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Then/Now

New Vocabulary

Key Concept:Operations with Functions

Example 1: Operations with Functions

Key Concept: Composition of Functions

Example 2: Compose Two Functions

Example 3: Find a Composite Function with a Restricted Domain

Example 4: Decompose a Composite Function

Example 5:Real-World Example: Compose Real-World Functions

http://connected.mcgraw-hill.com/connected/

Use the graph of y = x 2 to describe the graph of the

related function y = 0.5x 2.

A. The parent graph is translated up 0.5 units.

B. The parent graph is expanded horizontally by a factor of 0.5.

C. The parent graph is compressed vertically.

D. The parent graph is translated down 0.5 units.


Use the graph of y = x 2 to describe the graph of the

related function y = (x – 4)2 – 3.

A. The parent graph is translated left 3 units and up 4 units.

B. The parent graph is translated right 3 units and down 4 units.

C. The parent graph is translated left 4 units and down 3 units.

D. The parent graph is translated right 4 units and down 3 units.


A.

B.

C.

D.


Identify the parent function f (x) if

and describe how the graphs of g (x) and f (x) are

related.

A. f (x) = x; f (x) is translated left 4 units.

B. f(x) = |x|; f(x) is translated right 4 units.

C.

D.


You evaluated functions. (Lesson 1-1)

• Perform operations with functions.

• Find compositions of functions.


• composition


http://www.glencoe.com/apps/eGlossary612/grade.php


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 domain 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) = 29

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.

= 29 Simplify.


A. 2x 2 + 5; 4x

2 – 12x + 13; 23

B. 2x 2 + 11; 4x

2 – 12x + 13; 23

C. 2x 2 + 5; 4x

2 – 12x + 5; 23

D. 2x 2 + 5; 4x

2 – 12x + 13; 13

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 (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.


Compose Real-World Functions

A. COMPUTER ANIMATION An animator starts with an image of a circle with a radius of 25 pixels. The animator then increases the radius by 10 pixels per second. Find functions to model the data.

The length r of the radius increases at a rate of 10 pixels per second, so R(t) = 25 + 10t, where t is the time in seconds and t 0. The area of the circle is times the square of the radius. So, the area of the circle is A(R) = R

2.

So, the functions are R(t) = 25 + 10t and A(R) = R 2.

Answer: R(t) = 25 + 10t; A(R) = R 2



B. COMPUTER ANIMATION An animator starts with an image of a circle with a radius of 25 pixels. The animator then increases the radius by 10 pixels per second. Find A ○ R. What does the function represent?

A ○ R = A[R(t)] Definition of A ○ R

=A(25 + 10t)Replace R (t) with 25 + 10t.

= (25 + 10t)2

Substitute (25 + 10t) for R in A(R).

= 100t 2 + 500t +

625Simplify.


Answer: A ○ R = 100t 2 + 500t + 625 ; the function models the area of the circle as a function of time.


So, A ○ R = 100t 2 + 500t + 625. The composite

function models the area of the circle as a function of time.



C. COMPUTER ANIMATION An animator starts with an image of a circle with a radius of 25 pixels. The animator then increases the radius by 10 pixels per second. How long does it take for the circle to quadruple its original size?

The initial area of the circle is ● 25 2 = 625 pixels. The

circle will be four times its original size when [A ◦ R ](t) = 100t

2 + 500t + 625 = 4(625) = 2500. Solve for t to find that t = 2.5 or –7.5 seconds. Because a negative t-value is not part of the domain of R (t), it is also not part of the domain of the composite function. The area will quadruple after 2.5 seconds.

Answer: 2.5 seconds


BUSINESS A satellite television company offers a 20% discount on the installation of any satellite television system. The company also advertises $50 in coupons for the cost of equipment. Find [c ◦ d](x) and [d ◦ c](x). Which composition of the coupon and discount results in the lower price? Explain.

A. [c ◦ d](x) = 0.80x – 40; [d ◦ c](x) = 0.80x – 50; Sample answer: [d ◦ c](x) represents the cost of installation using the coupon and then the discount results in the lower cost.

B. [c ◦ d](x) = 0.80x – 40; [d ◦ c](x) = 0.80x – 50; Sample answer: [c ◦ d](x) represents the cost of installation using the discount and then the coupon results in the lower cost.

C. [c ◦ d](x) = 0.80x – 50; [d ◦ c](x) = 0.80x – 40; Sample answer: [c ◦ d](x) represents the cost of installation using the discount and then the coupon results in the lower cost.

D. [c ◦ d](x) = 0.80x – 50; [d ◦ c](x) = 0.80x – 40; Sample answer: [c ◦ d](x) represents the cost of installation using the coupon and then the discount results in the lower cost.



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