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

Vocabulary composition

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 (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 – 4Simplify. The domain of f and g are both so the domain of (f + g) is Answer:

Example 1 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 = 3x 2 – 2x – 1Simplify. The domain of f and h are both so the domain of (f – h) is Answer:

Example 1 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 + 8xSimplify. The domain of f and g are both so the domain of (f ● g) is Answer:

Example 1 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 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)  (0, 2)  (2, ∞). Answer: D = (–∞, 0)  (0, 2)  (2, ∞)

Example 1 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 A. B. C. D.

Key Concept 2

Example 2 Compose Two Functions A. Given f (x) = 2x 2 – 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 x + 17 Expand (x +3) 2 = 2(x 2 + 6x + 9) – 1 Simplify.= 2x x + 17

Example 2 Compose Two Functions B. Given f (x) = 2x 2 – 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 Answer: [g ○ f](x) = 2x 2 + 2

Example 2 Compose Two Functions 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) + 17Substitute 2 for x. = 49Simplify.

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

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

Example 3 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]  [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, ∞). Example 3 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.

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

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

Example 3 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 x 2  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).

Example 3 Find a Composite Function with a Restricted Domain

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

Example 3 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 Find a Composite Function with a Restricted Domain

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

Example 4 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 Decompose a Composite Function Sample answer: h

Example 4 Decompose a Composite Function h (x) = 3x 2 – 12x + 12Notice that h is factorable. = 3(x 2 – 4x + 4) or 3(x – 2) 2 Factor. 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. One way to write h (x) as a composition is to let f (x) = 3x 2 and g (x) = x – 2.

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

Example 4 A. B. C. D.

Example 5 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 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  Simplify.

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

Example 5 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  = 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 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.

End of the Lesson