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Section 7.6

Section 7.6. Function Operations. Function Operations. ALGEBRA 2 LESSON 7-6. (For help, go to Lesson 2-1.). Find the domain and range of each function. 1. {(0, –5), (2, –3), (4, –1)} 2. {(–1, 0), (0, 0), (1, 0)} 3. ƒ( x ) = 2 x – 12 4. g ( x ) = x 2

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Section 7.6

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  1. Section 7.6 Function Operations

  2. Function Operations ALGEBRA 2 LESSON 7-6 (For help, go to Lesson 2-1.) Find the domain and range of each function. 1. {(0, –5), (2, –3), (4, –1)} 2. {(–1, 0), (0, 0), (1, 0)} 3. ƒ(x) = 2x – 12 4.g(x) = x2 Evaluate each function for the given value of x. 5. Let ƒ(x) = 3x + 4. Find ƒ(2). 6. Let g(x) = 2x2 – 3x + 1. Find g(–3). Check Skills You’ll Need 7-6

  3. Function Operations ALGEBRA 2 LESSON 7-6 Solutions 1. {(0, –5), (2, –3), (4, –1)}The domain is {0, 2, 4}. The range is {–5, –3, –1}. 2. {(–1, 0), (0, 0), (1, 0)}The domain is {–1, 0, 1}. The range is {0}. 3. ƒ(x) = 2x – 12The domain is the set of all real numbers. Therange is the set of all real numbers. 4.g(x) = x2The domain is the set of all real numbers. Therange is the set of all nonnegative real numbers. 5. ƒ(x) = 3x + 4; ƒ(2) = 3(2) + 4 = 6 + 4 = 10 6.g(x) = 2x2 – 3x + 1; g(–3) = 2(–3)2 – 3(–3) + 1 =2(9) + 9 + 1 = 18 + 10 = 28 7-6

  4. Function Operations ALGEBRA 2 LESSON 7-6 Let ƒ(x) = –2x + 6 and g(x) = 5x – 7. Find ƒ + g and ƒ – g and their domains. (ƒ + g)(x) = ƒ(x) + g(x) (ƒ – g)(x) = ƒ(x) – g(x) = (–2x + 6) + (5x – 7) = (–2x + 6) – (5x – 7) = 3x – 1 = –7x + 13 The domains of ƒ + g and ƒ – g are the set of real numbers. Quick Check 7-6

  5. ƒ g ƒ g ƒ g (x) = ƒ(x) g(x) x2 + 1 x4 – 1 = 1 x2 – 1 x2 + 1 (x2 + 1)(x2 – 1) = = The domain of does not include 1 and –1 because g(1) and g(–1) = 0. Function Operations ALGEBRA 2 LESSON 7-6 Quick Check Let ƒ(x) = x2 + 1 and g(x) = x4 – 1. Find ƒ • g and and their domains. (ƒ • g)(x) = ƒ(x) • g(x) = (x2 + 1)(x4 – 1) = x6 + x4 – x2 – 1 The domains of ƒ and g are the set of real numbers, so the domain of ƒ • g is also the set of real numbers. 7-6

  6. Method 1: Method 2: (g°ƒ)(x) = g(ƒ(x)) = g(x3) = x6 + 7 (g°ƒ)(x) = g(ƒ(x)) = g(8) = 82 + 7 = 71 Function Operations ALGEBRA 2 LESSON 7-6 Let ƒ(x) = x3 and g(x) = x2 + 7. Find (g°ƒ)(2). (g°ƒ)(2) = (2)6 + 7 = 64 + 7 = 71 g(ƒ(2)) = g(23) Quick Check 7-6

  7. ƒ(x) ƒ(x) 1 3 = all reals except 2x2 3x – 1 g(x) g(x) Function Operations ALGEBRA 2 LESSON 7-6 Let ƒ(x) = 2x2 and g(x) = 3x – 1. Perform each function operation. Then find the domain. 1. ƒ(x) + g(x) 2. ƒ(x) – g(x) 3. ƒ(x) • g(x) 4. 5. (ƒ °g)(x) 6. A store is offering a 15% discount on all items. You have a coupon worth $2 off any item. Let x be the original cost of an item. Use a composition of functions to find a function c(x) that gives the final cost of an item if the discount is applied first and then the coupon. Then use this function to find the final cost of an item originally priced at $10. ƒ(x) + g(x) = 2x2 + 3x – 1; all reals ƒ(x) – g(x) = 2x2 – 3x + 1; all reals ƒ(x) • g(x) = 6x3 – 2x2; all reals (ƒ °g)(x) = 18x2 – 12x + 2; all reals c(x) = 0.85x – 2; $6.50 7-6

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