# Cos x x csc2 x cot x1 x csc2 x cot x cot x because

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Unformatted text preview: = 3. The function is y = 3x2 − x − 2. dy 58. Let P (x0 , y0 ) be the point where y = x2 + k is tangent to y = 2x. The slope of the curve is = 2x dx and the slope of the line is 2 thus at P , 2x0 = 2 so x0 = 1. But P is on the line, so y0 = 2x0 = 2. Because P is also on the curve we get y0 = x2 + k so k = y0 − x2 = 2 − (1)2 = 1. 0 0 59. The points (−1, 1) and (2, 4) are on the secant line so its slope is (4 − 1)/(2 + 1) = 1. The slope of the tangent line to y = x2 is y = 2x so 2x = 1, x = 1/2. 60. The √ points (1, 1) and (4, 2) are on the secant√ so its slope is 1/3. The slope of the tangent line to line √ √ y = x is y = 1/(2 x) so 1/(2 x) = 1/3, 2 x = 3, x = 9/4. 61. y = −2x, so at any point (x0 , y0 ) on y = 1 − x2 the tangent line is y − y0 = −2x0 (x − x0 ), or to y = −2x0 x + x2 + 1. The point (2, 0) is√ be on the line, so 0 = −4x0 + x2 + 1, x2 − 4x0 + 1 = 0. Use 0 0 0 √ 4 ± 16 − 4 = 2 ± 3. the quadratic formula to get x0 = 2 62. Let P1 (x1 , ax2 ) and P2 (x2 , ax2 ) be the points of tangency. y = 2ax so the tangent lines at P1 and P2 1 2 are y − ax2 = 2ax1 (x − x1 ) and y − ax2 = 2ax2 (x − x2 ). Solve for x to get x = 1 (x1 + x2 ) which is the 1 2 2 x-coordinate of a point on the vertical line halfway between P1 and P2 . 63. y = 3ax2 + b; the tangent line at x = x0 is y − y0 = (3ax2 + b)(x − x0 ) where y0 = ax3 + bx0 . Solve 0 0 with y = ax3 + bx to get (ax3 + bx) − (ax3 + bx0 ) = (3ax2 + b)(x − x0 ) 0 0 ax3 + bx − ax3 − bx0 = 3ax2 x − 3ax3 + bx − bx0 0 0 0 x3 − 3x2 x + 2x3 = 0 0 0 (x − x0 )(x2 + xx0 − 2x2 ) = 0 0 (x − x0 )2 (x + 2x0 ) = 0, so x = −2x0 . 64. Let (x0 , y0 ) be the point of tangency. Refer to the solution to Exercise 65 to see that the endpoints of the line segment are at (2x0 , 0) and (0, 2y0 ), so (x0 , y0 ) is the midpoint of the segment. 77 65. Chapter 3 1 1 x 2 ; the tangent line at x = x0 is y − y0 = − 2 (x − x0 ), or y = − 2 + . The tangent line x2 x0 x0 x0 1 crosses the x-axis at 2x0 , the y -axis at 2/x0 , so that the area of the triangle is (2/x0 )(2x0 ) = 2. 2 y =− 66. f (x) = 3ax2 + 2bx + c; there is a horizontal tangent where f (x) = 0. Use the quadratic formula on √ 3ax2 + 2bx + c = 0 to get x = (−b ± b2 − 3ac)/(3a) which gives two real solutions, one real solution, or none if (b) b2 − 3ac = 0 (c) b2 − 3ac &lt; 0 (a) b2 − 3ac &gt; 0 67. F = GmM r−2 , dF 2GmM = −2GmM r−3 = − dr r3 68. dR/dT = 0.04124 − 3.558 × 10−5 T which decreases as T increases from 0 to 700. When T = 0, dR/dT = 0.04124 Ω/◦ C; when T = 700, dR/dT = 0.01633 Ω/◦ C. The resistance is most sensitive to temperature changes at T = 0◦ C, least sensitive at T = 700◦ C. 69. f (x) = 1 + 1/x2 &gt; 0 for all x 6 -6 6 -6 70. x2 − 4 ; (x2 + 4)2 f (x) &gt; 0 when x2 &lt; 4, i.e. on −2 &lt; x &lt; 2 1.5 f (x) = −5 -5 5 -1.5 71. (f · g · h) = [(f · g ) · h] = (f · g )h + h(f · g ) = (f · g )h + h[f g + f g ] = f gh + f g h + f g h 72. (f1 f2 · · · fn ) = (f1 f2 · · · fn ) + (f1 f2 · · · fn ) + · · · + (f1 f2 · · · fn ) 73. (a) 2(1 + x−1 )(x−3 + 7) + (2x + 1)(−x−2 )(x−3 + 7) + (2x + 1)(1 + x−1 )(−3x−4 ) (b) (x7 + 2x − 3)3 = (x7 + 2x − 3)(x7 + 2x − 3)(x7 + 2x − 3) so d7 (x + 2x − 3)3 = (7x6 + 2)(x7 + 2x − 3)(x7 + 2x − 3) dx +(x7 + 2x − 3)(7x6 + 2)(x7 + 2x − 3) +(x7 + 2x − 3)(x7 + 2x − 3)(7x6 + 2) = 3(7x6 + 2)(x7 + 2x − 3)2 74. (a) −5x−6 (x2 + 2x)(4 − 3x)(2x9 + 1) + x−5 (2x + 2)(4 − 3x)(2x9 + 1) +x−5 (x2 + 2x)(−3)(2x9 + 1) + x−5 (x2 + 2x)(4 − 3x)(18x8 ) Exercise Set 3.3 (b) 75. 78 (x2 + 1)50 = (x2 + 1)(x2 + 1) · · · (x2 + 1), where (x2 + 1) occurs 50 times so d2 (x + 1)50 = [(2x)(x2 + 1) · · · (x2 + 1)] + [(x2 + 1)(2x) · · · (x2 + 1)] dx + · · · + [(x2 + 1)(x2 + 1) · · · (2x)] = 2x(x2 + 1)49 + 2x(x2 + 1)49 + · · · + 2x(x2 + 1)49 = 100x(x2 + 1)49 because 2x(x2 + 1)49 occurs 50 times. f is continuous at 1 because lim− f (x) = lim+ f (x) = f (1), also lim− f (x) = lim− 2x = 2 and x→1 x→1 1 1 lim+ f (x) = lim+ √ = so f is not diﬀerentiable at 1. x→1 x→1 2 x 2 x→1 x→1 76. f is continuous at 1/2 because lim − f (x) = lim + f (x) = f (1/2), also x→1/2 x→1/2 2 lim f (x) = lim − 3x = 3/4 and lim + f (x) = lim + 3x/2 = 3/4 so f (1/2) = 3/4. x→1/2− x→1/2 x→1/2 x→1/2 77. If f is diﬀerentiable at x = 1, then f is continuous there; lim+ f (x) = lim− f (x) = f (1) = 3, a + b = 3; lim+ f (x) = a and x→1 x→1 x→1 lim− f (x) = 6 so a = 6 and b = 3 − 6 = −3. x→1 78. (a) lim f (x) = lim− 2x = 0 and lim+ f (x) = lim+ 2x = 0; f (0) does not exist because f is not x→0− x→0 x→0 x→0 continuous at x = 0. (b) lim− f (x) = lim+ f (x) = 0 and f is continuous at x = 0, so f (0) = 0; x→0 x→0 lim− f (x) = lim− (2) = 2 and lim+ f (x) = lim+ 6x = 0, so f (0) does not exist. x→0 79. (a) x→0 x→0 x→0 f (x) = 3x − 2 if x ≥ 2/3, f (x) = −3x + 2 if x &lt; 2/3 so f is diﬀerentiable everywhere except perhaps at 2/3. f is continuous at 2/3, also lim − f (x) = lim − (−3) = −3 and lim + f (x) = x→2/3 x→2/3 x→2/3 lim + (3) = 3 so f is not diﬀerentiable at x = 2/3. x→2/3 (b) f (x) = x2 − 4 if |x| ≥ 2, 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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