X 1 1 0 x x1 thus on 1 the function f starts negative

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: - 1 ) 2 (0, -1) 15. 16. x3 − 1 1 = ; x x 3 2x + 1 y= , x2 31 y = 0 when x = − ≈ −0.8; 2 3 2(x − 1) y= x3 y = x2 − 2x2 − 1 ; x2 2 y = 3; x 6 y =− 4 x y ≈(-0.8, 1.9) x (1, 0) y y= y=2 x Exercise Set 5.3 17. 156 x3 − 1 ; x3 + 1 6x2 ; y= 3 (x + 1)2 12x(1 − 2x3 ) y= (x3 + 1)3 y y= y=1 x = -1 18. 19. 8 ; 4 − x2 16x y= ; (4 − x2 )2 16(3x2 + 4) y= (4 − x2 )3 y y= x−1 ; x2 − 4 2 x − 2x + 4 y =− (x2 − 4)2 x 3 (1/ 2 , -1/3) (0, -1) x = -2 x=2 1 x y y= x = -2 4 x x=2 20. 4 4 −; x x2 4(x + 2) ; y= x3 8(x + 3) y =− x4 y =3− y (-3, 35 ) 9 (-2, 4) y=3 x 21. (x − 1)2 ; x2 2(x − 1) y= ; x3 2(3 − 2x) y= x4 y= y y=1 x (1, 0) () 3, 1 29 157 22. Chapter 5 3 1 −; x x3 3(1 − x2 ) y= ; x4 2 6(x − 2) y= x5 y y =2+ (1, 4) ( 2, 2 + 5 4 2) y=2 x (− (b) I (c) III 2, 2 – 5 4 2) (–1, 0) 23. (a) VI (d) V (e) IV (f ) II 24. (a) When n is even the function is defined only for x ≥ 0; as n increases the graph approaches the line y = 1 for x > 0. y x (b) When n is odd the graph is symmetric with respect to the origin; as n increases the graph approaches the line y = 1 for x > 0 and the line y = −1 for x < 0. y x 25. 26. √ x2 − 1; x y =√ ; 2−1 x 1 y =− 2 (x − 1)3/2 y y= x2 − 4; 2x ; y= 2 − 4)2/3 3(x 2(3x2 + 4) y =− 9(x2 − 4)5/3 y= 3 -1 x 1 y 3 3 (-2, 0) (2, 0) -2 2 (0, -2) x Exercise Set 5.3 27. 158 y y = 2x + 3x2/3 ; y = 2 + 2x−1/3 ; 2 y = − x−4/3 3 5 4 (0, 0) 28. y = 4x − 3x4/3 ; x y y = 4 − 4x1/3 ; 4 y = − x−2/3 3 3 (1, 1) x 1 29. 30. y y = x(3 − x)1/2 ; 3(2 − x) ; y= √ 2 3−x 3(x − 4) y= 4(3 − x)3/2 (2, 2) x y = x1/3 (4 − x); 4(1 − x) ; y= 3x2/3 4(x + 2) y =− 9x5/3 y 10 (1, 3) x 3 (-2, -6 2 ) -10 √ 8( x − 1) ; 31. y = x√ 4(2 − x) ; y= x √2 2(3 x − 8) y= x3 y 4 (4, 2) (64 , 15 ) 98 x 15 159 32. 33. Chapter 5 √ 1+ x √; 1− x 1 √; y= √ 2 x(1− x) √ 3 x−1 √ y = 3/2 2x (1 − x)3 y y= x=1 2 (1 , 2) 9 x -2 y = -1 y = x + sin x; y y = 1 + cos x, y = 0 when x = π + 2nπ ; y = − sin x; y = 0 when x = nπ c n = 0, ±1, ±2, . . . 34. c c y = x − cos x; x x y y = 1 + sin x; y = 0 when x = −π/2 + 2nπ ; c y = cos x; y = 0 when x = π/2 + nπ n = 0, ±1, ±2, . . . 35. y y = sin x + cos x; y = cos x − sin x; y = 0 when x = π/4 + nπ ; y = − sin x − cos x; 2 x o -o -2 y = 0 when x = 3π/4 + nπ 36. √ 3 cos x + sin x; √ y = − 3 sin x + cos x; y y= y = 0 when x = π/6 + nπ ; √ y = − 3 cos x − sin x; y = 0 when x = 2π/3 + nπ 2 o -2 x Exercise Set 5.3 37. 160 y = sin2 x, 0 ≤ x ≤ 2π ; y y = 2 sin x cos x = sin 2x; 1 y = 2 cos 2x c 38. y = x tan x, −π/2 < x < π/2; x o y y = x sec2 x + tan x; y = 0 when x = 0; y = 2 sec2 x(x tan x + 1), which is always positive for −π/2 < x < π/2 39. (a) x -6 6 lim xex = +∞, lim xex = 0 x→+∞ y x→−∞ (b) y = xex ; y = (x + 1)ex ; y = (x + 2)ex 1 -5 -3 -1 x (-2, -0.27) -1 (-1, -0.37) 40. (a) (b) lim xe−2x = 0, lim xe−2x = −∞ x→+∞ y x→−∞ y = xe−2x ; y = −2 x − 1 2 0.3 e−2x ; y = 4(x − 1)e−2x (0.5, 0.18) (1, 0.14) 0.1 x -3 -1 -0.1 1 3 -0.3 41. x2 x2 = 0, lim 2x = +∞ x→+∞ e2x x→−∞ e 2 2x (b) y = x /e = x2 e−2x ; y = 2x(1 − x)e−2x ; y = 2(2x2 − 4x + 1)e−2x ; 2 y = 0 if √x2 − 4x + 1 = 0, when √ 4 ± 16 − 8 x= = 1 ± 2/2 ≈ 0.29, 1.71 4 (a) lim y 0.3 (1, 0.14) (1.71, 0.10) (0, 0) 1 2 (0.29, 0.05) 42. (a) lim x2 e2x = +∞, lim x2 e2x = 0. x→+∞ x→−∞ 3 x 161 Chapter 5 (b) = x2 e2x ; = 2x(x + 1)e2x ; = 2(2x2 + 4x + 1)e2x ; = 0 if 2√ + 4x + 1 = 0, when x2 √ −4 ± 16 − 8 = −1 ± 2/2 ≈ −0.29, −1.71 x= 4 y y y y y 0.3 0.2 (-1, 0.14) (-1.71, 0.10) x -3 -2 -1 (0, 0) (-0.29, 0.05) 43. (a) lim f (x) = +∞, lim f (x) = −∞ x→+∞ x→−∞ x2 y (b) y = xe ; 2 y = (1 + 2x2 )ex ; 2 x2 y = 2x(3 + 2x )e no relative extrema, inflection point at (0, 0) 100 (0,0) -2 2 x -100 44. (a) lim f (x) = 1 x→±∞ . (b) f (x) = 2x−3 e−1/x so f (x) < 0 for x < 0 and f (x) > 0 for x > 0. By L’Hˆpital’s Rule limx→0 f (x) = 0, so (by the first o derivative test) f (x) has a minimum at x = 0. 2 f (x) = (−6x−4 + 4x−6 )e−1/x , so f (x) has points of inflection at x = ± 2/3 2 y 1 0.8 0.4 (− 2/3, e −3/2) -5 (0, 0) -10 45. (a) lim+ y = lim+ x ln x = lim+ x→0 x→0 lim y = +∞ x→0 ( ln x 1/x = 0; = lim+ 1/x x→0 −1/x2 2/3, e −3/2) 5 10 x y x→+∞ (b) y = x ln x, y = 1 + ln x, y = 1/x, y = 0 when x = e−1 x 1 (e-1, -e-1) 46. (a) ln x 1/x = lim+ = 0, x→0 −2/x3 1/x2 lim y = +∞ lim y = lim+ x→0+ y x→0 0.2 x→+∞ (b) y = x2 ln x, y = x(1 + 2 ln x), y = 3 + 2 ln x, y = 0 if x = e−1/2 , y = 0 if x = e−3/2 , lim+ y = 0 x→0 0.1 (e -3/2, - 3 e-3) 2 x 1 -0.1 -0.2 (e -1/2, - 1 e-1) 2 Exercise Set 5.3 47. 48. 162 ln x = −∞; x2 ln x 1/x =0 lim y = lim = lim 2 x→+∞ x→+∞ x x→+∞ 2x ln x 1 − 2 ln x (b) y = 2 , y = , x x3 6 ln x − 5 , y= x4 y = 0 if x = e1/2 , y = 0 if x = e5/6 (a) (a) lim y = lim+ x→0+ y x→0 0.4 0.3 0.2 0.1 x x→+∞ 1 -0.1 -0.2 -...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online