3 108 1 dy dy 1 y2 2 0 so 2 2 dx y x dx x 15 16 2x

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Unformatted text preview: 0 on [x1 , x2 ], so f (x1 ) = f (x2 ) if x1 &lt; −1 &lt; x2 then f (x1 ) &gt; 1 &gt; f (x2 ) since f (x) &gt; 1 on (−∞, −1) and f (x) &lt; 1 on (−1, +∞) f (x) = 13. y = f −1 (x), x = f (y ) = y 5 , y = x1/5 = f −1 (x) 14. y = f −1 (x), x = f (y ) = 6y , y = 15. y = f −1 (x), x = f (y ) = 7y − 6, y = 16. y = f −1 (x), x = f (y ) = 1 x = f −1 (x) 6 1 (x + 6) = f −1 (x) 7 x+1 y+1 , xy − x = y + 1, (x − 1)y = x + 1, y = = f −1 (x) y−1 x−1 101 Chapter 4 17. y = f −1 (x), x = f (y ) = 3y 3 − 5, y = 18. y = f −1 (x), x = f (y ) = √ 5 19. y = f −1 (x), x = f (y ) = √ 3 20. y = f −1 (x), x = f (y ) = 21. y = f −1 (x), x = f (y ) = 3/y 2 , y = − 3/x = f −1 (x) 22. y = f −1 (x), x = f (y ) = 23. y = f −1 (x), x = f (y ) = 3 4y + 2, y = (x + 5)/3 = f −1 (x) 15 (x − 2) = f −1 (x) 4 2y − 1, y = (x3 + 1)/2 = f −1 (x) 5−x = f −1 (x) x 5 ,y= y2 + 1 2y, y ≤ 0 , y2 , y &gt; 0 x/2, x ≤ 0 √ x, x &gt; 0 y = f −1 (x) = 5/2 − y, y &lt; 2 , y = f −1 (x) = 1/y, y ≥ 2 5/2 − x, x &gt; 1/2 1/x, 0 &lt; x ≤ 1/2 24. y = p−1 (x), x = p(y ) = y 3 − 3y 2 + 3y − 1 = (y − 1)3 , y = x1/3 + 1 = p−1 (x) 25. y = f −1 (x), x = f (y ) = (y + 2)4 for y ≥ 0, y = f −1 (x) = x1/4 − 2 for x ≥ 16 26. y = f −1 (x), x = f (y ) = 27. √ y + 3 for y ≥ −3, y = f −1 (x) = x2 − 3 for x ≥ 0 √ y = f −1 (x), x = f (y ) = − 3 − 2y for y ≤ 3/2, y = f −1 (x) = (3 − x2 )/2 for x ≤ 0 28. y = f −1 (x), x = f (y ) = 3y 2 + 5y − 2 for y ≥ 0, 3y 2 + 5y − 2 − x = 0 for y ≥ 0, √ y = f −1 (x) = (−5 + 12x + 49)/6 for x ≥ −2 29. y = f −1 (x), x = f (y ) = y − 5y 2 for y ≥ 1, 5y 2 − y + x = 0 for y ≥ 1, √ y = f −1 (x) = (1 + 1 − 20x)/10 for x ≤ −4 30. (a) (b) 5 (F − 32) 9 how many degrees Celsius given the Fahrenheit temperature C= (c) 32. (a) y = f (x) = (6.214 × 10−4 )x (c) 31. C = −273.15◦ C is equivalent to F = −459.67◦ F, so the domain is F ≥ −459.67, the range is C ≥ −273.15 how many meters in y miles 104 y 6.214 f and f −1 are continuous so f (3) = lim f (x) = 7; then f −1 (7) = 3, and x→3 lim f x→7 33. (b) x = f −1 (y ) = (a) −1 (x) = f −1 lim x = f x→7 −1 (7) = 3 √ f (g (x)) = f ( x) √ = ( x)2 = x, x &gt; 1; g (f (x)) = g (x2 ) √ = x2 = x, x &gt; 1 (b) y y = f (x) y = g(x) x Exercise Set 4.1 (c) 34. 102 no, because f (g (x)) = x for every x in the domain of g is not satisﬁed (the domain of g is x ≥ 0) y = f −1 (x), x = f (y ) = ay 2 + by + c, ay 2 + by + c − x = 0, use the quadratic formula to get −b ± b2 − 4a(c − x) ; y= 2a −b + b2 − 4a(c − x) −b − b2 − 4a(c − x) (b) f −1 (x) = (a) f −1 (x) = 2a 2a 3−x 1 − x = 3 − 3x − 3 + x = x so f = f −1 35. (a) f (f (x)) = 3−x 1−x−3+x 1− 1−x (b) symmetric about the line y = x 3− 36. y = m(x − x0 ) is an equation of the line. The graph of the inverse of f (x) = m(x − x0 ) will be the reﬂection of this line about y = x. Solve y = m(x − x0 ) for x to get x = y/m + x0 = f −1 (y ) so y = f −1 (x) = x/m + x0 . 37. f (x) = x3 − 3x2 + 2x = x(x − 1)(x − 2) so f (0) = f (1) = f (2) = 0 thus f is not one-to-one. √ √ 6 ± 36 − 24 = 1 ± 3/3. f (x) &gt; 0 (f is increasing) (b) f (x) = 3x2 − 6x + 2, f (x) = 0 when x = 6√ √ √ if x &lt; 1 − 3/3,√ (x) &lt; 0 (f is decreasing) √ 1 − 3/3 &lt; x &lt; 1 + 3/3, so f (x) takes on values f if √ less than f (1 − 3/3) on both sides of 1 − 3/3 thus 1 − 3/3 is the largest value of k . 38. (a) (a) f (x) = x3 (x − 2) so f (0) = f (2) = 0 thus f is not one to one. (b) f (x) = 4x3 − 6x2 = 4x2 (x − 3/2), f (x) = 0 when x = 0 or 3/2; f is decreasing on (−∞, 3/2] and increasing on [3/2, +∞) so 3/2 is the smallest value of k . 39. if f −1 (x) = 1, then x = f (1) = 2(1)3 + 5(1) + 3 = 10 40. if f −1 (x) = 2, then x = f (2) = (2)3 /[(2)2 + 1] = 8/5 41. 42. 6 -2 10 -5 6 10 -2 43. -5 44. 3 0 3 0 6 0 6 0 45. f (f (x)) = x thus f = f −1 so the graph is symmetric about y = x. 46. (a) Suppose x1 = x2 where x1 and x2 are in the domain of g and g (x1 ), g (x2 ) are in the domain of f then g (x1 ) = g (x2 ) because g is one-to-one so f (g (x1 )) = f (g (x2 )) because f is one-to-one thus f ◦ g is one-to-one because (f ◦ g )(x1 ) = (f ◦ g )(x2 ) if x1 = x2 . 103 Chapter 4 (b) f , g , and f ◦ g all have inverses because they are all one-to-one. Let h = (f ◦ g )−1 then (f ◦ g )(h(x)) = f [g (h(x))] = x, apply f −1 to both sides to get g (h(x)) = f −1 (x), then apply g −1 to get h(x) = g −1 (f −1 (x)) = (g −1 ◦ f −1 )(x), so h = g −1 ◦ f −1 47. y x 48. Suppose that g and h are both inverses of f then f (g (x)) = x, h[f (g (x))] = h(x), but h[f (g (x))] = g (x) because h is an inverse of f so g (x) = h(x). 49. F (x) = 2f (2g (x))g (x) so F (3) = 2f (2g (3))g (3). By inspection f (1) = 3, so g (3) = f −1 (3) = 1 and g (3) = (f −1 ) (3) = 1/f (f −1 (3)) = 1/f (1) = 1/7 because f (x...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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