graphing Calculators and Computers

1. f(x) =x3 5x2

(a) [5, 5] by [5, 5](There is no graph shown.)

(b) [0, 10] by [0, 2] (c) [0, 10] by [0, 10]

The most appropriate graph is produced in viewing rectangle (c).

3. Since the graph of f(x) = 5 + 20x x2 is aparabola opening downward, an appropriate viewing

rectangle should include the maximum point.

5. f(x) = 481 x4 is defined when 81 x4 0

x4 81 |x| 3, so the domain of f is [3, 3]. Also0 481 x4 481 = 3, so the range is [0, 3].

7. The graph of f(x) = x3 225x is symmetric with respect to the origin.Since f(x) = x3 225x = x(x2 225) = x(x+ 15)(x 15), thereare x-intercepts at 0, 15, and 15. f(20) = 3500.

9. The period of g(x) = sin(1000x) is 21000

0.0063 and its range is[1, 1]. Since f(x) = sin2(1000x) is the square of g, its range is[0, 1] and a viewing rectangle of [0.01, 0.01] by [0, 1.1] seemsappropriate.

11. The domain of y =x is x 0, so the domain of f(x) = sinx is [0,)

and the range is [1, 1]. With a little trial-and-error experimentation, we findthat an Xmax of 100 illustrates the general shape of f , so an appropriate

viewing rectangle is [0, 100] by [1.5, 1.5].


13. The first term, 10 sinx, has period 2 and range [10, 10]. It will be the dominant term in any large graph of

y = 10 sinx+ sin 100x, as shown in the first figure. The second term, sin 100x, has period 2100

= 50

and range [1, 1].

It causes the bumps in the first figure and will be the dominant term in any small graph, as shown in the view near the

origin in the second figure.

15. We must solve the given equation for y to obtain equations for the upper andlower halves of the ellipse.

4x2 + 2y2 = 1 2y2 = 1 4x2 y2 = 1 4x2

2

y = 1 4x2

2

17. From the graph of y = 3x2 6x+ 1and y = 0.23x 2.25 in the viewingrectangle [1, 3] by [2.5, 1.5], it isdifficult to see if the graphs intersect.

If we zoom in on the fourth quadrant,

we see the graphs do not intersect.

19. From the graph of f(x) = x3 9x2 4, we see that there is one solutionof the equation f(x) = 0 and it is slightly larger than 9. By zooming in or

using a root or zero feature, we obtain x 9.05.

21. We see that the graphs of f(x) = x2 and g(x) = sinx intersect twice. One

solution is x = 0. The other solution of f = g is the x-coordinate of thepoint of intersection in the first quadrant. Using an intersect feature or

zooming in, we find this value to be approximately 0.88. Alternatively, we

could find that value by finding the positive zero of h(x) = x2 sinx.

Note: After producing the graph on a TI-83 Plus, we can find the approximate value 0.88 by using the following keystrokes:

. The 1 is just a guess for 0.88.

SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS 25

23. g(x) = x3/10 is larger than f(x) = 10x2 whenever x > 100.

25. We see from the graphs of y = |sinx x| and y = 0.1 that there aretwo solutions to the equation |sinx x| = 0.1: x 0.85 andx 0.85. The condition |sinx x| < 0.1 holds for any x lyingbetween these two values, that is, 0.85 < x < 0.85.

27. (a) The root functions y =x,

y = 4x and y = 6

x

(b) The root functions y = x,

y = 3x and y = 5

x

(c) The root functions y =x, y = 3

x,

y = 4x and y = 5

x

(d) For any n, the nth root of 0 is 0 and the nth root of 1 is 1; that is, all nth root functions pass through the points (0, 0)and (1, 1).

For odd n, the domain of the nth root function is R, while for even n, it is {x R | x 0}. Graphs of even root functions look similar to that ofx, while those of odd root functions resemble that of 3x. As n increases, the graph of nx becomes steeper near 0 and flatter for x > 1.

29. f(x) = x4 + cx2 + x. If c < 1.5, there are three humps: two minimum pointsand a maximum point. These humps get flatter as c increases, until at c = 1.5two of the humps disappear and there is only one minimum point. This single

hump then moves to the right and approaches the origin as c increases.

31. y = xn2x. As n increases, the maximum of the

function moves further from the origin, and gets

larger. Note, however, that regardless of n, the

function approaches 0 as x.


33. y2 = cx3 + x2. If c < 0, the loop is to the right of the origin, and if c is positive,

it is to the left. In both cases, the closer c is to 0, the larger the loop is. (In the

limiting case, c = 0, the loop is infinite, that is, it doesnt close.) Also, the

larger |c| is, the steeper the slope is on the loopless side of the origin.

35. The graphing window is 95 pixels wide and we want to start with x = 0 and end with x = 2. Since there are 94 gaps

between pixels, the distance between pixels is 2094 . Thus, the x-values that the calculator actually plots are x = 0 +294 n,

where n = 0, 1, 2, . . . , 93, 94. For y = sin 2x, the actual points plotted by the calculator are 294 n, sin 2 2

94 n for

n = 0, 1, . . . , 94. For y = sin 96x, the points plotted are 294 n, sin 96 2

94 n for n = 0, 1, . . . , 94. But

sin 96 294 n = sin 94 2

94 n+ 2 2

94 n = sin 2n+ 2 2

94 n

= sin 2 294 n [by periodicity of sine], n = 0, 1, . . . , 94

So the y-values, and hence the points, plotted for y = sin 96x are identical to those plotted for y = sin 2x.Note: Try graphing y = sin 94x. Can you see why all the y-values are zero?

1.5 Exponential Functions

1. (a) f(x) = ax, a > 0 (b) R (c) (0,) (d) See Figures 4(c), 4(b), and 4(a), respectively.3. All of these graphs approach 0 as x, all of them pass through the point(0, 1), and all of them are increasing and approach as x. The larger thebase, the faster the function increases for x > 0, and the faster it approaches 0 as

x.Note: The notation x can be thought of as x becomes large at this point.More details on this notation are given in Chapter 2.

5. The functions with bases greater than 1 (3x and 10x) are increasing, while those

with bases less than 1 13

x and 110

x are decreasing. The graph of 13

x is the

reflection of that of 3x about the y-axis, and the graph of 110

x is the reflection of

that of 10x about the y-axis. The graph of 10x increases more quickly than that of

3x for x > 0, and approaches 0 faster as x.

7. We start with the graph of y = 4x (Figure 3) and then

shift 3 units downward. This shift doesnt affect the

domain, but the range of y = 4x 3 is (3,) .There is a horizontal asymptote of y = 3.

y = 4x y = 4x 3

SECTION 1.5 EXPONENTIAL FUNCTIONS 27

9. We start with the graph of y = 2x (Figure 3),

reflect it about the y-axis, and then about the

x-axis (or just rotate 180 to handle both

reflections) to obtain the graph of y = 2x.In each graph, y = 0 is the horizontal

asymptote.y = 2x y = 2x y = 2x

11. We start with the graph of y = ex (Figure 13) and reflect about the y-axis to get the graph of y = ex. Then we compress the

graph vertically by a factor of 2 to obtain the graph of y = 12ex and then reflect about the x-axis to get the graph of

y = 12ex. Finally, we shift the graph upward one unit to get the graph of y = 1 1

2ex.

13. (a) To find the equation of the graph that results from shifting the graph of y = ex 2 units downward, we subtract 2 from the

original function to get y = ex 2.

(b) To find the equation of the graph that results from shifting the graph of y = ex 2 units to the right, we replace x with x 2in the original function to get y = e(x2).

(c) To find the equation of the graph that results from reflecting the graph of y = ex about the x-axis, we multiply the original

function by 1 to get y = ex.

(d) To find the equation of the graph that results from reflecting the graph of y = ex about the y-axis, we replace x with x inthe original function to get y = ex.

(e) To find the equation of the graph that results from reflecting the graph of y = ex about the x-axis and then about the

y-axis, we first multiply the original function by 1 (to get y = ex) and then replace x with x in this equation toget y = ex.

15. (a) The denominator 1 + ex is never equal to zero because ex > 0, so the domain of f(x) = 1/(1 + ex) is R.

(b) 1 ex = 0 ex = 1 x = 0, so the domain of f(x) = 1/(1 ex) is (, 0) (0,).

17. Use y = Cax with the points (1, 6) and (3, 24). 6 = Ca1 C = 6a

and 24 = Ca3 24 = 6a

a3

4 = a2 a = 2 [since a > 0] and C = 62= 3. The function is f(x) = 3 2x.

19. If f(x) = 5x, then f(x+ h) f(x)h

=5x+h 5x

h=5x5h 5x

h=5x 5h 1

h= 5x

5h 1h

.


21. 2 ft = 24 in, f(24) = 242 in = 576 in = 48 ft. g(24) = 224 in = 224/(12 5280) mi 265 mi

23. The graph of g finally surpasses that of f at x 35.8.

25. (a) Fifteen hours represents 5 doubling periods (one doubling period is three hours). 100 25 = 3200

(b) In t hours, there will be t/3 doubling periods. The initial population is 100,

so the population y at time t is y = 100 2t/3.

(c) t = 20 y = 100 220/3 10,159

(d) We graph y1 = 100 2x/3 and y2 = 50,000. The two curves intersect atx 26.9, so the population reaches 50,000 in about 26.9 hours.

27. An exponential model is y = abt, where a = 3.154832569 1012

and b = 1.017764706. This model gives y(1993) 5498 million andy(2010) 7417 million.

29. From the graph, it appears that f is an odd function (f is undefined for x = 0).

To prove this, we must show that f(x) = f(x).

f(x) = 1 e1/(x)

1 + e1/(x)=1 e(1/x)1 + e(1/x)

=1 1

e1/x

1 +1

e1/x

e1/x

e1/x=

e1/x 1e1/x + 1

= 1 e1/x

1 + e1/x= f(x)

so f is an odd function.

1.6 Inverse Functions and Logarithms

1. (a) See Definition 1.

(b) It must pass the Horizontal Line Test.

3. f is not one-to-one because 2 6= 6, but f(2) = 2.0 = f(6).

5. No horizontal line intersects the graph of f more than once. Thus, by the Horizontal Line Test, f is one-to-one.

SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS 29

7. The horizontal line y = 0 (the x-axis) intersects the graph of f in more than one point. Thus, by the Horizontal Line Test,

f is not one-to-one.

9. The graph of f(x) = x2 2x is a parabola with axis of symmetry x = b2a= 2

2(1)= 1. Pick any x-values equidistant

from 1 to find two equal function values. For example, f(0) = 0 and f(2) = 0, so f is not one-to-one.

11. g(x) = 1/x. x1 6= x2 1/x1 6= 1/x2 g (x1) 6= g (x2), so g is one-to-one.Geometric solution: The graph of g is the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test,

so g is one-to-one.

13. A football will attain every height h up to its maximum height twice: once on the way up, and again on the way down. Thus,

even if t1 does not equal t2, f(t1) may equal f(t2), so f is not 1-1.

15. Since f(2) = 9 and f is 1-1, we know that f1(9) = 2. Remember, if the point (2, 9) is on the graph of f , then the point

(9, 2) is on the graph of f1.

17. First, we must determine x such that g(x) = 4. By inspection, we see that if x = 0, then g(x) = 4. Since g is 1-1 (g is an

increasing function), it has an inverse, and g1(4) = 0.

19. We solve C = 59 (F 32) for F : 95C = F 32 F = 95C + 32. This gives us a formula for the inverse function, thatis, the Fahrenheit temperature F as a function of the Celsius temperature C. F 459.67 9

5C + 32 459.67 95C 491.67 C 273.15, the domain of the inverse function.

21. f(x) =10 3x y = 10 3x (y 0) y2 = 10 3x 3x = 10 y2 x = 1

3y2 + 10

3.

Interchange x and y: y = 13x

2 + 103 . So f1(x) = 13x2 + 103 . Note that the domain of f1 is x 0.

23. y = f(x) = ex3 ln y = x3 x = 3ln y. Interchange x and y: y = 3lnx. So f1(x) = 3lnx.

25. y = f(x) = ln (x+ 3) x+ 3 = ey x = ey 3. Interchange x and y: y = ex 3. So f1(x) = ex 3.

27. y = f(x) = x4 + 1 y 1 = x4 x = 4y 1 (not since

x 0). Interchange x and y : y = 4x 1. So f1(x) = 4x 1. The

graph of y = 4x 1 is just the graph of y = 4x shifted right one unit.

From the graph, we see that f and f1 are reflections about the line y = x.

29. Reflect the graph of f about the line y = x. The points (1,2), (1,1),(2, 2), and (3, 3) on f are reflected to (2,1), (1, 1), (2, 2), and (3, 3)on f1.


31. (a) It is defined as the inverse of the exponential function with base a, that is, loga x = y ay = x.(b) (0,) (c) R (d) See Figure 11.

33. (a) log5 125 = 3 since 53 = 125. (b) log31

27= 3 since 33 = 1

33=1

27.

35. (a) log2 6 log2 15 + log2 20 = log2( 615 ) + log2 20 [by Law 2]= log2(

615 20) [by Law 1]

= log2 8, and log2 8 = 3 since 23 = 8.

(b) log3 100 log3 18 log3 50 = log3 10018 log3 50 = log3 1001850= log3(

19), and log3 19 = 2 since 32 = 19 .

37. ln 5 + 5 ln 3 = ln 5 + ln 35 [by Law 3]

= ln(5 35) [by Law 1]= ln1215

39. ln(1 + x2) + 12 lnx ln sinx = ln(1 + x2) + lnx1/2 ln sinx = ln[(1 + x2)x ] ln sinx = ln (1 + x

2)x

sinx

41. To graph these functions, we use log1.5 x =lnx

ln 1.5and log50 x =

lnx

ln 50.

These graphs all approach as x 0+, and they all pass through thepoint (1, 0). Also, they are all increasing, and all approach as x.The functions with larger bases increase extremely slowly, and the ones with

smaller bases do so somewhat more quickly. The functions with large bases

approach the y-axis more closely as x 0+.43. 3 ft = 36 in, so we need x such that log2 x = 36 x = 236 = 68,719,476,736. In miles, this is

68,719,476,736 in 1 ft12 in

1 mi5280 ft

1,084,587.7 mi.

45. (a) Shift the graph of y = log10 x five units to the left to

obtain the graph of y = log10(x+5). Note the vertical

asymptote of x = 5.

y = log10 x y = log10(x+ 5)

(b) Reflect the graph of y = lnx about the x-axis to obtain

the graph of y = lnx.

y = lnx y = lnx

47. (a) 2 lnx = 1 lnx = 12 x = e1/2 = e

(b) ex = 5 x = ln 5 x = ln 549. (a) 2x5 = 3 log2 3 = x 5 x = 5 + log2 3.

Or: 2x5 = 3 ln 2x5 = ln3 (x 5) ln 2 = ln 3 x 5 = ln 3ln 2

x = 5 + ln 3ln 2

SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS 31

(b) lnx+ ln(x 1) = ln(x(x 1)) = 1 x(x 1) = e1 x2 x e = 0. The quadratic formula (with a = 1,

b = 1, and c = e) gives x = 1211 + 4e , but we reject the negative root since the natural logarithm is not

defined for x < 0. So x = 121 +

1 + 4e .

51. (a) ex < 10 ln ex < ln 10 x < ln 10 x (, ln 10)(b) lnx > 1 elnx > e1 x > e1 x (1/e,)

53. (a) For f(x) =3 e2x, we must have 3 e2x 0 e2x 3 2x ln 3 x 12 ln 3. Thus, the domain

of f is (, 12 ln 3].

(b) y = f(x) =3 e2x [note that y 0] y2 = 3 e2x e2x = 3 y2 2x = ln(3 y2)

x = 12ln(3 y2). Interchange x and y: y = 1

2ln(3 x2). So f1(x) = 1

2ln(3 x2). For the domain of f1, we must

have 3 x2 > 0 x2 < 3 |x| < 3 3 < x < 3 0 x < 3 since x 0. Note that the

domain of f1, [0,3 ), equals the range of f .

55. We see that the graph of y = f(x) =x3 + x2 + x+ 1 is increasing, so f is 1-1.

Enter x = y3 + y2 + y + 1 and use your CAS to solve the equation for y.

Using Derive, we get two (irrelevant) solutions involving imaginary expressions,

as well as one which can be simplified to the following:

y = f1(x) = 346

3D 27x2 + 20 3D + 27x2 20 + 32

where D = 3327x4 40x2 + 16.

Maple and Mathematica each give two complex expressions and one real expression, and the real expression is equivalent

to that given by Derive. For example, Maples expression simplifies to 16

M2/3 8 2M1/32M1/3

, where

M = 108x2 + 1248 120x2 + 81x4 80.

57. (a) n = 100 2t/3 n100

= 2t/3 log2n

100=

t

3 t = 3 log2

n

100. Using formula (10), we can write

this as t = f1(n) = 3 ln(n/100)ln 2

. This function tells us how long it will take to obtain n bacteria (given the number n).

(b) n = 50,000 t = f1(50,000) = 3 ln50,000100

ln 2= 3

ln 500

ln 2 26.9 hours

59. (a) sin132

= 3 since sin

3 =

32 and

3 is in 2 , 2 .

(b) cos1(1) = since cos = 1 and is in [0, ].

61. (a) arctan 1 = 4 since tan4 = 1 and

4 is in 2 , 2 .

(b) sin1 12= 4 since sin

4 =

12

and 4 is in 2 , 2 .


63. (a) In general, tan(arctanx) = x for any real number x. Thus, tan(arctan 10) = 10.

(b) sin1 sin 73

= sin1 sin 3= sin1

32=

3since sin

3=32

and 3

is in 2, 2

.

[Recall that 73 =

3 + 2 and the sine function is periodic with period 2.]

65. Let y = sin1 x. Then 2 y

2 cos y 0, so cos(sin1 x) = cos y = 1 sin2 y = 1 x2.

67. Let y = tan1 x. Then tan y = x, so from the triangle we see that

sin(tan1 x) = sin y =x1 + x2

.

69. The graph of sin1 x is the reflection of the graph of

sinx about the line y = x.

71. g(x) = sin1(3x+ 1).

Domain (g) = {x | 1 3x+ 1 1} = {x | 2 3x 0} = x | 23 x 0 = 2

3, 0 .

Range (g) = y | 2 y

2=

2, 2

.

73. (a) If the point (x, y) is on the graph of y = f(x), then the point (x c, y) is that point shifted c units to the left. Since f is

1-1, the point (y, x) is on the graph of y = f1(x) and the point corresponding to (x c, y) on the graph of f is

(y, x c) on the graph of f1. Thus, the curves reflection is shifted down the same number of units as the curve itself is

shifted to the left. So an expression for the inverse function is g1(x) = f1(x) c.

(b) If we compress (or stretch) a curve horizontally, the curves reflection in the line y = x is compressed (or stretched)

vertically by the same factor. Using this geometric principle, we see that the inverse of h(x) = f(cx) can be expressed as

h1(x) = (1/c) f1(x).

CHAPTER 1 REVIEW 33

1 Review

1. (a) A function f is a rule that assigns to each element x in a set A exactly one element, called f(x), in a set B. The set A is

called the domain of the function. The range of f is the set of all possible values of f(x) as x varies throughout the

domain.

(b) If f is a function with domain A, then its graph is the set of ordered pairs {(x, f(x)) | x A}.(c) Use the Vertical Line Test on page 16.

2. The four ways to represent a function are: verbally, numerically, visually, and algebraically. An example of each is given

below.Verbally: An assignment of students to chairs in a classroom (a description in words)

Numerically: A tax table that assigns an amount of tax to an income (a table of values)

Visually: A graphical history of the Dow Jones average (a graph)

Algebraically: A relationship between distance, rate, and time: d = rt (an explicit formula)

3. (a) An even function f satisfies f(x) = f(x) for every number x in its domain. It is symmetric with respect to the y-axis.

(b) An odd function g satisfies g(x) = g(x) for every number x in its domain. It is symmetric with respect to the origin.

4. A function f is called increasing on an interval I if f(x1) < f(x2) whenever x1 < x2 in I.

5. Amathematical model is a mathematical description (often by means of a function or an equation) of a real-world

phenomenon.

6. (a) Linear function: f(x) = 2x+ 1, f(x) = ax+ b 7.

(b) Power function: f(x) = x2, f(x) = xa

(c) Exponential function: f(x) = 2x, f(x) = ax

(d) Quadratic function: f(x) = x2 + x+ 1, f(x) = ax2 + bx+ c

(e) Polynomial of degree 5: f(x) = x5 + 2

(f ) Rational function: f(x) = xx+ 2

, f(x) = P (x)Q(x)

where P (x) and

Q(x) are polynomials

8. (a) (b)


(c) (d) (e)

(f ) (g) (h)

9. (a) The domain of f + g is the intersection of the domain of f and the domain of g; that is, A B.(b) The domain of fg is also A B.(c) The domain of f/g must exclude values of x that make g equal to 0; that is, {x A B | g(x) 6= 0}.

10. Given two functions f and g, the composite function f g is defined by (f g) (x) = f(g (x)). The domain of f g is theset of all x in the domain of g such that g(x) is in the domain of f .

11. (a) If the graph of f is shifted 2 units upward, its equation becomes y = f(x) + 2.

(b) If the graph of f is shifted 2 units downward, its equation becomes y = f(x) 2.

(c) If the graph of f is shifted 2 units to the right, its equation becomes y = f(x 2).

(d) If the graph of f is shifted 2 units to the left, its equation becomes y = f(x+ 2).

(e) If the graph of f is reflected about the x-axis, its equation becomes y = f(x).

(f ) If the graph of f is reflected about the y-axis, its equation becomes y = f(x).

(g) If the graph of f is stretched vertically by a factor of 2, its equation becomes y = 2f(x).

(h) If the graph of f is shrunk vertically by a factor of 2, its equation becomes y = 12f(x).

(i) If the graph of f is stretched horizontally by a factor of 2, its equation becomes y = f 12x .

(j) If the graph of f is shrunk horizontally by a factor of 2, its equation becomes y = f(2x).

12. (a) A function f is called a one-to-one function if it never takes on the same value twice; that is, if f(x1) 6= f(x2) wheneverx1 6= x2. (Or, f is 1-1 if each output corresponds to only one input.)

Use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more thanonce.

(b) If f is a one-to-one function with domain A and range B, then its inverse function f1 has domain B and range A and is

defined by

f1(y) = x f(x) = y

for any y in B. The graph of f1 is obtained by reflecting the graph of f about the line y = x.

CHAPTER 1 REVIEW 35

13. (a) The inverse sine function f(x) = sin1 x is defined as follows:

sin1 x = y sin y = x and 2 y

2

Its domain is 1 x 1 and its range is 2 y

2.

(b) The inverse cosine function f(x) = cos1 x is defined as follows:

cos1 x = y cos y = x and 0 y Its domain is 1 x 1 and its range is 0 y .

(c) The inverse tangent function f(x) = tan1 x is defined as follows:

tan1 x = y tan y = x and 2< y 6. D = (6,)Range: x+ 6 > 0, so ln(x+ 6) takes on all real numbers and, hence, the range is R.

R = (,)

9. (a) To obtain the graph of y = f(x) + 8, we shift the graph of y = f(x) up 8 units.

(b) To obtain the graph of y = f(x+ 8), we shift the graph of y = f(x) left 8 units.

(c) To obtain the graph of y = 1 + 2f(x), we stretch the graph of y = f(x) vertically by a factor of 2, and then shift the

resulting graph 1 unit upward.

(d) To obtain the graph of y = f(x 2) 2, we shift the graph of y = f(x) right 2 units (for the 2 inside theparentheses), and then shift the resulting graph 2 units downward.

(e) To obtain the graph of y = f(x), we reflect the graph of y = f(x) about the x-axis.

(f ) To obtain the graph of y = f1(x), we reflect the graph of y = f(x) about the line y = x (assuming f is oneto-one).

11. y = sin 2x: Start with the graph of y = sinx, compress horizontally by a factor of 2, and reflect about the x-axis.

13. y = 12(1 + ex):

Start with the graph of y = ex,

shift 1 unit upward, and compress

vertically by a factor of 2.

15. f(x) = 1x+ 2

:

Start with the graph of f(x) = 1/x

and shift 2 units to the left.

CHAPTER 1 REVIEW 37

17. (a) The terms of f are a mixture of odd and even powers of x, so f is neither even nor odd.

(b) The terms of f are all odd powers of x, so f is odd.

(c) f(x) = e(x)2 = ex2 = f(x), so f is even.

(d) f(x) = 1 + sin(x) = 1 sinx. Now f(x) 6= f(x) and f(x) 6= f(x), so f is neither even nor odd.

19. f(x) = lnx, D = (0,); g(x) = x2 9, D = R.

(a) (f g)(x) = f(g(x)) = f(x2 9) = ln(x2 9).Domain: x2 9 > 0 x2 > 9 |x| > 3 x (,3) (3,)

(b) (g f)(x) = g(f(x)) = g(lnx) = (lnx)2 9. Domain: x > 0, or (0,)

(c) (f f)(x) = f(f(x)) = f(lnx) = ln(lnx). Domain: lnx > 0 x > e0 = 1, or (1,)

(d) (g g)(x) = g(g(x)) = g(x2 9) = (x2 9)2 9. Domain: x R, or (,)

21. Many models appear to be plausible. Your choice depends on whether you

think medical advances will keep increasing life expectancy, or if there is

bound to be a natural leveling-off of life expectancy. A linear model,

y = 0.2493x 423.4818, gives us an estimate of 77.6 years for theyear 2010.

23. We need to know the value of x such that f(x) = 2x+ lnx = 2. Since x = 1 gives us y = 2, f1(2) = 1.

25. (a) e2 ln 3 = eln 3 2 = 32 = 9

(b) log10 25 + log10 4 = log10(25 4) = log10 100 = log10 102 = 2

(c) tan arcsin 12= tan

6= 1

3

(d) Let = cos1 45 , so cos =45 . Then sin cos

1 45= sin =

1 cos2 = 1 45

2= 925 =

35 .

27. (a) The population would reach 900 in about 4.4 years.

(b) P = 100,000100 + 900et

100P + 900Pet = 100,000 900Pet = 100,000 100P

et =100,000 100P

900P t = ln 1000 P

9P t = ln 1000 P

9P, or ln 9P

1000 P ; this is the time

required for the population to reach a given number P .

(c) P = 900 t = ln 9 9001000 900 = ln 81 4.4 years, as in part (a).

PRINCIPLES OF PROBLEM SOLVING1. By using the area formula for a triangle, 1

2(base) (height), in two ways, we see that

12(4) (y) = 1

2(h) (a), so a = 4y

h. Since 42 + y2 = h2, y =

h2 16, and

a =4h2 16h

.

3. |2x 1| =2x 1 if x 1

2

1 2x if x < 12

and |x+ 5| =x+ 5 if x 5x 5 if x < 5

Therefore, we consider the three cases x < 5, 5 x < 12

, and x 12

.

If x < 5, we must have 1 2x (x 5) = 3 x = 3, which is false, since we are considering x < 5.If 5 x < 1

2 , we must have 1 2x (x+ 5) = 3 x = 73 .If x 1

2 , we must have 2x 1 (x+ 5) = 3 x = 9.So the two solutions of the equation are x = 7

3 and x = 9.

5. f(x) = x2 4 |x|+ 3 . If x 0, then f(x) = x2 4x+ 3 = |(x 1)(x 3)|.Case (i): If 0 < x 1, then f(x) = x2 4x+ 3.Case (ii): If 1 < x 3, then f(x) = (x2 4x+ 3) = x2 + 4x 3.Case (iii): If x > 3, then f(x) = x2 4x+ 3.

This enables us to sketch the graph for x 0. Then we use the fact that f is an evenfunction to reflect this part of the graph about the y-axis to obtain the entire graph. Or, we

could consider also the cases x < 3, 3 x < 1, and 1 x < 0.

7. Remember that |a| = a if a 0 and that |a| = a if a < 0. Thus,

x+ |x| =2x if x 00 if x < 0

and y + |y| =2y if y 00 if y < 0

We will consider the equation x+ |x| = y + |y| in four cases.

(1) x 0, y 02x = 2y

x = y

(2) x 0, y < 02x = 0

x = 0

(3) x < 0, y 00 = 2y

0 = y

(4) x < 0, y < 00 = 0

Case 1 gives us the line y = x with nonnegative x and y.

Case 2 gives us the portion of the y-axis with y negative.

Case 3 gives us the portion of the x-axis with x negative.

Case 4 gives us the entire third quadrant.

39

40 CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING

9. |x|+ |y| 1. The boundary of the region has equation |x|+ |y| = 1. In quadrantsI, II, III, and IV, this becomes the lines x+ y = 1, x+ y = 1, x y = 1, andx y = 1 respectively.

11. (log2 3)(log3 4)(log4 5) (log31 32) =ln 3

ln 2

ln 4

ln 3

ln 5

ln 4 ln 32

ln 31=ln 32

ln 2=ln 25

ln 2=5 ln 2

ln 2= 5

13. ln x2 2x 2 0 x2 2x 2 e0 = 1 x2 2x 3 0 (x 3)(x+ 1) 0 x [1, 3].

Since the argument must be positive, x2 2x 2 > 0 x 13 x 1 +3 > 0

x , 13 1 +3, . The intersection of these intervals is 1, 13 1 +3, 3 .

15. Let d be the distance traveled on each half of the trip. Let t1 and t2 be the times taken for the first and second halves of the trip.

For the first half of the trip we have t1 = d/30 and for the second half we have t2 = d/60. Thus, the average speed for the

entire trip is total distancetotal time

=2d

t1 + t2=

2dd

30+

d

60

6060=

120d

2d+ d=120d

3d= 40. The average speed for the entire trip

is 40 mi/h.

17. Let Sn be the statement that 7n 1 is divisible by 6. S1 is true because 71 1 = 6 is divisible by 6. Assume Sk is true, that is, 7k 1 is divisible by 6. In other words, 7k 1 = 6m for some positive integer m. Then7k+1 1 = 7k 7 1 = (6m+ 1) 7 1 = 42m+ 6 = 6(7m+ 1), which is divisible by 6, so Sk+1 is true.

Therefore, by mathematical induction, 7n 1 is divisible by 6 for every positive integer n.

19. f0(x) = x2 and fn+1(x) = f0(fn(x)) for n = 0, 1, 2, . . ..

f1(x) = f0(f0(x)) = f0 x2 = x2

2= x4, f2(x) = f0(f1(x)) = f0(x4) = (x4)2 = x8,

f3(x) = f0(f2(x)) = f0(x8) = (x8)2 = x16, . . .. Thus, a general formula is fn(x) = x2

n+1

.

2 LIMITS AND DERIVATIVES2.1 The Tangent and Velocity Problems

1. (a) Using P (15, 250), we construct the following table:

t Q slope = mPQ

5 (5, 694) 694250515 = 44410 = 44.4

10 (10, 444) 4442501015 = 1945 = 38.8

20 (20, 111) 1112502015 = 1395 = 27.8

25 (25, 28) 282502515 = 22210 = 22.2

30 (30, 0) 02503015 = 25015 = 16.6

(b) Using the values of t that correspond to the points

closest to P (t = 10 and t = 20), we have

38.8 + (27.8)2

= 33.3

(c) From the graph, we can estimate the slope of the

tangent line at P to be 3009

= 33.3.

3. (a)x Q mPQ

(i) 0.5 (0.5, 0.333333) 0.333333

(ii) 0.9 (0.9, 0.473684) 0.263158

(iii) 0.99 (0.99, 0.497487) 0.251256

(iv) 0.999 (0.999, 0.499750) 0.250125

(v) 1.5 (1.5, 0.6) 0.2

(vi) 1.1 (1.1, 0.523810) 0.238095

(vii) 1.01 (1.01, 0.502488) 0.248756

(viii) 1.001 (1.001, 0.500250) 0.249875

(b) The slope appears to be 14 .

(c) y 12= 1

4(x 1) or y = 1

4x+ 1

4.

5. (a) y = y(t) = 40t 16t2. At t = 2, y = 40(2) 16(2)2 = 16. The average velocity between times 2 and 2 + h is

vave =y(2 + h) y(2)(2 + h) 2 =

40(2 + h) 16(2 + h)2 16h

=24h 16h2

h= 24 16h, if h 6= 0.

(i) [2, 2.5]: h = 0.5, vave = 32 ft/s (ii) [2, 2.1]: h = 0.1, vave = 25.6 ft/s(iii) [2, 2.05]: h = 0.05, vave = 24.8 ft/s (iv) [2, 2.01]: h = 0.01, vave = 24.16 ft/s

(b) The instantaneous velocity when t = 2 (h approaches 0) is 24 ft/s.41

42 CHAPTER 2 LIMITS AND DERIVATIVES

7. (a) (i) On the interval [1, 3], vave =s(3) s(1)3 1 =

10.7 1.42

=9.3

2= 4.65 m/s.

(ii) On the interval [2, 3], vave =s(3) s(2)3 2 =

10.7 5.11

= 5.6 m/s.

(iii) On the interval [3, 5], vave =s(5) s(3)5 3 =

25.8 10.72

=15.1

2= 7.55 m/s.

(iv) On the interval [3, 4], vave =s(4) s(3)4 3 =

17.7 10.71

= 7 m/s.

(b) Using the points (2, 4) and (5, 23) from the approximate tangent

line, the instantaneous velocity at t = 3 is about 23 45 2 6.3 m/s.

9. (a) For the curve y = sin(10/x) and the point P (1, 0):

x Q mPQ

2 (2, 0) 0

1.5 (1.5, 0.8660) 1.7321

1.4 (1.4,0.4339) 1.08471.3 (1.3,0.8230) 2.74331.2 (1.2, 0.8660) 4.3301

1.1 (1.1,0.2817) 2.8173

x Q mPQ

0.5 (0.5, 0) 0

0.6 (0.6, 0.8660) 2.16510.7 (0.7, 0.7818) 2.60610.8 (0.8, 1) 50.9 (0.9,0.3420) 3.4202

As x approaches 1, the slopes do not appear to be approaching any particular value.

(b) We see that problems with estimation are caused by the frequent

oscillations of the graph. The tangent is so steep at P that we need to

take x-values much closer to 1 in order to get accurate estimates of

its slope.

(c) If we choose x = 1.001, then the point Q is (1.001,0.0314) and mPQ 31.3794. If x = 0.999, then Q is(0.999, 0.0314) and mPQ = 31.4422. The average of these slopes is 31.4108. So we estimate that the slope of thetangent line at P is about 31.4.

SECTION 2.2 THE LIMIT OF A FUNCTION 43

2.2 The Limit of a Function

1. As x approaches 2, f(x) approaches 5. [Or, the values of f(x) can be made as close to 5 as we like by taking x sufficiently

close to 2 (but x 6= 2).] Yes, the graph could have a hole at (2, 5) and be defined such that f(2) = 3.

3. (a) limx3

f(x) = means that the values of f(x) can be made arbitrarily large (as large as we please) by taking x

sufficiently close to 3 (but not equal to3).(b) lim

x4+f(x) = means that the values of f(x) can be made arbitrarily large negative by taking x sufficiently close to 4

through values larger than 4.

5. (a) f(x) approaches 2 as x approaches 1 from the left, so limx1

f(x) = 2.

(b) f(x) approaches 3 as x approaches 1 from the right, so limx1+

f(x) = 3.

(c) limx1

f(x) does not exist because the limits in part (a) and part (b) are not equal.

(d) f(x) approaches 4 as x approaches 5 from the left and from the right, so limx5

f(x) = 4.

(e) f(5) is not defined, so it doesnt exist.

7. (a) limt0

g(t) = 1 (b) limt0+

g(t) = 2

(c) limt0

g(t) does not exist because the limits in part (a) and part (b) are not equal.

(d) limt2

g(t) = 2 (e) limt2+

g(t) = 0

(f ) limt2

g(t) does not exist because the limits in part (d) and part (e) are not equal.

(g) g(2) = 1 (h) limt4

g(t) = 3

9. (a) limx7

f(x) = (b) limx3

f(x) = (c) limx0

f(x) =

(d) limx6

f(x) = (e) limx6+

f(x) =

(f ) The equations of the vertical asymptotes are x = 7, x = 3, x = 0, and x = 6.11. (a) lim

x0f(x) = 1

(b) limx0+

f(x) = 0

(c) limx0

f(x) does not exist because the limits

in part (a) and part (b) are not equal.

13. limx1

f(x) = 2, limx1+

f(x) = 2, f(1) = 2 15. limx3+

f(x) = 4, limx3

f(x) = 2, limx2

f(x) = 2,

f(3) = 3, f(2) = 1


17. For f(x) = x2 2x

x2 x 2 :

x f(x)

2.5 0.714286

2.1 0.677419

2.05 0.672131

2.01 0.667774

2.005 0.667221

2.001 0.666778

x f(x)

1.9 0.655172

1.95 0.661017

1.99 0.665552

1.995 0.666110

1.999 0.666556

It appears that limx2

x2 2xx2 x 2 = 0.6 =

23

.

19. For f(x) = ex 1 x

x2:

x f(x)

1 0.718282

0.5 0.594885

0.1 0.517092

0.05 0.508439

0.01 0.501671

x f(x)

1 0.3678790.5 0.4261230.1 0.4837420.05 0.4917700.01 0.498337


ex 1 xx2

= 0.5 = 12 .

21. For f(x) =x+ 4 2

x:

x f(x)

1 0.236068

0.5 0.242641

0.1 0.248457

0.05 0.249224

0.01 0.249844

x f(x)

1 0.2679490.5 0.2583430.1 0.2515820.05 0.2507860.01 0.250156


x+ 4 2

x= 0.25 = 1

4 .

23. For f(x) = x6 1

x10 1 :

x f(x)

0.5 0.985337

0.9 0.719397

0.95 0.660186

0.99 0.612018

0.999 0.601200

x f(x)

1.5 0.183369

1.1 0.484119

1.05 0.540783

1.01 0.588022

1.001 0.598800


x6 1x10 1 = 0.6 =

35 .

25. limx3+

x+ 2

x+ 3= since the numerator is negative and the denominator approaches 0 from the positive side as x 3+.

27. limx1

2 x(x 1)2 = since the numerator is positive and the denominator approaches 0 through positive values as x 1.

29. Let t = x2 9. Then as x 3+, t 0+, and limx3+

ln(x2 9) = limt0+

ln t = by (3).

31. limx2

x cscx = limx2

x

sinx= since the numerator is positive and the denominator approaches 0 through negative

values as x 2.

33. (a) f(x) = 1x3 1 .

From these calculations, it seems that

limx1

f(x) = and limx1+

f(x) =.

x f(x)

0.5 1.140.9 3.690.99 33.70.999 333.70.9999 3333.70.99999 33,333.7

x f(x)

1.5 0.42

1.1 3.02

1.01 33.0

1.001 333.0

1.0001 3333.0

1.00001 33,333.3

SECTION 2.2 THE LIMIT OF A FUNCTION 45

(b) If x is slightly smaller than 1, then x3 1 will be a negative number close to 0, and the reciprocal of x3 1, that is, f(x),will be a negative number with large absolute value. So lim

x1f(x) = .

If x is slightly larger than 1, then x3 1 will be a small positive number, and its reciprocal, f(x), will be a large positivenumber. So lim

x1+f(x) =.

(c) It appears from the graph of f that

limx1

f(x) = and limx1+

f(x) =.

35. (a) Let h(x) = (1 + x)1/x.

x h(x)

0.001 2.719640.0001 2.718420.00001 2.718300.000001 2.718280.000001 2.71828

0.00001 2.71827

0.0001 2.71815

0.001 2.71692


(1 + x)1/x 2.71828, which is approximately e.In Section 3.6 we will see that the value of the limit is exactly e.

(b)

37. For f(x) = x2 (2x/1000):(a)

x f(x)

1 0.998000

0.8 0.638259

0.6 0.358484

0.4 0.158680

0.2 0.038851

0.1 0.008928

0.05 0.001465


f(x) = 0.

(b)x f(x)

0.04 0.000572

0.02 0.0006140.01 0.0009070.005 0.0009780.003 0.0009930.001 0.001000


f(x) = 0.001.


39. No matter how many times we zoom in toward the origin, the graphs of f(x) = sin(/x) appear to consist of almost-vertical

lines. This indicates more and more frequent oscillations as x 0.

41. There appear to be vertical asymptotes of the curve y = tan(2 sinx) at x 0.90and x 2.24. To find the exact equations of these asymptotes, we note that thegraph of the tangent function has vertical asymptotes at x =

2+ n. Thus, we

must have 2 sinx = 2+ n, or equivalently, sinx =

4+

2n. Since

1 sinx 1, we must have sinx = 4

and so x = sin1 4

(corresponding

to x 0.90). Just as 150 is the reference angle for 30, sin1 4 is the

reference angle for sin1 4 . So x = sin1 4 are also equations of

vertical asymptotes (corresponding to x 2.24).

2.3 Calculating Limits Using the Limit Laws

1. (a) limx2

[f(x) + 5g(x)] = limx2

f(x) + limx2

[5g(x)] [Limit Law 1]

= limx2

f(x) + 5 limx2

g(x) [Limit Law 3]

= 4 + 5(2) = 6

(b) limx2

[g(x)]3 = limx2

g(x)3

[Limit Law 6]

= (2)3 = 8

(c) limx2

f(x) = limx2

f(x) [Limit Law 11]

=4 = 2

SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS 47

(d) limx2

3f(x)

g(x)=limx2

[3f(x)]

limx2

g(x)[Limit Law 5]

=3 limx2

f(x)

limx2

g(x)[Limit Law 3]

=3(4)

2 = 6

(e) Because the limit of the denominator is 0, we cant use Limit Law 5. The given limit, limx2

g(x)

h(x), does not exist because the

denominator approaches 0 while the numerator approaches a nonzero number.

(f ) limx2

g(x)h(x)

f(x)=limx2

[g(x)h(x)]

limx2

f(x)[Limit Law 5]

=limx2

g(x) limx2

h(x)

limx2

f(x)[Limit Law 4]

=2 04

= 0

3. limx2

(3x4 + 2x2 x+ 1) = limx2

3x4 + limx2

2x2 limx2

x+ limx2

1 [Limit Laws 1 and 2]

= 3 limx2

x4 + 2 limx2

x2 limx2

x+ limx2

1 [3]

= 3(2)4 + 2(2)2 (2) + (1) [9, 8, and 7]= 48 + 8 + 2 + 1 = 59

5. limx8

(1 + 3x ) (2 6x2 + x3) = lim

x8(1 + 3

x ) lim

x8(2 6x2 + x3) [Limit Law 4]

= limx8

1 + limx8

3x lim

x82 6 lim

x8x2 + lim

x8x3 [1, 2, and 3]

= 1 + 38 2 6 82 + 83 [7, 10, 9]

= (3)(130) = 390

7. limx1

1 + 3x

1 + 4x2 + 3x4

3

= limx1

1 + 3x

1 + 4x2 + 3x4

3

[6]

=limx1

(1 + 3x)

limx1

(1 + 4x2 + 3x4)

3

[5]

=limx1

1 + 3 limx1

x

limx1

1 + 4 limx1

x2 + 3 limx1

x4

3

[2, 1, and 3]

=1 + 3(1)

1 + 4(1)2 + 3(1)4

3

=4

8

3

=1

2

3

=1

8[7, 8, and 9]

9. limx4

16 x2= lim

x4(16 x2) [11]

= limx4

16 limx4

x2 [2]

= 16 (4)2 = 0 [7 and 9]


11. limx2

x2 + x 6x 2 = limx2

(x+ 3)(x 2)x 2 = limx2(x+ 3) = 2 + 3 = 5

13. limx2

x2 x+ 6x 2 does not exist since x 2 0 but x

2 x+ 6 8 as x 2.

15. limt3

t2 92t2 + 7t+ 3

= limt3

(t+ 3)(t 3)(2t+ 1)(t+ 3)

= limt3

t 32t+ 1

=3 32(3) + 1 =

65 =

6

5

17. limh0

(4 + h)2 16h

= limh0

(16 + 8h+ h2) 16h

= limh0

8h+ h2

h= lim

h0h(8 + h)

h= lim

h0(8 + h) = 8 + 0 = 8

19. By the formula for the sum of cubes, we have

limx2

x+ 2

x3 + 8= lim

x2x+ 2

(x+ 2)(x2 2x+ 4) = limx21

x2 2x+ 4 =1

4 + 4 + 4=1

12.

21. limt9

9 t3t = limt9

3 +t 3t3t = limt9 3 +

t = 3 +

9 = 6

23. limx7

x+ 2 3x 7 = limx7

x+ 2 3x 7

x+ 2 + 3x+ 2 + 3

= limx7

(x+ 2) 9(x 7) x+ 2 + 3

= limx7

x 7(x 7) x+ 2 + 3 = limx7

1x+ 2+ 3

=19 + 3

=1

6

25. limx4

1

4+1

x4 + x

= limx4

x+ 4

4x4 + x

= limx4

x+ 4

4x(4 + x)= lim

x41

4x=

1

4(4) = 1

16

27. limx16

4x16x x2 = limx16

(4x )(4 +x )(16x x2)(4 +x ) = limx16

16 xx(16 x)(4 +x )

= limx16

1

x(4 +x )

=1

16 4 +16

=1

16(8)=

1

128

29. limt0

1

t1 + t

1t

= limt0

11 + tt1 + t

= limt0

11 + t 1 +1 + ttt+ 1 1 +

1 + t

= limt0

tt1 + t 1 +

1 + t

= limt0

11 + t 1 +

1 + t

=1

1 + 0 1 +1 + 0

= 12

31. (a)

limx0

x1 + 3x 1

2

3

(b)x f(x)

0.001 0.66616630.0001 0.66661670.00001 0.66666170.000001 0.66666620.000001 0.66666720.00001 0.66667170.0001 0.66671670.001 0.6671663

The limit appears to be 23

.

SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS 49

(c) limx0

x1 + 3x 1

1 + 3x+ 11 + 3x+ 1

= limx0

x1 + 3x+ 1

(1 + 3x) 1 = limx0x1 + 3x+ 1

3x

=1

3limx0

1 + 3x+ 1 [Limit Law 3]

=1

3limx0

(1 + 3x) + limx0

1 [1 and 11]

=1

3limx0

1 + 3 limx0

x+ 1 [1, 3, and 7]

=1

3

1 + 3 0 + 1 [7 and 8]

=1

3(1 + 1) =

2

3

33. Let f(x) = x2, g(x) = x2 cos 20x and h(x) = x2. Then1 cos 20x 1 x2 x2 cos 20x x2 f(x) g(x) h(x).So since lim

x0f(x) = lim

x0h(x) = 0, by the Squeeze Theorem we have

limx0

g(x) = 0.

35. We have limx4

(4x 9) = 4(4) 9 = 7 and limx4

x2 4x+ 7 = 42 4(4) + 7 = 7. Since 4x 9 f(x) x2 4x+ 7

for x 0, limx4

f(x) = 7 by the Squeeze Theorem.

37. 1 cos(2/x) 1 x4 x4 cos(2/x) x4. Since limx0

x4 = 0 and limx0

x4 = 0, we have

limx0

x4 cos(2/x) = 0 by the Squeeze Theorem.

39. |x 3| =x 3 if x 3 0(x 3) if x 3 < 0 =

x 3 if x 33 x if x < 3

Thus, limx3+

(2x+ |x 3|) = limx3+

(2x+ x 3) = limx3+

(3x 3) = 3(3) 3 = 6 and

limx3

(2x+ |x 3|) = limx3

(2x+ 3 x) = limx3

(x+ 3) = 3 + 3 = 6. Since the left and right limits are equal,

limx3

(2x+ |x 3|) = 6.

41. 2x3 x2 = x2(2x 1) = x2 |2x 1| = x2 |2x 1|

|2x 1| =2x 1 if 2x 1 0(2x 1) if 2x 1 < 0 =

2x 1 if x 0.5(2x 1) if x < 0.5

So 2x3 x2 = x2[(2x 1)] for x < 0.5.

Thus, limx0.5

2x 1|2x3 x2| = limx0.5

2x 1x2[(2x 1)] = limx0.5

1x2

=1(0.5)2

=10.25

= 4.


43. Since |x| = x for x < 0, we have limx0

1

x 1|x| = limx0

1

x 1x = limx0

2

x, which does not exist since the

denominator approaches 0 and the numerator does not.

45. (a) (b) (i) Since sgnx = 1 for x > 0, limx0+

sgnx = limx0+

1 = 1.

(ii) Since sgnx = 1 for x < 0, limx0

sgn x = limx0

1 = 1.

(iii) Since limx0

sgnx 6= limx0+

sgnx, limx0

sgnx does not exist.

(iv) Since |sgnx| = 1 for x 6= 0, limx0

|sgnx| = limx0

1 = 1.

47. (a) (i) limx1+

F (x) = limx1+

x2 1|x 1| = limx1+

x2 1x 1 = limx1+ (x+ 1) = 2 (c)

(ii) limx1

F (x) = limx1

x2 1|x 1| = limx1

x2 1 (x 1) = limx1 (x+ 1) = 2

(b) No, limx1

F (x) does not exist since limx1+

F (x) 6= limx1

F (x).

49. (a) (i) [[x]] = 2 for 2 x < 1, so limx2+

[[x]] = limx2+

(2) = 2

(ii) [[x]] = 3 for 3 x < 2, so limx2

[[x]] = limx2

(3) = 3.

The right and left limits are different, so limx2

[[x]] does not exist.

(iii) [[x]] = 3 for 3 x < 2, so limx2.4

[[x]] = limx2.4

(3) = 3.

(b) (i) [[x]] = n 1 for n 1 x < n, so limxn

[[x]] = limxn

(n 1) = n 1.

(ii) [[x]] = n for n x < n+ 1, so limxn+

[[x]] = limxn+

n = n.

(c) limxa

[[x]] exists a is not an integer.

51. The graph of f(x) = [[x]] + [[x]] is the same as the graph of g(x) = 1 with holes at each integer, since f(a) = 0 for anyinteger a. Thus, lim

x2f(x) = 1 and lim

x2+f (x) = 1, so lim

x2f(x) = 1. However,

f(2) = [[2]] + [[2]] = 2 + (2) = 0, so limx2

f(x) 6= f(2).

53. Since p(x) is a polynomial, p(x) = a0 + a1x+ a2x2 + + anxn. Thus, by the Limit Laws,

limxa

p(x) = limxa

a0 + a1x+ a2x2 + + anxn = a0 + a1 lim

xax+ a2 lim

xax2 + + an lim

xaxn

= a0 + a1a+ a2a2 + + anan = p(a)

Thus, for any polynomial p, limxa

p(x) = p(a).

SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 51

55. limx1

[f(x) 8] = limx1

f(x) 8x 1 (x 1) = limx1

f(x) 8x 1 limx1(x 1) = 10 0 = 0.

Thus, limx1

f(x) = limx1

{[f(x) 8] + 8} = limx1

[f(x) 8] + limx1

8 = 0 + 8 = 8.

Note: The value of limx1

f(x) 8x 1 does not affect the answer since its multiplied by 0. Whats important is that limx1

f(x) 8x 1

exists.

57. Observe that 0 f(x) x2 for all x, and limx0

0 = 0 = limx0

x2. So, by the Squeeze Theorem, limx0

f(x) = 0.

59. Let f(x) = H(x) and g(x) = 1H(x), where H is the Heaviside function defined in Exercise 1.3.57.

Thus, either f or g is 0 for any value of x. Then limx0

f(x) and limx0

g(x) do not exist, but limx0

[f(x)g(x)] = limx0

0 = 0.

61. Since the denominator approaches 0 as x 2, the limit will exist only if the numerator also approaches

0 as x 2. In order for this to happen, we need limx2

3x2 + ax+ a+ 3 = 0

3(2)2 + a(2) + a+ 3 = 0 12 2a+ a+ 3 = 0 a = 15. With a = 15, the limit becomes

limx2

3x2 + 15x+ 18

x2 + x 2 = limx23(x+ 2)(x+ 3)

(x 1)(x+ 2) = limx23(x+ 3)

x 1 =3(2 + 3)2 1 =

3

3 = 1.

2.4 The Precise Definition of a Limit

1. On the left side of x = 2, we need |x 2| < 107 2 = 4

7. On the right side, we need |x 2| < 10

3 2 = 4

3. For both of

these conditions to be satisfied at once, we need the more restrictive of the two to hold, that is, |x 2| < 47

. So we can choose

= 47 , or any smaller positive number.

3. The leftmost question mark is the solution ofx = 1.6 and the rightmost,

x = 2.4. So the values are 1.62 = 2.56 and

2.42 = 5.76. On the left side, we need |x 4| < |2.56 4| = 1.44. On the right side, we need |x 4| < |5.76 4| = 1.76.

To satisfy both conditions, we need the more restrictive condition to hold namely, |x 4| < 1.44. Thus, we can choose

= 1.44, or any smaller positive number.

5. From the graph, we find that tanx = 0.8 when x 0.675, so4 1 0.675 1 4 0.675 0.1106. Also, tanx = 1.2when x 0.876, so

4+ 2 0.876 2 = 0.876 4 0.0906.

Thus, we choose = 0.0906 (or any smaller positive number) since this is

the smaller of 1 and 2.


7. For = 1, the definition of a limit requires that we find such that 4 + x 3x3 2 < 1 1 < 4 + x 3x3 < 3

whenever 0 < |x 1| < . If we plot the graphs of y = 1, y = 4 + x 3x3 and y = 3 on the same screen, we see that weneed 0.86 x 1.11. So since |1 0.86| = 0.14 and |1 1.11| = 0.11, we choose = 0.11 (or any smaller positive

number). For = 0.1, we must find such that 4 + x 3x3 2 < 0.1 1.9 < 4 + x 3x3 < 2.1 whenever

0 < |x 1| < . From the graph, we see that we need 0.988 x 1.012. So since |1 0.988| = 0.012 and|1 1.012| = 0.012, we choose = 0.012 (or any smaller positive number) for the inequality to hold.

9. (a) From the graph, we find that y = tan2 x = 1000 when x 1.539 andx 1.602 for x near

2. Thus, we get 1.602

2 0.031 for

M = 1000.

(b) From the graph, we find that y = tan2 x = 10,000 when x 1.561 andx 1.581 for x near

2. Thus, we get 1.581

2 0.010 for

M = 10,000.

11. (a) A = r2 and A = 1000 cm2 r2 = 1000 r2 = 1000

r = 1000

(r > 0) 17.8412 cm.

(b) |A 1000| 5 5 r2 1000 5 1000 5 r2 1000 + 5

995 r 1005

17.7966 r 17.8858. 1000

995

0.04466 and 1005

1000

0.04455. So

if the machinist gets the radius within 0.0445 cm of 17.8412, the area will be within 5 cm2 of 1000.

(c) x is the radius, f(x) is the area, a is the target radius given in part (a), L is the target area (1000), is the tolerance in the

area (5), and is the tolerance in the radius given in part (b).

13. (a) |4x 8| = 4 |x 2| < 0.1 |x 2| < 0.14

, so = 0.14= 0.025.

(b) |4x 8| = 4 |x 2| < 0.01 |x 2| < 0.014

, so = 0.014

= 0.0025.

SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 53

15. Given > 0, we need > 0 such that if 0 < |x 1| < , then

|(2x+ 3) 5| < . But |(2x+ 3) 5| <

|2x 2| < 2 |x 1| < |x 1| < /2.

So if we choose = /2, then 0 < |x 1| <

|(2x+ 3) 5| < . Thus, limx1

(2x+ 3) = 5 by the definition of a limit.

17. Given > 0, we need > 0 such that if 0 < |x (3)| < , then

|(1 4x) 13| < . But |(1 4x) 13| <

|4x 12| < |4| |x+ 3| < |x (3)| < /4.

So if we choose = /4, then 0 < |x (3)| <

|(1 4x) 13| < . Thus, limx3

(1 4x) = 13 by the definition of

a limit.x

19. Given > 0, we need > 0 such that if 0 < |x 3| < , then x5 35

< 15|x 3| < |x 3| < 5.

So choose = 5. Then 0 < |x 3| < |x 3| < 5 |x 3|5

< x5 35

< . By the definition

of a limit, limx3

x

5=3

5.

21. Given > 0, we need > 0 such that if 0 < |x 2| < , then x2 + x 6x 2 5 <

(x+ 3)(x 2)x 2 5 < |x+ 3 5| < [x 6= 2] |x 2| < . So choose = .

Then 0 < |x 2| < |x 2| < |x+ 3 5| < (x+ 3)(x 2)x 2 5 < [x 6= 2]

x2 + x 6x 2 5 < . By the definition of a limit, limx2

x2 + x 6x 2 = 5.

23. Given > 0, we need > 0 such that if 0 < |x a| < , then |x a| < . So = will work.

25. Given > 0, we need > 0 such that if 0 < |x 0| < , then x2 0 < x2 < |x| < . Take = .

Then 0 < |x 0| < x2 0 < . Thus, limx0

x2 = 0 by the definition of a limit.

27. Given > 0, we need > 0 such that if 0 < |x 0| < , then |x| 0 < . But |x| = |x|. So this is true if we pick = .

Thus, limx0

|x| = 0 by the definition of a limit.


29. Given > 0, we need > 0 such that if 0 < |x 2| < , then x2 4x+ 5 1 < x2 4x+ 4 <

(x 2)2 < . So take = . Then 0 < |x 2| < |x 2| < (x 2)2 < . Thus,

limx2

x2 4x+ 5 = 1 by the definition of a limit.

31. Given > 0, we need > 0 such that if 0 < |x (2)| < , then x2 1 3 < or upon simplifying we need

x2 4 < whenever 0 < |x+ 2| < . Notice that if |x+ 2| < 1, then 1 < x+ 2 < 1 5 < x 2 < 3

|x 2| < 5. So take = min {/5, 1}. Then 0 < |x+ 2| < |x 2| < 5 and |x+ 2| < /5, so

x2 1 3 = |(x+ 2)(x 2)| = |x+ 2| |x 2| < (/5)(5) = . Thus, by the definition of a limit, limx2

(x2 1) = 3.

33. Given > 0, we let = min 2, 8

. If 0 < |x 3| < , then |x 3| < 2 2 < x 3 < 2

4 < x+ 3 < 8 |x+ 3| < 8. Also |x 3| < 8

, so x2 9 = |x+ 3| |x 3| < 8 8= . Thus, lim

x3x2 = 9.

35. (a) The points of intersection in the graph are (x1, 2.6) and (x2, 3.4)

with x1 0.891 and x2 1.093. Thus, we can take to be thesmaller of 1 x1 and x2 1. So = x2 1 0.093.

(b) Solving x3 + x+ 1 = 3 + gives us two nonreal complex roots and one real root, which is

x() =216 + 108+ 12

336 + 324+ 812

2/3 126 216 + 108+ 12

336 + 324+ 812

1/3. Thus, = x() 1.

(c) If = 0.4, then x() 1.093 272 342 and = x() 1 0.093, which agrees with our answer in part (a).

37. 1. Guessing a value for Given > 0, we must find > 0 such that |xa| < whenever 0 < |x a| < . But

|xa| = |x a|x+

a< (from the hint). Now if we can find a positive constant C such that

x+

a > C then

|x a|x+

a 12a+

a, and so

C = 12a+a is a suitable choice for the constant. So |x a| < 12a+

a . This suggests that we let

= min 12a, 1

2a+

a .

2. Showing that works Given > 0, we let = min 12a, 1

2a+

a . If 0 < |x a| < , then

|x a| < 12a x+a > 1

2a+

a (as in part 1). Also |x a| < 1

2a+

a , so

|xa | = |x a|x+

a 0 such that 0 < |x| < |f(x) L| < 12 . Take any rational

number r with 0 < |r| < . Then f(r) = 0, so |0 L| < 12

, so L |L| < 12

. Now take any irrational number s with

0 < |s| < . Then f(s) = 1, so |1 L| < 12

. Hence, 1 L < 12

, so L > 12

. This contradicts L < 12

, so limx0

f(x) does not

exist.

41. 1(x+ 3)4

> 10,000 (x+ 3)4 < 110,000

|x+ 3| < 1410,000

|x (3)| < 110

43. Given M < 0 we need > 0 so that lnx < M whenever 0 < x < ; that is, x = eln x < eM whenever 0 < x < . This

suggests that we take = eM . If 0 < x < eM , then lnx < ln eM =M . By the definition of a limit, limx0+

lnx = .

2.5 Continuity

1. From Definition 1, limx4

f(x) = f(4).

3. (a) The following are the numbers at which f is discontinuous and the type of discontinuity at that number: 4 (removable),2 ( jump), 2 ( jump), 4 (infinite).

(b) f is continuous from the left at 2 since limx2

f(x) = f(2). f is continuous from the right at 2 and 4 since

limx2+

f(x) = f(2) and limx4+

f(x) = f(4). It is continuous from neither side at 4 since f(4) is undefined.

5. The graph of y = f(x) must have a discontinuity at x = 3 and must show that limx3

f(x) = f(3).

7. (a) (b) There are discontinuities at times t = 1, 2, 3, and 4. A person

parking in the lot would want to keep in mind that the charge will

jump at the beginning of each hour.

9. Since f and g are continuous functions,

limx3

[2f(x) g(x)] = 2 limx3

f(x) limx3

g(x) [by Limit Laws 2 and 3]

= 2f(3) g(3) [by continuity of f and g at x = 3]= 2 5 g(3) = 10 g(3)

Since it is given that limx3

[2f(x) g(x)] = 4, we have 10 g(3) = 4, so g(3) = 6.


11. limx1

f(x) = limx1

x+ 2x34= lim

x1x+ 2 lim

x1x3

4

= 1 + 2(1)3 4 = (3)4 = 81 = f(1).

By the definition of continuity, f is continuous at a = 1.

13. For a > 2, we have

limxa

f(x) = limxa

2x+ 3

x 2 =limxa

(2x+ 3)

limxa

(x 2) [Limit Law 5]

=2 limxa

x+ limxa

3

limxa

x limxa

2[1, 2, and 3]

=2a+ 3

a 2 [7 and 8]

= f(a)

Thus, f is continuous atx = a for every a in (2,); that is, f is continuous on (2,).

15. f(x) = ln |x 2| is discontinuous at 2 since f(2) = ln 0 is not defined.

17. f(x) =ex if x < 0

x2 if x 0The left-hand limit of f at a = 0 is lim

x0f(x) = lim

x0ex = 1. The

right-hand limit of f at a = 0 is limx0+

f(x) = limx0+

x2 = 0. Since these

limits are not equal, limx0

f(x) does not exist and f is discontinuous at 0.

19. f(x) =

cosx if x < 0

0 if x = 0

1 x2 if x > 0

limx0

f(x) = 1, but f(0) = 0 6= 1, so f is discontinuous at 0.

21. F (x) = xx2 + 5x+ 6

is a rational function. So by Theorem 5 (or Theorem 7), F is continuous at every number in its domain,

x | x2 + 5x+ 6 6= 0 = {x | (x+ 3)(x+ 2) 6= 0} = {x | x 6= 3, 2} or (,3) (3,2) (2,).

23. By Theorem 5, the polynomials x2 and 2x 1 are continuous on (,). By Theorem 7, the root functionx iscontinuous on [0,). By Theorem 9, the composite function2x 1 is continuous on its domain, [ 1

2,).

By part 1 of Theorem 4, the sum R(x) = x2 +2x 1 is continuous on [ 1

2,).

SECTION 2.5 CONTINUITY 57

25. By Theorem 7, the exponential function e5t and the trigonometric function cos 2t are continuous on (,).

By part 4 of Theorem 4, L(t) = e5t cos 2t is continuous on (,).

27. By Theorem 5, the polynomial t4 1 is continuous on (,). By Theorem 7, lnx is continuous on its domain, (0,).

By Theorem 9, ln t4 1 is continuous on its domain, which is

t | t4 1 > 0 = t | t4 > 1 = {t | |t| > 1} = (,1) (1,)

29. The function y = 11 + e1/x

is discontinuous at x = 0 because the

left- and right-hand limits at x = 0 are different.

31. Because we are dealing with root functions, 5 +x is continuous on [0,),x+ 5 is continuous on [5,), so the

quotient f(x) = 5 +x

5 + xis continuous on [0,). Since f is continuous at x = 4, lim

x4f(x) = f(4) = 7

3.

33. Because x2 x is continuous on R, the composite function f(x) = ex2x is continuous on R, so

limx1

f(x) = f(1) = e1 1 = e0 = 1.

35. f(x) =x2 if x < 1x if x 1

By Theorem 5, since f(x) equals the polynomial x2 on (, 1), f is continuous on (, 1). By Theorem 7, since f(x)equals the root function

x on (1,), f is continuous on (1,). At x = 1, lim

x1f(x) = lim

x1x2 = 1 and

limx1+

f(x) = limx1+

x = 1. Thus, lim

x1f(x) exists and equals 1. Also, f(1) =

1 = 1. Thus, f is continuous at x = 1.

We conclude that f is continuous on (,).

37. f(x) =

1 + x2 if x 02 x if 0 < x 2(x 2)2 if x > 2

f is continuous on (, 0), (0, 2), and (2,) since it is a polynomial oneach of these intervals. Now lim

x0f(x) = lim

x0(1 + x2) = 1 and lim

x0+f(x) = lim

x0+(2 x) = 2, so f is

discontinuous at 0. Since f(0) = 1, f is continuous from the left at 0. Also, limx2

f(x) = limx2

(2 x) = 0,

limx2+

f(x) = limx2+

(x 2)2 = 0, and f(2) = 0, so f is continuous at 2. The only number at which f is discontinuous is 0.


39. f(x) =

x+ 2 if x < 0

ex if 0 x 12 x if x > 1

f is continuous on (, 0) and (1,) since on each of these intervals

it is a polynomial; it is continuous on (0, 1) since it is an exponential.

Now limx0

f(x) = limx0

(x+ 2) = 2 and limx0+

f(x) = limx0+

ex = 1, so f is discontinuous at 0. Since f(0) = 1, f is

continuous from the right at 0. Also limx1

f(x) = limx1

ex = e and limx1+

f(x) = limx1+

(2 x) = 1, so f is discontinuous

at 1. Since f(1) = e, f is continuous from the left at 1.

41. f(x) =cx2 + 2x if x < 2

x3 cx if x 2f is continuous on (, 2) and (2,). Now lim

x2f(x) = lim

x2cx2 + 2x = 4c+ 4 and

limx2+

f(x) = limx2+

x3 cx = 8 2c. So f is continuous 4c+4 = 8 2c 6c = 4 c = 23

. Thus, for f

to be continuous on (,), c = 23 .

43. (a) f(x) = x4 1x 1 =

(x2 + 1)(x2 1)x 1 =

(x2 + 1)(x+ 1)(x 1)x 1 = (x

2 + 1)(x+ 1) [or x3 + x2 + x+ 1]

for x 6= 1. The discontinuity is removable and g(x) = x3 + x2 + x+ 1 agrees with f for x 6= 1 and is continuous on R.

(b) f(x) = x3 x2 2xx 2 =

x(x2 x 2)x 2 =

x(x 2)(x+ 1)x 2 = x(x+ 1) [or x

2 + x] for x 6= 2. The discontinuity

is removable and g(x) = x2 + x agrees with f for x 6= 2 and is continuous on R.

(c) limx

f(x) = limx

[[sinx]] = limx

0 = 0 and limx+

f(x) = limx+

[[sinx]] = limx+

(1) = 1, so limx

f(x) does not

exist. The discontinuity at x = is a jump discontinuity.

45. f(x) = x2 + 10 sinx is continuous on the interval [31, 32], f(31) 957, and f(32) 1030. Since 957 < 1000 < 1030,there is a number c in (31, 32) such that f(c) = 1000 by the Intermediate Value Theorem. Note: There is also a number c in

(32,31) such that f(c) = 1000.

47. f(x) = x4 + x 3 is continuous on the interval [1, 2], f(1) = 1, and f(2) = 15. Since 1 < 0 < 15, there is a number c

in (1, 2) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation x4 + x 3 = 0 in theinterval (1, 2).

49. f(x) = cosx x is continuous on the interval [0, 1], f(0) = 1, and f(1) = cos 1 1 0.46. Since 0.46 < 0 < 1, thereis a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation

cosx x = 0, or cosx = x, in the interval (0, 1).

SECTION 2.5 CONTINUITY 59

51. (a) f(x) = cosx x3 is continuo

Stewart solution 6ed

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