Key Concept 1. Example 1 Operations with Functions A. Given f (x) = x 2 – 2x, g (x) = 3x – 4, and h (x) = –2x 2 + 1, find the function and domain for

Operations with Functions

A. Given f (x) = x 2 – 2x, g (x) = 3x – 4, and

h (x) = –2x 2 + 1, find the function and domain for

(f + g)(x).

(f + g)(x) = f(x) + g(x) Definition of sum oftwo functions

= (x 2 – 2x) + (3x – 4)

f (x) = x 2 – 2x;

g (x) = 3x – 4

= x 2 + x – 4

Simplify.The domain of f and g are both so the domain of (f + g) is

Answer:


B. Given f (x) = x 2 – 2x, g (x) = 3x – 4, and


(f – h)(x).

(f – h)(x) = f(x) – h(x) Definition of difference of two functions

= (x 2 – 2x) – (–2x

2 + 1) f(x) = x

2 – 2x; h(x) = –2x

2 + 1

= 3x 2 – 2x – 1

Simplify. The domain of f and h are both so the domain of (f – h) is

Answer:


C. Given f (x) = x 2 – 2x, g(x) = 3x – 4, and


(f ● g)(x).(f ● g)(x) = f (x) ● g(x) Definition of product of

two functions

= (x 2 – 2x)(3x – 4)

f (x) = x 2 – 2x;

g (x) = 3x – 4

= 3x 3 – 10x

2 + 8xSimplify.

The domain of f and g are both so the domain of (f ● g) is

Answer:


D. Given f (x) = x 2 – 2x, g (x) = 3x – 4, and


Definition of quotient of two functions

f(x) = x 2 – 2x; h(x) = –2x

2 + 1


The domains of h and f are both (–∞, ∞), but x = 0 or

x = 2 yields a zero in the denominator of . So, the

domain of (–∞, 0) (0, 2) (2, ∞).

Answer: D = (–∞, 0) (0, 2) (2, ∞)

Find (f + g)(x), (f – g)(x), (f ● g)(x), and for

f (x) = x 2 + x, g (x) = x – 3. State the domain of each

new function.

A.

B.

C.

D.

Compose Two Functions

A. Given f (x) = 2x2 – 1 and g (x) = x + 3, find [f ○ g](x).

Replace g (x) with x + 3= f (x + 3)

Substitute x + 3 for x in f (x).

= 2(x + 3)2 – 1

Answer: [f ○ g](x) = 2x 2 + 12x + 17

Expand (x +3)2= 2(x 2 + 6x + 9) – 1

Simplify.= 2x 2 + 12x + 17


B. Given f (x) = 2x2 – 1 and g (x) = x + 3, find [g ○ f](x).

Substitute 2x 2 – 1 for

x in g (x).= (2x

2 – 1) + 3

Simplify= 2x 2 + 2

Answer: [g ○ f](x) = 2x 2 + 2


Evaluate the expression you wrote in part A for x = 2.

Answer: [f ○ g](2) = 49

C. Given f (x) = 2x 2 – 1 and g (x) = x + 3, find [f ○ g](2).

[f ○ g](2) = 2(2)2 + 12(2) + 17 Substitute 2 for x.

= 49 Simplify.

A. 2x 2 + 11; 4x

2 – 12x + 13; 23

B. 2x 2 + 11; 4x

2 – 12x + 5; 23

C. 2x 2 + 5; 4x

2 – 12x + 5; 23

D. 2x 2 + 5; 4x

2 – 12x + 13; 23

Find for f (x) = 2x – 3 and g (x) = 4 + x

2.

Find a Composite Function with a Restricted Domain

A. Find .


To find , you must first be able to find g(x) = (x – 1)

2,

which can be done for all real numbers. Then you must

be able to evaluate for each of these

g (x)-values, which can only be done when g (x) > 1.

Excluding from the domain those values for which

0 < (x – 1) 2 <1, namely when 0 < x < 1, the domain of

f ○ g is (–∞, 0] [2, ∞). Now find [f ○ g](x).

Notice that is not defined for 0 < x < 2.

Because the implied domain is the same as the

domain determined by considering the domains of

f and g, we can write the composition as

for (–∞, 0] [2, ∞).


Replace g (x) with (x – 1)2.

Substitute (x – 1)2 for x in

f (x).

Simplify.


Answer: for (–∞, 0] [2, ∞).


B. Find f ○ g.


To find f ○ g, you must first be able to find ,

which can be done for all real numbers x such that x2 1.

Then you must be able to evaluate for each of

these g (x)-values, which can only be done when g (x) 0.

Excluding from the domain those values for which

0 > x 2 – 1, namely when –1 < x < 1, the domain of f ○ g is

(–∞, –1) (1, ∞). Now find [f ○ g](x).



Answer:


Check Use a graphing calculator to check this result.

Enter the function as . The graph appears

to have asymptotes at x = –1 and x = 1. Use the

TRACE feature to help determine that the domain of

the composite function does not include any values in

the interval [–1, 1].


Find f ○ g.

A. D = (–∞, –1) (–1, 1) (1, ∞);

B. D = [–1, 1];

C. D = (–∞, –1) (–1, 1) (1, ∞);

D. D = (0, 1);

Decompose a Composite Function

A. Find two functions f and g such that

when . Neither function may be the

identity function f (x) = x.


Sample answer:

h


h (x) = 3x2 – 12x + 12 Notice that h is factorable.

= 3(x2 – 4x + 4) or 3(x – 2)

2 Factor.

B. Find two functions f and g such that

when h (x) = 3x 2 – 12x + 12. Neither function may

be the identity function f (x) = x.

One way to write h (x) as a composition is to let f (x) = 3x2 and g (x) = x – 2.

Sample answer: g (x) = x – 2 and f (x) = 3x 2


A.

B.

C.

D.

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Key Concept 1. Example 1 Operations with Functions A. Given f (x) = x 2 – 2x, g (x) = 3x – 4, and h (x) = –2x 2 + 1, find the function and domain for