5_1 - Chapter 5 Exponential and Logarithmic Functions 5.1 1...

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Chapter 5 623 Exponential and Logarithmic Functions 5.1 Composite Functions 1. f 3 ( 29 = - 4 3 ( 29 2 + 5 3 ( 29 = - 4 9 ( 29 + 15 = - 36 + 15 = - 11 2. f 3 x ( 29 = 4 - 2 3 x ( 29 2 = 4 - 2 9 x 2 ( 29 = 4 - 18 x 2 3. f ( x ) = x 2 - 1 x 2 - 4 4. g o f ( 29 ( x ) or f o g ( 29 ( x ) x 2 - 4 0 x + 2 ( 29 x - 2 ( 29 0 x ≠ - 2, x 2 Domain: x x ≠ - 2, x 2 { } . 5. False 6. False 7. (a) f o g ( 29 1 ( 29 = f g 1 ( 29 ( 29 = f 0 ( 29 = - 1 (b) f o g ( 29 - 1 ( 29 = f g - 1 ( 29 ( 29 = f 0 ( 29 = - 1 (c) g o f ( 29 - 1 ( 29 = g f - 1 ( 29 ( 29 = g - 3 ( 29 = 8 (d) g o f ( 29 0 ( 29 = g f 0 ( 29 ( 29 = g - 1 ( 29 = 0 (e) g o g ( 29 - 2 ( 29 = g g - 2 ( 29 ( 29 = g 3 ( 29 = 8 (f) f o f ( 29 - 1 ( 29 = f f - 1 ( 29 ( 29 = f - 3 ( 29 = - 7 8. (a) f o g ( 29 1 ( 29 = f g 1 ( 29 ( 29 = f 0 ( 29 = 5 (b) f o g ( 29 2 ( 29 = f g 2 ( 29 ( 29 = f - 3 ( 29 = 11 (c) g o f ( 29 2 ( 29 = g f 2 ( 29 ( 29 = g 1 ( 29 = 0 (d) g o f ( 29 3 ( 29 = g f 3 ( 29 ( 29 = g - 1 ( 29 = 0 (e) g o g ( 29 1 ( 29 = g g 1 ( 29 ( 29 = g 0 ( 29 = 1 (f) f o f ( 29 3 ( 29 = f f 3 ( 29 ( 29 = f - 1 ( 29 = 7
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Chapter 5 Exponential and Logarithmic Functions 624 9. f ( x ) = 2 x g ( x ) = 3 x 2 + 1 (a) ( f o g )(4) = f ( g (4)) = f 3(4) 2 + 1 ( 29 = f (49) = 2(49) = 98 (b) ( g o f )(2) = g ( f (2)) = g (2 2) = g (4) = 3(4) 2 + 1 = 48 + 1 = 49 (c) ( f o f )(1) = f ( f (1)) = f (2(1)) = f (2) = 2(2) = 4 (d) ( g o g )(0) = g ( g (0)) = g 3(0) 2 + 1 ( 29 = g (1) = 3(1) 2 + 1 = 4 10. f ( x ) = 3 x + 2 g ( x ) = 2 x 2 - 1 (a) ( f o g )(4) = f ( g (4)) = f 2(4) 2 - 1 ( 29 = f (31) = 3(31) + 2 = 95 (b) ( g o f )(2) = g ( f (2)) = g (3(2) + 2) = g (8) = 2(8) 2 - 1 = 128 - 1 = 127 (c) ( f o f = f ( f = f + 2 ( 29 = f (5) = 3(5) + 2 = 17 (d) ( g o g )(0) = g ( g (0)) = g 2(0) 2 - 1 ( 29 = g ( - 1) = 2( - 1) 2 - 1 = 1 11. f ( x ) = 4 x 2 - 3 g ( x ) = 3 - 1 2 x 2 (a) ( f o g )(4) = f ( g (4)) = f 3 - 1 2 (4) 2  = f ( - 5) = 4( - 5) 2 - 3 = 97 (b) ( g o f )(2) = g ( f (2)) = g (4(2) 2 - 3) = g (13) = 3 - 1 2 (13) 2 = 3 - 169 2 = - 163 2 (c) ( f o f = f ( f = f (4(1) 2 - 3) = f = 4(1) 2 - 3 = 1 (d) ( g o g )(0) = g ( g (0)) = g 3 - 1 2 (0) 2  = g (3) = 3 - 1 2 (3) 2 = 3 - 9 2 = - 3 2 12. f ( x ) = 2 x 2 g ( x ) = 1 - 3 x 2 (a) ( f o g )(4) = f ( g (4)) = f 1 - 3(4) 2 ( 29 = f ( - 47) = 2( - 47) 2 = 4418 (b) ( g o f )(2) = g ( f (2)) = g (2(2) 2 ) = g (8) = 1 - 3(8) 2 = 1 - 192 = - 191 (c) ( f o f = f ( f = f 2(1) 2 ( 29 = f (2) = 2(2) 2 = 8 (d) ( g o g )(0) = g ( g (0)) = g 1 - 3(0) 2 ( 29 = g = 1 - 2 = 1 - 3 = - 2 13. f ( x ) = x g ( x ) = 2 x (a) ( f o g )(4) = f ( g (4)) = f 2(4) ( 29 = f (8) = 8 = 2 2 (b) ( g o f )(2) = g ( f (2)) = g 2 ( 29 = 2 2 (c) ( f o f = f ( f = f 1 ( 29 = f = 1 = 1 (d) ( g o g )(0) = g ( g (0)) = g 2(0) ( 29 = g (0) = 2(0) = 0 14. f ( x ) = x + 1 g ( x ) = 3 x (a) ( f o g )(4) = f ( g (4)) = f 3(4) ( 29 = f (12) = 12 + 1 = 13 (b) ( g o f )(2) = g ( f (2)) = g 2 + 1 ( 29 = g 3 ( 29 = 3 3 (c) ( f o f = f ( f = f 1 + 1 ( 29 = f 2 ( 29 = 2 + 1 (d) ( g o g )(0) = g ( g (0)) = g 3(0) ( 29 = g (0) = 3(0) = 0
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Section 5.1 Composite Functions 625 15. f ( x ) = x g ( x ) = 1 x 2 + 1 (a) ( f o g )(4) = f ( g (4)) = f 1 4 2 + 1  = f 1 17  = 1 17 = 1 17 (b) ( g o f )(2) = g ( f (2)) = g 2 ( 29 = g (2) = 1 2 2 + 1 = 1 5 (c) ( f o f )(1) = f ( f (1)) = f 1 ( 29 = f (1) = 1 = 1 (d) ( g o g )(0) = g ( g (0)) = g 1 0 2 + 1  = g = 1 1 2 + 1 = 1 2 16. f ( x ) = x - 2 g ( x ) = 3 x 2 + 2 (a) ( f o g )(4) = f ( g (4)) = f 3 4 2 + 2  = f 3 18  = f 1 6  = 1 6 - 2 = - 11 6 = 11 6 (b) ( g o f )(2) = g ( f (2)) = g 2 - 2 ( 29 = g (0) = 3 0 2 + 2 = 3 2 (c) ( f o f = f ( f = f 1 - 2 ( 29 = f = 1 - 2 = 1 (d) ( g o g )(0) = g ( g (0)) = g 3 0 2 + 2  = g 3 2  = 3 3 2 2 + 2 = 3 17 4 = 12 17 17. f ( x ) = 3 x + 1 g ( x ) = x 3 (a) ( f o g )(4) = f ( g (4)) = f 4 3 ( 29 = 3 4 3 + 1 (b) ( g o f )(2) = g ( f (2)) = g 3 2 + 1  = g 3 3  = 1 3 = 1 (c) ( f o f = f ( f = f 3 1 + 1  = f 3 2  = 3 3 2 + 1 = 3 5 2 = 6 5 (d) ( g o g )(0) = g ( g (0)) = g 0 ( 29 = g (0) = 0 = 0 18. f ( x ) = x 3/2 g ( x ) = 2 x + 1 (a) ( f o g )(4) = f ( g (4)) = f 2 4 + 1  = f 2 5  = 2 5 = 2 5 3 = 8 125 = 2 2 5 5 5 5 = 2 10 25 (b) ( g o f )(2) = g ( f (2)) = g 2 ( 29 = g 2 2 ( 29 = 2 2 2 + 1 (c) ( f o f = f ( f = f 1 ( 29 = f 1 ( 29 = 1 = 1 (d) ( g o g )(0) = g ( g (0)) = g 2 0 + 1  = g (2) = 2 2 + 1 = 2 3
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Chapter 5 Exponential and Logarithmic Functions 626 19. The domain of g is
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5_1 - Chapter 5 Exponential and Logarithmic Functions 5.1 1...

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