sol14-4 - SOLUTIONS CHAPTER 14.4 MATH 132 WI01 2 y =-2x2 6x...

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Unformatted text preview: SOLUTIONS CHAPTER 14.4 MATH 132 WI01 2. y = -2x2 + 6x + 12. Find first critical numbers, by means of finding zeroes of first derivative: y = -2 2x + 6 = -4x + 6 = 0 6 = 4x x = 6 = 1.5. To check 4 now whether it is or not maximum or minimum, we plug it in the second derivative y = -4: y (1.5) = -4 negative ( ) 1.5 is a relative maximum. 5. y = x3 - 27x + 1. y = 0 3x2 - 27 = 0 3x2 = 27 x2 = 9 x = 3. Plug these values in the second derivative y = 3 2x = 6x: y (-3) = 6 (-3) = -18 negative ( ) -3 is a relative maximum; y (3) = 6 3 = 18 positive ( ) 3 is a relative minimum. 7. y = -x3 + 3x2 + 1. y = 0 -3x2 + 3 2x = 0 3x(-x + 2) = 0 x = 0, -x + 2 = 0 x = 2. Plug these values in second derivative y = -3 2x + 3 2 = 6(-x + 1): y (0) = 6(-0 + 1) = 6 positive ( ) 0 is a relative minimum; y (2) = 6(-2+1) = 6(-1) = -6 negative ( ) 2 is a relative maximum. 1 11. y = 81x5 - 5x. y = 0 81 5x4 - 5 = 0 81 5x4 = 5 x4 = 81 x = 3 = 0.33. Plug these values in the second derivative y = 81 5 4x = 1620x3 : y (-0.33) = -60 negative ( ) -0.33 is a relative maximum; y (0.33) = 60 positive ( ) 0.33 is a relative minimum. 13. y = (x2 + 7x + 10)2 . y = 0 2(x2 + 7x + 10) (2x + 7) = 0 2 (x + 2)(x + 5)(2x + 7) = 0 x + 2 = 0, x = -2; x + 5 = 0, x = -5; 2x + 7 = 0, 2x = 7 -7, x = - 2 = -3.5. Plug these values in the second derivative y = [2(x + 2)(x + 5)(2x + 7)] = 2[(x2 + 7x + 10)(2x + 7)] = 2[(2x + 7)(2x + 7) + (x2 + 7x + 10) 2] = 2(4x2 + 28x + 49 + 2x2 + 14x + 20) = 2(6x2 + 42x + 69): y (-2) = 2(6(-2)2 +42(-2)+69) = 2(24-84+69) = 18 positive ( ) x=-2 is a relative minimum; y (-5) = 2(6(-5)2 +42(-5)+69) = 2(150-210+69) = 18 positive ( ) x=-5 is also a relative minimum; y (-3.5) = 2(6(-3.5)2 + 42(-3.5) + 69) = -9 negative ( ) x=-3.5 is a relative maximum. 1 3 Date: 02/06/2001. 1 ...
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