1. Splash Screen

2. Key Concept: Operations with Functions
Five-Minute Check
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
Lesson Menu

3. A. The parent graph is translated up 0.5 units.
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 compressed horizontally by a factor of 0.5.
C. The parent graph is compressed vertically by a factor of 0.5.
D. The parent graph is translated down 0.5 units.
5–Minute Check 1

4. A. The parent graph is translated up 0.5 units.
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 compressed horizontally by a factor of 0.5.
C. The parent graph is compressed vertically by a factor of 0.5.
D. The parent graph is translated down 0.5 units.
5–Minute Check 1

5. A. The parent graph is translated left 3 units and up 4 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.
5–Minute Check 2

6. A. The parent graph is translated left 3 units and up 4 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.
5–Minute Check 2

7. A. B. C. D.
5–Minute Check 3

8. A. B. C. D.
5–Minute Check 3

9. A. f (x) = x; g(x) is f(x) translated left 4 units.
Identify the parent function f (x) if and describe how the graphs of g (x) and f (x) are related.
A. f (x) = x; g(x) is f(x) translated left 4 units.
B. f(x) = |x|; g(x) is f(x) translated right 4 units.
C g(x) is f(x) translated right 4 units.
D g(x) is f(x) translated left 4 units.
5–Minute Check 4

10. A. f (x) = x; g(x) is f(x) translated left 4 units.
Identify the parent function f (x) if and describe how the graphs of g (x) and f (x) are related.
A. f (x) = x; g(x) is f(x) translated left 4 units.
B. f(x) = |x|; g(x) is f(x) translated right 4 units.
C g(x) is f(x) translated right 4 units.
D g(x) is f(x) translated left 4 units.
5–Minute Check 4

11. You evaluated functions. (Lesson 1-1)
Perform operations with functions.
Find compositions of functions.
Then/Now

12. composition
Vocabulary

13. Key Concept 1

14. (f + g)(x) = f(x) + g(x) Definition of sum of two functions
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 of two 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:
Example 1

15. (f + g)(x) = f(x) + g(x) Definition of sum of two functions
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 of two 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:
Example 1

16. (f – h)(x) = f(x) – h(x) Definition of difference of two functions
Operations with Functions
B. Given f (x) = x 2 – 2x, g (x) = 3x – 4, and h (x) = –2x 2 + 1, find the function and domain for (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:
Example 1

17. (f – h)(x) = f(x) – h(x) Definition of difference of two functions
Operations with Functions
B. Given f (x) = x 2 – 2x, g (x) = 3x – 4, and h (x) = –2x 2 + 1, find the function and domain for (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:
Example 1

18. (f ● g)(x) = f (x) ● g(x) Definition of product of two functions
Operations with Functions
C. 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 product of two functions
= (x 2 – 2x)(3x – 4) f (x) = x 2 – 2x; g (x) = 3x – 4
= 3x 3 – 10x 2 + 8x Simplify.
The domain of f and g are both so the domain of (f ● g) is
Answer:
Example 1

19. (f ● g)(x) = f (x) ● g(x) Definition of product of two functions
Operations with Functions
C. 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 product of two functions
= (x 2 – 2x)(3x – 4) f (x) = x 2 – 2x; g (x) = 3x – 4
= 3x 3 – 10x 2 + 8x Simplify.
The domain of f and g are both so the domain of (f ● g) is
Answer:
Example 1

20. Definition of quotient of two functions
Operations with Functions
D. Given f (x) = x 2 – 2x, g (x) = 3x – 4, and h (x) = –2x 2 + 1, find the function and domain for
Definition of quotient of two functions
f(x) = x 2 – 2x; h(x) = –2x 2 + 1
Example 1

21. Operations with Functions
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) È (0, 2) È (2, ∞).
Answer:
Example 1

22. Answer: D = (–∞, 0) È (0, 2) È (2, ∞)
Operations with Functions
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) È (0, 2) È (2, ∞).
Answer: D = (–∞, 0) È (0, 2) È (2, ∞)
Example 1

23. 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.
Example 1

24. A. B. C. D.
Example 1

25. A. B. C. D.
Example 1

26. Key Concept 2

27. A. Given f (x) = 2x2 – 1 and g (x) = x + 3, find [f ○ g](x).
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
Expand (x +3)2
= 2(x 2 + 6x + 9) – 1
Simplify.
= 2x x + 17
Answer:
Example 2

28. A. Given f (x) = 2x2 – 1 and g (x) = x + 3, find [f ○ g](x).
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
Expand (x +3)2
= 2(x 2 + 6x + 9) – 1
Simplify.
= 2x x + 17
Answer: [f ○ g](x) = 2x x + 17
Example 2

29. B. Given f (x) = 2x2 – 1 and g (x) = x + 3, find [g ○ f](x).
Compose Two Functions
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:
Example 2

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

31. C. Given f (x) = 2x 2 – 1 and g (x) = x + 3, find [f ○ g](2).
Compose Two Functions
C. Given f (x) = 2x 2 – 1 and g (x) = x + 3, find [f ○ g](2).
Evaluate the expression you wrote in part A for x = 2.
[f ○ g](2) = 2(2)2 + 12(2) + 17 Substitute 2 for x.
= 49 Simplify.
Answer:
Example 2

32. C. Given f (x) = 2x 2 – 1 and g (x) = x + 3, find [f ○ g](2).
Compose Two Functions
C. Given f (x) = 2x 2 – 1 and g (x) = x + 3, find [f ○ g](2).
Evaluate the expression you wrote in part A for x = 2.
[f ○ g](2) = 2(2)2 + 12(2) + 17 Substitute 2 for x.
= 49 Simplify.
Answer: [f ○ g](2) = 49
Example 2

33. Find for f (x) = 2x – 3 and g (x) = 4 + x 2.
A. 2x ; 4x 2 – 12x + 13; 23
B. 2x ; 4x 2 – 12x + 5; 23
C. 2x 2 + 5; 4x 2 – 12x + 5; 23
D. 2x 2 + 5; 4x 2 – 12x + 13; 23
Example 2

34. Find for f (x) = 2x – 3 and g (x) = 4 + x 2.
A. 2x ; 4x 2 – 12x + 13; 23
B. 2x ; 4x 2 – 12x + 5; 23
C. 2x 2 + 5; 4x 2 – 12x + 5; 23
D. 2x 2 + 5; 4x 2 – 12x + 13; 23
Example 2

35. Find a Composite Function with a Restricted Domain
A. Find
Example 3

36. Find a Composite Function with a Restricted Domain
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] È [2, ∞). Now find [f ○ g](x).
Example 3

37. Substitute (x – 1)2 for x in f (x).
Find a Composite Function with a Restricted Domain
Replace g (x) with (x – 1)2.
Substitute (x – 1)2 for x in
f (x).
Simplify.
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] È [2, ∞).
Example 3

38. Find a Composite Function with a Restricted Domain
Answer:
Example 3

39. Find a Composite Function with a Restricted Domain
Answer: for (–∞, 0] È [2, ∞).
Example 3

40. B. Find f ○ g. Find a Composite Function with a Restricted Domain
Example 3

41. Find a Composite Function with a Restricted Domain
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) È (1, ∞). Now find [f ○ g](x).
Example 3

42. Find a Composite Function with a Restricted Domain
Example 3

43. Find a Composite Function with a Restricted Domain
Answer:
Example 3

44. Find a Composite Function with a Restricted Domain
Answer:
Example 3

45. Find a Composite Function with a Restricted Domain
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].
Example 3

46. Find a Composite Function with a Restricted Domain
Example 3

47. Find f ○ g. A. D = (–∞, –1)  (–1, 1)  (1, ∞); B. D = [–1, 1];
C. D = (–∞, –1)  (–1, 1)  (1, ∞);
D. D = (0, 1);
Example 3

48. Find f ○ g. A. D = (–∞, –1)  (–1, 1)  (1, ∞); B. D = [–1, 1];
C. D = (–∞, –1)  (–1, 1)  (1, ∞);
D. D = (0, 1);
Example 3

49. A. Find two functions f and g such that
Decompose a Composite Function
A. Find two functions f and g such that
when Neither function may be the
identity function f (x) = x.
Example 4

50. Decompose a Composite Functionh
Sample answer:
Example 4

51. Decompose a Composite Functionh
Sample answer:
Example 4

52. B. Find two functions f and g such that
Decompose a Composite Function
B. Find two functions f and g such that
when h (x) = 3x 2 – 12x Neither function may be the identity function f (x) = x.
h (x) = 3x2 – 12x + 12 Notice that h is factorable.
= 3(x2 – 4x + 4) or 3(x – 2) 2 Factor.
One way to write h (x) as a composition is to let f (x) = 3x2 and g (x) = x – 2.
Example 4

53. Decompose a Composite Function
Sample answer:
Example 4

54. Sample answer: g (x) = x – 2 and f (x) = 3x 2
Decompose a Composite Function
Sample answer: g (x) = x – 2 and f (x) = 3x 2
Example 4

55. A.
B.

C.

D.

Example 4

56. A.
B.

C.

D.

Example 4

57. So, the functions are R(t) = 25 + 10t and A(R) = R 2.
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) = t, 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) = t and A(R) = R 2.
Answer:
Example 5

58. So, the functions are R(t) = 25 + 10t and A(R) = R 2.
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) = t, 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) = t and A(R) = R 2.
Answer: R(t) = t; A(R) = R 2
Example 5

59. A ○ R = A[R(t)] Definition of A ○ R
Compose Real-World Functions
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( t) Replace R (t) with t.
= ( t)2 Substitute ( t) for R in A(R).
= 100t t + 625 Simplify.
Example 5

60. Compose Real-World Functions
So, A ○ R = 100t t + 625. The composite function models the area of the circle as a function of time.
Answer:
Example 5

61. Compose Real-World Functions
So, A ○ R = 100t t + 625. The composite function models the area of the circle as a function of time.
Answer: A ○ R = 100t t + 625 ; the function models the area of the circle as a function of time.
Example 5

62. Compose Real-World Functions
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) = 100t t + 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:
Example 5

63. Compose Real-World Functions
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) = 100t t + 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
Example 5

64. 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.
Example 5

65. 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.
Example 5

66. End of the Lesson