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1.
05_ch04b_pre-calculas12_wncp_solution.qxd
5/18/12 7:03 PM Page 23 Home Quit Lesson 4.4 Exercises, pages 314–321 A 3. For each function below, determine possible functions f and g so that y ϭ f (g(x)). √ a) y ϭ (x ϩ 4)2 b) y ϭ x + 5 Sample solution: Sample solution: √ Let f(g(x)) ( ؍x ؉ 4)2 Let f(g(x)) ؍x ؉ 5 Replace x ؉ 4 with x. Replace x ؉ 5 with x. √ Then, g(x) ؍x ؉ 4 and f(x) ؍x2 Then, g(x) ؍x ؉ 5 and f(x) ؍x 1 c) y ϭ x - 2 d) y ϭ (6 Ϫ x)3 Sample solution: Sample solution: Let f(g(x)) ؍ 1 Let f(g(x)) ؊ 6( ؍x)3 x؊2 Replace 6 ؊ x with x. Replace x ؊ 2 with x. 1 Then, g(x) ؊ 6 ؍x and f(x) ؍x3 Then, g(x) ؍x ؊ 2 and f(x) ؍x ©P DO NOT COPY. 4.4 Determining Restrictions on Composite Functions—Solutions 23
2.
05_ch04b_pre-calculas12_wncp_solution.qxd
5/18/12 7:03 PM Page 24 Home Quit B 4. Given f(x) ϭ x ϩ 3 and g(x) ϭ x2 ϩ 1, sketch the graph of each composite function below then state its domain and range. a) y ϭ f ( f(x)) b) y ϭ f (g(x)) y y 8 8 6 6 y ϭ f(f(x)) y ϭ f(g(x)) 4 4 2 2 x x Ϫ6 Ϫ4 Ϫ2 0 2 Ϫ4 Ϫ2 0 2 4 Make a table of values for the functions. x f(x) f (f(x)) g(x) f (g(x)) g(f(x)) g(g(x)) ؊4 ؊1 2 17 20 2 290 ؊3 0 3 10 13 1 101 ؊2 1 4 5 8 2 26 ؊1 2 5 2 5 5 5 0 3 6 1 4 10 2 1 4 7 2 5 17 5 2 5 8 5 8 26 26 a) Graph the points with coordinates (x, f(f(x))) that fit on the grid. Draw a line through the points for the graph of y ؍f (f(x)). From the graph, the domain is x ç ޒand the range is y ç .ޒ b) Graph the points with coordinates (x, f( g(x))) that fit on the grid. Draw a smooth curve through the points for the graph of y ؍f( g(x)). From the graph, the domain is x ç ޒand the range is y » 4. c) y ϭ g( f(x)) d) y ϭ g(g(x)) y ϭ g(f(x)) y y 30 4 20 2 10 x y ϭ g(g(x)) x Ϫ2 0 2 4 Ϫ4 Ϫ2 0 2 4 c) Graph the points with coordinates (x, g(f(x))) that fit on the grid. Draw a smooth curve through the points for the graph of y ؍g(f(x)). From the graph, the domain is x ç .ޒFrom the table, the range is y » 1. d) Graph the points with coordinates (x, g(g(x))) that fit on the grid. Draw a smooth curve through the points for the graph of y ؍g(g(x)). From the graph, the domain is x ç .ޒFrom the table, the range is y » 2. 24 4.4 Determining Restrictions on Composite Functions—Solutions DO NOT COPY. ©P
3.
05_ch04b_pre-calculas12_wncp_solution.qxd
5/18/12 7:03 PM Page 25 Home Quit 5. Consider the function h(x) ϭ (x Ϫ 1)(x ϩ 5). a) Why is it incorrect to write h(x) ϭ f (g(x)), where f(x) ϭ x Ϫ 1 and g(x) ϭ x ϩ 5? It is incorrect because, as written, h(x) is the product of f(x) and g(x), not their composition. b) For what functions f(x) and g(x) is h(x) a composite function? Expand: h(x) ( ؍x ؊ 1)(x ؉ 5) h(x) ؍x2 ؉ 4x ؊ 5 Complete the square: h(x) ( ؍x2 ؉ 4x ؉ 4) ؊ 9 h(x) ( ؍x ؉ 2)2 ؊ 9 Possible functions are: f(x) ؍x2 ؊ 9 and g(x) ؍x ؉ 2 for h(x) ؍f(g(x)) 6. For each pair of functions below: i) Determine an explicit equation for the indicated composite function. ii) State the domain of the composite function, and explain any restrictions on the variable. √ a) f(x) ϭ x + 1 and g(x) ϭ x 2 Ϫ x Ϫ 6; g ( f(x)) √ i) In g(x) ؍x2 ؊ x ؊ 6, replace x with x ؉ 1. √ √ g(f(x)) ( ؍x ؉ 1)2 ؊ x ؉ 1 ؊ 6 √ g(f(x)) ؍x ؉ 1 ؊ x ؉ 1 ؊ 6 √ g(f(x)) ؍x ؊ 5 ؊ x ؉ 1 √ ii) The domain of f(x) ؍x ؉ 1 is x » ؊1. The domain of g(x) ؍x2 ؊ x ؊ 6 is x ç .ޒ So, the domain of g(f(x)) is x » ؊1. The variable x is restricted because the square root of a real number is only defined for numbers that are greater than or equal to 0. √ 1 b) f(x) ϭ x - 1 and g(x) ϭ ; g( f(x)) x + 3 1 √ i) In g(x) ؍ , replace x with x ؊ 1 . x؉3 1 g(f(x)) √ ؍ x؊1؉3 √ ii) The domain of f(x) ؍ x ؊ 1 is x » 1. 1 The domain of g(x) ؍ is x ؊3. x؉3 ؊3 is not in the range of f(x). So, the domain of g(f(x)) is x » 1. The variable x is restricted because the square root of a real number is only defined for numbers that are greater than or equal to 0. ©P DO NOT COPY. 4.4 Determining Restrictions on Composite Functions—Solutions 25
4.
05_ch04b_pre-calculas12_wncp_solution.qxd
5/18/12 7:03 PM Page 26 Home Quit √ c) f(x) ϭ x + 3 and g(x) ϭ 2x Ϫ 1; f (g(x)) √ i) In f(x) ؍x ؉ 3, replace x with 2x ؊ 1. √ f(g(x)) 2 ؍x ؊ 1 ؉ 3 √ f(g(x)) 2 ؍x ؉ 2 ii) The domain of g(x) 2 ؍x ؊ 1 is x ç .ޒ √ The domain of f(x) ؍x ؉ 3 is x » ؊3. So, g(x) » ؊3 2x ؊ 1 » ؊3 2x » ؊2 x » ؊1 So, the domain of f(g(x)) is x » ؊1. The variable x is restricted because the square root of a real number is only defined for numbers that are greater than or equal to 0. 1 d) f(x) ϭ x - 1 and g(x) ϭ x 2 + 2x; f ( f(x)) 1 1 i) In f(x) ؍ , replace x with . x؊1 x؊1 1 x؊1 f(f(x)) ؍ , which simplifies to f (f(x)) ؍ ,x 1 1 2؊x ؊1 x؊1 1 ii) The domain of f(x) ؍ is x 1. x؊1 Also, 2 ؊ x 0 x 2 So, the domain of f(f(x)) is x 1 and x 2. The variable x is restricted because the denominator of a fraction can never be 0. 7. For each function below i) Determine possible functions f and g so that y ϭ f (g(x)). ii) Determine possible functions f, g, and h so that y ϭ f (g(h(x))). a) y ϭ x2 Ϫ 6x + 5 b) y ϭ Ϫ3x2 Ϫ 30x Ϫ 40 Sample solution: Sample solution: y ؍x2 ؊ 6x ؉ 5 y 3؊ ؍x2 ؊ 30x ؊ 40 y ( ؍x2 ؊ 6x ؉ 9) ؊ 4 y (3؊ ؍x2 ؉ 10x ؉ 25) ؉ 75 ؊ 40 y ( ؍x ؊ 3)2 ؊ 4 y (3؊ ؍x ؉ 5)2 ؉ 35 Let f(g(x)) ( ؍x ؊ 3)2 ؊ 4 Let f(g(x)) (3؊ ؍x ؉ 5)2 ؉ 35 i) Replace x ؊ 3 with x. i) Replace x ؉ 5 with x. Then, g(x) ؍x ؊ 3 and Then, g(x) ؍x ؉ 5 and f(x) ؍x2 ؊ 4 f(x) 3؊ ؍x2 ؉ 35 ii) Replace x ؊ 3 with x. ii) Replace x ؉ 5 with x. Then, h(x) ؍x ؊ 3, g(x) ؍x2, Then, h(x) ؍x ؉ 5, g(x) ؍x2, and f(x) ؍x ؊ 4 and f(x) 3؊ ؍x ؉ 35 26 4.4 Determining Restrictions on Composite Functions—Solutions DO NOT COPY. ©P
5.
05_ch04b_pre-calculas12_wncp_solution.qxd
5/18/12 7:03 PM Page 27 Home Quit √ √ c) y ϭ (x - 2)2 + 3 d) y = x2 + 4x + 3 Sample solution: √ Sample solution: √2 Let f(g(x)) ( ؍x ؊ 2)2 ؉ 3 y؍ i) Replace x ؊ 2 with x. √x 2؉ 4x ؉ 3 y ( ؍x ؉ 4x ؉ 4) ؊ 1 Then, g(x) ؍x ؊ 2 and √ √ y؍ (x ؉ 2)2 ؊ 1 f(x) ؍x2 ؉ 3 √ Let f(g(x)) ؍ (x ؉ 2)2 ؊ 1 ii) Replace x ؊ 2 with x. i) Replace x ؉ 2 with x. Then, h(x) ؍x ؊ 2, Then, g(x) ؍x ؉ 2 and √ g(x) ؍x2, and √ f(x) ؍x2 ؊ 1 f(x) ؍x ؉ 3 ii) Replace x ؉ 2 with x. Then, h(x) ؍x ؉ 2, g(x) ؍x2, √ and f(x) ؍x ؊ 1 8. Create composite functions using either or both functions in each pair of functions below. In each case, how many different composite functions could you create? Justify your answer. 1 a) f(x) ϭ ƒ x ƒ and g(x) ϭ x f (f(x)) ͦͦ ؍xͦͦ , which simplifies to f (f(x)) ͦ ؍xͦ f (g(x)) ` ؍x ` , which simplifies to f (g(x)) ؍ 1 1 ͦxͦ 1 g (f(x)) ؍ ͦxͦ 1 g (g(x)) , ؍which simplifies to g(g(x)) ؍x, x 0 1 x There are only 3 different composite functions, because f (g(x)) ؍g (f(x)). √ b) f(x) ϭ x and g(x) ϭ ƒ x ƒ √√ f (f(x)) ؍ x √ f (g(x)) ͦ ؍xͦ √ √ g(f(x)) ͦ ؍xͦ , which simplifies to g (f(x)) ؍x g(g(x)) ͦͦ ؍xͦͦ , which simplifies to g (g(x)) ͦ ؍xͦ There are 4 different composite functions. 1 c) f(x) ϭ x3 and g(x) ϭ x f (f(x)) ( ؍x3)3, which simplifies to f(f(x)) ؍x9 3 f (g(x)) ؍axb , which simplifies to f (g(x)) 3 ؍ 1 1 x 1 g(f(x)) ؍ x3 1 g(g(x)) , ؍which simplifies to g(g(x)) ؍x, x 0 1 x There are only 3 different composite functions, because f (g(x)) ؍g(f(x)). ©P DO NOT COPY. 4.4 Determining Restrictions on Composite Functions—Solutions 27
6.
05_ch04b_pre-calculas12_wncp_solution.qxd
5/18/12 7:03 PM Page 28 Home Quit x 9. Given the function y ϭ √ , determine possible functions: x - 3 f(x) a) f and g so that y ϭ g(x) Sample solution: √ f(x) ؍x and g(x) ؍x ؊ 3 f(x) b) f, g, and h so that y ϭ g1h(x)2 Sample solution: Replace x ؊ 3 with x. √ Let h(x) ؍x ؊ 3, then g(x) ؍x, and f(x) ؍x. c) f and g so that y ϭ f (g(x)) Sample solution: When g(x) replaces x in f(x), the numerator must be x and the denominator √ x؉3 must be x ؊ 3. So, g(x) ؍x ؊ 3 and f(x) √ ؍ x √ 2 10. Given the functions f(x) ϭ x, g(x) ϭ x2 Ϫ x ϩ 6, and k(x) ϭ x , write an explicit equation for each combination. a) h(x) ϭ f (g(x)) ϩ k(x) b) h(x) ϭ g( f(x)) Ϫ f (g(x)) For f(g(x)), replace x in For g(f(x)), replace x in √ √ f(x) ؍x with x2 ؊ x ؉ 6. g(x) ؍x2 ؊ x ؉ 6 with x . √ √ 2 √ Then, f(g(x)) ؍x2 ؊ x ؉ 6 Then, g(f(x)) ( ؍x) ؊ x ؉ 6 √ √ So, h(x) ؍ x2 ؊ x ؉ 6 ؉ x , 2 Or, g(f(x)) ؍x ؊ x ؉ 6, x » 0 √ So, h(x) ؍x ؊ x ؉ 6 ؊ x 0 √ x2 ؊ x ؉ 6 , x »0 c) h(x) ϭ k(g(x)) ϩ k( f(x)) d) h(x) ϭ f (g(x)) и k(x) For k(g(x)), replace x in From part √ a, 2 f (g(x)) ؍x2 ؊ x ؉ 6 k(x) ؍x with x 2 ؊ x ؉ 6. √ x2 ؊ x ؉ 6 # axb , x 2 2 Then, k(g(x)) 2 ؍ So, h(x) ؍ 0 x ؊x؉6 For k(f(x)), replace x in 2 √ k(x) ؍x with f(x) ؍x 2 Then, k(f(x)) , √ ؍x>0 x 2 2 So, h(x) 2 ؍ ؉ √ , x>0 x ؊x؉6 x 28 4.4 Determining Restrictions on Composite Functions—Solutions DO NOT COPY. ©P
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05_ch04b_pre-calculas12_wncp_solution.qxd
5/18/12 7:03 PM Page 29 Home Quit √ 11. Given the function y ϭ (x2 - 9) x + 2, determine possible functions in each case: a) functions f and g so that y ϭ f(x) и g(x) Sample solution: √ f(x) ؍x2 ؊ 9 and g(x) ؍x ؉ 2 b) functions f, g, and h so that y ϭ f(x) и g(h(x)) Sample solution: f(x) ؍x2 ؊ 9 √ For g(h(x)), let h(x) ؍x ؉ 2, then g(x) ؍x c) functions f, g, h, and k so that y ϭ f(x) и k(x) и g(h(x)) Sample solution: √ From part b, for g(h(x)), let h(x) ؍x ؉ 2, then g(x) ؍x Factor: x2 ؊ 9 ( ؍x ؉ 3)(x ؊ 3) Then, f(x) ؍x ؉ 3 and k(x) ؍x ؊ 3 12. Is there a function f(x) such that each relationship is true? Justify your answer. a) f ( f(x)) ϭ f(x) b) f ( f(x)) ϭ f(x) ϩ f(x) Yes, when f(x) ؍x, then Yes, when f(x) 2 ؍x, then f(f(x)) 4 ؍x f(f(x)) ؍x and f(x) ؉ f(x) 2 ؍x ؉ 2x, or 4x C 1 13. Given f(x) ϭ x - 2, g(x) is a quadratic function, and h(x) ϭ f (g(x)), determine an explicit equation for g(x) for each situation below. Explain your strategies. a) The domain of h(x) is x H .ޒ Sample solution: The denominator of h(x) must never be 0. 1 When g(x) ؍x2 ؉ 3, then f(g(x)) ؍ , which simplifies to x2 ؉ 3 ؊ 2 1 f(g(x)) ؍ . x2 ؉ 1 b) The domain of h(x) is x Z a and x Z b, where a and b are real numbers. Sample solution: There must be exactly two values of x that make the denominator of h(x) equal to 0. When g(x) ؍x2 ؉ 1, then 1 1 f(g(x)) ؍ , which simplifies to f(g(x)) 2 ؍ . x2 ؉ 1 ؊ 2 x ؊1 So, a 1 ؍and b 1؊ ؍ ©P DO NOT COPY. 4.4 Determining Restrictions on Composite Functions—Solutions 29
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05_ch04b_pre-calculas12_wncp_solution.qxd
5/18/12 7:03 PM Page 30 Home Quit c) The domain of h(x) is x Z c, where c is a real number. Sample solution: There must be exactly one value of x that makes the denominator of h(x) equal to 0. When g(x) ؍x2 ؉ 2, then 1 1 f(g(x)) ؍ , which simplifies to f(g(x)) . 2 ؍So, c 0 ؍ x2 ؉ 2 ؊ 2 x 1 - x 14. Use f(x) ϭ . 1 + x a) Determine an explicit equation for f ( f(x)), then state the domain of the function. 1؊x 1؊x In f(x) ؍ , replace x with 1؉x 1؉x 1؊x 1؊ 1؉x f(f(x)) ؍ 1؊x 1؉ 1؉x 1 ؉ x ؊ (1 ؊ x) 1؉x ؍ 1 ؉ x ؉ (1 ؊ x) 1؉x ؍x, x ؊1 The domain of the function is: x ؊1 b) What is the inverse of f(x)? Explain. Since f(f(x)) ؍x, x ؊1, then f(x) is its own inverse. 1؊x So, the inverse of f(x) is f؊1(x) ؍ . 1؉x 30 4.4 Determining Restrictions on Composite Functions—Solutions DO NOT COPY. ©P
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