# 2 04 x 06 lim f x 0 lim f x 0 x0 x0 critical

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Unformatted text preview: x &gt; 3, f (3) does not exist; critical point x = 3, relative min of 6 at x = 3 35. f (x) = 2 cos 2x if sin 2x &gt; 0, f (x) = −2 cos 2x if sin 2x &lt; 0, f (x) does not exist when x = π/2, π, 3π/2; critical points x = π/4, 3π/4, 5π/4, 7π/4, π/2, π, 3π/2 relative min of 0 at x = π/2, π, 3π/2; relative max of 1 at x = π/4, 3π/4, 5π/4, 7π/4 1 0 0 36. √ f (x) = 3 + 2 cos x; critical points x = 5π/6, 7π/6 √ relative min of 7 3π/6 − 1 at x = 7π/6; relative max of √ 5 3π/6 + 1 at x = 5π/6 12 0 0 37. f (x) = − sin 2x; critical points x = π/2, π, 3π/2 relative min of 0 at x = π/2, 3π/2; relative max of 1 at x = π 0 f (x) = (2 cos x − 1)/(2 − cos x)2 ; critical points x = π/3, 5π/3 √ relative max of 3/3 at x = π/3, relative min of √ − 3/3 at x = 5π/3 o 1 0 38. o o 0.8 0 -0.8 o Exercise Set 5.2 39. f (x) = ln x + 1, f (x) = 1/x; f (1/e) = 0, f (1/e) &gt; 0; relative min of −1/e at x = 1/e 150 2.5 0 2.5 -0.5 40. ex − e−x = 0 when x = 0. By the ﬁrst derivative test (ex + e−x )2 f (x) &gt; 0 for x &lt; 0 and f (x) &lt; 0 for x &gt; 0; relative max of 1 at x=0 f (x) = −2 1 -2 2 0 41. f (x) = 2x(1 − x)e−2x = 0 at x = 0, 1. f (x) = (4x2 − 8x + 2)e−2x ; f (0) &gt; 0 and f (1) &lt; 0, so a relative min of 0 at x = 0 and a relative max of 1/e2 at x = 1. 0.14 -0.3 4 0 42. f (x) = 10/x − 1 = 0 at x = 10; f (x) = −10/x2 &lt; 0; relative max of 10(ln(10) − 1) ≈ 13.03 at x = 10 14 0 20 -4 43. Relative minima at x = −3.58, 3.33; relative max at x = 0.25 250 -5 5 -250 44. Relative min at x = −0.84; relative max at x = 0.84 1.2 -6 6 -1.2 151 Chapter 5 45. relative max at x = 0.255 46. relative max at x = 0.845 47. Relative min at x = −1.20 and a relative max at x = 1.80 y f''(x) 1 f'(x) -4 -2 2 2 x 4 4 -1 -2 48. Relative max at x = −0.78 and a relative min at x = 1.55 y f''(x) 5 -4 -2 f'(x) 49. (a) x -5 k k 2x3 − k Let f (x) = x2 + , then f (x) = 2x − 2 = . f has a relative extremum when 2x3 − k = 0, x x x2 so k = 2x3 = 2(3)3 = 54. k − x2 x , then f (x) = 2 . f has a relative extremum when k − x2 = 0, so x2 + k (x + k )2 k = x2 = 32 = 9. (b) Let f (x) = 50. (a) one relative maximum, located at x = n 0.3 0 14 0 (b) 51. f (x) = cxn−1 (−x + n)e−x = 0 at x = n. Since f (x) &gt; 0 for x &lt; n and f (x) &lt; 0 for x &gt; n it’s a maximum. (a) f (x) = −xf (x). Since f (x) is always positive, f (x) = 0 at x = 0, f (x) &gt; 0 for x &lt; 0 and f (x) &lt; 0 for x &gt; 0, so x = 0 is a maximum. (b) y 1 2π ( µ, µ 1 2π ) x Exercise Set 5.3 52. (a) 152 relative minima at x = ±0.6436, relative max at x = 0 (b) x = ±0.6436, 0 y 2 1.8 1.6 1.4 1.2 1 x -1.5 -1 -0.5 0.5 1 1.5 53. f (x) = 3ax2 + 2bx + c and f (x) has roots at x = 0, 1, so f (x) must be of the form f (x) = 3ax(x − 1); thus c = 0 and 2b = −3a, b = −3a/2. f (x) = 6ax + 2b = 6ax − 3a, so f (0) &gt; 0 and f (1) &lt; 0 provided a &lt; 0. Finally f (0) = d, so d = 0; and f (1) = a + b + c + d = a + b = −a/2 so a = −2. Thus f (x) = −2x3 + 3x2 . 54. (a) Because h and g have relative maxima at x0 , h(x) ≤ h(x0 ) for all x in I1 and g (x) ≤ g (x0 ) for all x in I2 , where I1 and I2 are open intervals containing x0 . If x is in I1 ∩ I2 then both inequalities are true and by addition so is h(x) + g (x) ≤ h(x0 ) + g (x0 ) which shows that h + g has a relative maximum at x0 . (b) By counterexample; both h(x) = −x2 and g (x) = −2x2 have relative maxima at x = 0 but h(x) − g (x) = x2 has a relative minimum at x = 0 so in general h − g does not necessarily have a relative maximum at x0 . 55. (a) (b) y ( x0 ) x f (x0 ) is not an extreme value. (c) y ( x0 ) y ( x y = x2 − 2x − 3; y y = 2(x − 1); y =2 x 1 -4 ) x f (x0 ) is a relative minimum. f (x0 ) is a relative maximum. EXERCISE SET 5.3 1. x0 (1, -4) 153 2. Chapter 5 y = 1 + x − x2 ; y y = −2(x − 1/2); y = −2 (1 , 5) 24 1 x 1 3. y = x3 − 3x + 1; y y = 3(x2 − 1); (-1, 3) y = 6x (0, 1) x (1, -1) 4. y = 2x3 − 3x2 + 12x + 9; y y = 6(x − x + 2); 2 y = 12(x − 1/2) ( 1 , 29 ) 22 10 x 1 5. y = x4 + 2x3 − 1; y 2 y = 4x (x + 3/2); y = 12x(x + 1) x (0, -1) (-1, -2) (- 3 , - 43 ) 2 16 6. y = x4 − 2x2 − 12; y y = 4x(x2 − 1); 10 y = 12(x2 − 1/3) 1 (1, -13) (-1, -13) ( - 1 3 ,- 113 9 x (0, -12) ) ( 13 , - 113 ) 9 Exercise Set 5.3 7. 154 y = x3 (3x2 − 5); y = 15x2 (x2 − 1); y (- 13 , 783 ) (-1, 2) y = 30x(2x − 1) 2 (0, 0) x (1, -2) ( 13 , - 78 3 ) 8. y = 3x3 (x + 4/3); y y = 12x2 (x + 1); y = 36x(x + 2/3) (0, 0) (-1, -1) 9. y = x(x − 1)3 ; x (- 2 , - 16 ) 3 27 y y = (4x − 1)(x − 1)2 ; y = 6(2x − 1)(x − 1) 0.1 x (1, 0) ( -0.1 ( 10. 1, 4 1 ,- 1 2 16 27 - 256 ) ) y = x4 (x + 5); y = 5x3 (x + 4); y (-4, 256) 300 2 y = 20x (x + 3) x (-3, 162) 11. y = 2x/(x − 3); (0, 0) y y = −6/(x − 3)2 ; y = 12/(x − 3)3 y=2 x x=3 155 12. Chapter 5 x ; x2 − 1 x2 + 1 ; y =− 2 (x − 1)2 2x(x2 + 3) y= (x2 − 1)3 y y= x = -1 1 x (0, 0) x=1 13. x2 ; −1 2x y =− 2 ; (x − 1)2 2 2(3x + 1) y= (x2 − 1)3 y= y x2 y=1 (0, 0) x = -1 14. x2 − 1 ; x2 + 1 4x y= 2 ; (x + 1)2 4(1 − 3x2 ) y= (x2 + 1)3 x x=1 y y= y=1 x (- 13 , - 1 ) 2 ( 13 ,...
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