L1 Functions and Its Operations

8/13/2019 L1 Functions and Its Operations

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Functions: Domain and Operations

Mathematics 100

Institute of Mathematics

Math 100 (Institute of Mathematics) Functions and Its Operations 1 / 30



Outline

1 FunctionsDomain of Some Functions

Operations on Functions

2 Lecture Exercise




Functions

"Intuitive" Definition

A function can be thought of as a correspondence from a set X of realnumbers

x to a set

Y of real numbers

y, where

y is unique for a specific value

of x.

We say that "y is a function of x", and write this symbolically as

y = f (x), (read y equals f of x).




Functions

The function can be pictured by an arrow diagram.

Remark:

Observe that a function can have the same value at two different valuesof x from X , but each x is assigned to a single element y of Y .




Functions

Definition

Let X and Y be nonempty sets.A function f from X to Y , denoted f : X

→ Y , is a rule that assigns to each

element x ∈ X a unique element y ∈ Y .The set of all admissible values of x is called the domain of the function,written dom f .

The set of all resulting values of y is called the range of the function,written ran f .




Remarks

A function may be written as y = f (x) where x is called the independent

variable while y is the dependent variable.

Alternatively, a function f is a set of ordered pairs (x, y) in which no twodistinct ordered pairs have the same first coordinate. Moreover, (x, y) ∈ f if and only if y = f (x).




Functions

Example

The equation x + y = 1 defines a function.

To see this, we isolate the dependent variable on the left hand side so that

y = −x + 1.

Note that for any given value of x, there is exactly one value for y.




Functions

Example

The equation |y| = −x + 1 does not define a function.

Observe that (x, y) = (−1, 2) and (x, y) = (−1, −2) both satisfy the givenequation.

So there are two distinct ordered pairs having the same x-coordinate thatsatisfy the given equation.




Functions

Example

Is x2 + y = 3 a function?

Solution: It is equivalent to

y = −x2 + 3.

Any value of x will give exactly one resulting value of y.

Therefore, the given equation defines y as a function of x.




Functions

Example

Is x2 + y2 = 4 a function?

Solution: Solve for y in terms of x:

y = ±

4 − x2.

So any value of x will correspond to two values of y.

Therefore, it is not a function.




Functions

Example

The equation y = √ 4 − x2 defines a function.

Recall: y denotes the principal square root of 4 − x2, which is unique,whenever it is defined.

Example

The expression y ≥ 3x + 2 does not define a function.

Note that the ordered pairs (x, y) = (1, 5) and (x, y) = (1, 6) satisfy the given

expression.

The examples which do not define a function are called relations.




Domain of Some Functions

Polynomial Functions - functions of the form

f (x) = anxn + an−1xn−1 + . . . + a1x + a0,

where an, an−1, . . . , a1, a0 ∈ with an = 0 and n is a nonnegative integer

dom f =


S




Rational Functions - functions of the form f (x) = h(x)

g(x), where h and g are

polynomial functions, and g is not the constant zero function

dom f = {x ∈ : g(x) = 0}


D i f S F i




Functions involving radicals - functions of the form f (x) = n

g(x)

If n is a positive even integer, then dom f = {x ∈ : g(x) ≥ 0}.

If n is a positive odd integer, then dom f = {x ∈ : g(x) ∈ }.


D i f S F ti




Example

Consider the function f (x) = x − 3.

dom f = , since f is a polynomial function

Example

Consider the quadratic function defined by g(x) = x2 + 5x + 6.

dom g =


D i f S F ti




ExampleFind the domain of h(x) =

√ 4 − x2.

Solution: Since h is a radical function with even index, then

dom h = {x ∈ : 4 − x2 ≥ 0}.

Solve the inequality:

4 − x2

≥ 0(2 − x)(2 + x) ≥ 0


D i f S F ti




Solution cont’d.:

Table of Signs:

(−∞, −2) (−2, 2) (2, +∞)Test number

−3 0 3

2 − x + + −2 + x − + +

(2 − x)(2 + x) − + −

Note that equality holds when x =

−2 or x = 2.

Therefore, dom h = [−2, 2].






Example

Find the domain of f (x) = 4

x2 − 9.

Solution: Since f is a rational function,

dom f = {x ∈ : x2 − 9 = 0} = {x ∈ : x = ±3} = \ {±3}.






Example

Find the domain of g(x) = |x + 3| − 4.

Solution: Since g is a radical function,

dom g = {x ∈

: |x + 3| − 4 ≥ 0}= {x ∈ : |x + 3| ≥ 4}

Next, we solve the inequality |x + 3| ≥ 4:

x + 3 ≥

4

x ≥ 4 − 3

x ≥ 1

or x + 3 ≤ −4x ≤ −7

Therefore, dom g = (−∞, −7] ∪ [1, +∞).






Example

Find the domain of h(x) =3√

x + 2

x2 − 1 .

Solution:

Since h is written as a fraction, we want all values of x for which itsdenominator is nonzero. At the same time, the numerator must be defined.

numerator - always defined for any real number value of x

denominator -

dom h = {x ∈ : x2 − 1 = 0} = \ {±1}.






DefinitionIf f and g are functions, then their

sum, denoted by f + g, is the function defined by(f + g)(x) = f (x) + g(x);

difference, denoted by f −

g, is the function defined by(f − g)(x) = f (x) − g(x);

product, denoted by f g, is the function defined by (f g)(x) = f (x) · g(x);

quotient, denoted by f /g, is the function defined by

f g (x) = f (x)

g(x) , g(x)

= 0;

with dom (f + g) = dom (f − g) = dom (f g) = dom f ∩ dom g,and dom (f /g) = (dom f ∩ dom g) \ {x ∈ : g(x) = 0}.






ExampleLet f (x) =

√ 4 − x and g(x) =

√ 3 + x. Find the functions f + g, f − g, f g and

f /g, and find their respective domains.

Solution:

(f + g)(x) = f (x) + g(x) =√

4 − x +√

3 + x,

(f − g)(x) = f (x) − g(x) =√

4 − x − √ 3 + x,

(f g)(x) = f (x) · g(x) =√

4 − x · √ 3 + x =

(4 − x)(3 + x),

f g

(x) = f (x)

g(x) = √ 4 − x√

3 + x=

4 − x3 + x

.






Solution cont’d.:

The domains of f and g are:

dom f = {x ∈ : x ≤ 4} = (−∞, 4],

dom g = {x ∈ : x ≥ −3} = [−3, +∞).

The intersection of these domains is (−∞, 4] ∩ [−3, +∞) = [−3, 4]. Thus,

dom (f + g) = dom (f − g) = dom (f g) = [−3, 4].

Since g(

−3) = 0, x =

−3 must be excluded from the domain of the quotient

function. Hence,dom (f /g) = (−3, 4].


Composition of Two Functions




Definition

If f and g are functions, their composite function, denoted by f ◦ g, isdefined by

(f

◦g)(x) = f (g(x)).

The domain of f ◦ g is the set of all real numbers x in the domain of g suchthat g(x) is in the domain of f . That is,

dom (f ◦ g) = {x ∈ : x ∈ dom g and g(x) ∈ dom f }.






Example

Find f ◦ g and its domain given f (x) =

√ 4 − x

2

and g(x) =

√ 3 − x.

Solution: We first state the domains of f and g:

dom f =

{x

∈ : 4

−x2

≥ 0

} = [

−2, 2],

dom g = {x ∈ : 3 − x ≥ 0} = (−∞, 3].

Now, let us find the composition

(f ◦ g)(x) = f (g(x)) = f (√

3 − x)

=

4 − (√ 3 − x)2

=

4 − (3 − x)

=√

1 + x.






Solution cont’d.:

Even though√

1 + x is defined for all x ≥ −1, we must restrict the domain off ◦

g to those values that are also in the domain of g. Thus,

dom (f ◦ g) = {x ∈ : x ≥ −1 and x ≤ 3} = [−1, 3].






Example

Given f (x) = |

x + 3| −

2 and g(x) = 1

√ x. Find the composite functions f

◦g

and g ◦ f , and their respective domains.

Solution: Note that dom f = and dom g = (0, +∞).For f ◦ g:

(f ◦ g)(x) = f (g(x)) = f

1√

x

=

1√ x

+ 3

− 2

= 1√ x + 3 − 2

= 1 +

√ x√

x

Thus, dom (f

◦g) = (0, +

∞).






Solution cont’d.:

For g ◦ f :

(g ◦ f )(x) = g(f (x)) = g (|x + 3| − 2)

= 1 |x + 3| − 2

Now, |x + 3| − 2 > 0 when x ∈ (−∞, −5) ∪ (−1, +∞).Since we need f (x) > 0 in order for g(f (x)) to be defined, we have

dom(g◦

f ) = (−∞

,−

5)∪

(−

1, +∞

).


Lecture Exercise



Let f (x) =√

1

−x and g(x) =

x − 1

x + 1. Find the composite function f

◦g

and its domain.


L1 Functions and Its Operations

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