# 2.9 Prelim Answers - Math 106 Prelim 2 Answers 1 a x3 3x1/2...

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Math 106 Prelim 2 Answers March 29, 2007 R. Durrett 1. a. x 3 + 3 x 1 / 2 + x - 3 dx = x 4 4 + 3 x 3 / 2 3 / 2 + x - 2 - 2 b. Let u = x 3 - 3 x 2 + 3 x , du = 3 x 2 - 6 x + 3 dx . 3( x 2 - 2 x + 1) e x 3 - 3 x 2 +3 x dx = e u du = e u = e x 3 - 3 x 2 +3 x c. Let u = x 5 + 27, du = 5 x 4 dx 5 x 13 + 27 x 8 dx = 5 x 4 x 5 + 27 dx = u du = u 3 / 2 3 / 2 = 2 3 x 5 + 27 d. x 2 x - 2 dx . Let u = x - 2, du = dx , and hence x = u + 2 = ( u +2) 2 u du = u + 4 + 4 u du = u 2 2 + 4 u + 4 ln | u | = ( x - 2) 2 2 + 4( x - 2) + 4 ln | x - 2 | 2. f ( x ) = x 3 + ax 2 + bx + c . f ( x ) = 3 x 2 + 2 ax + b . If 1 and 3 are to be critical points we must have 3 x 2 + 2 ax + b = 3( x - 1)( x - 3) = 3( x 2 - 4 x + 3) and hence a = - 6, b = 9. f ( x ) = 3(2 x - 4) so the inflection point is at 2. 0 = f (2) = 8 - 6 · 4 + 9 · 2 + c = 26 - 24 + c , so c = - 2. 3. g ( x ) = x - 5 ln x - 4 x . g ( x ) = 1 - 5 x + 4 x 2 = 1 x 2 ( x 2 - 5 x + 4) = 1 x 2 ( x - 1)( x - 4) g ( x ) = 5 x 2 - 8 x 3 (a) Critical points at 1 and 4. (b) g (1) = 5 - 8 < 0 so 1 is a local maximum; g (4) = (5 / 16)
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Unformatted text preview: g ± ( x ) < 0 when 1 < x < 4 (d) g ±± ( x ) > when 5 /x 2 > 8 /x 3 or x > 8 / 5. 4. Minimize P = x + y + ³ x 2 + y 2 subject to xy = 4 P ( x ) = x + 4 x + ´ x 2 + 16 x 2 P ± ( x ) = 1-4 x 2 + 1 2 µ x 2 + 16 x 2 ¶-1 / 2 µ 2 x-32 x 3 ¶ P ± (2) = 0 since when x = 2, (1-4 /x 2 ) = 0 and 2 x-32 x 3 = 0. When x < 2, P ± ( x ) < 0 since (1-4 /x 2 ) < 0 and 2 x-32 x 3 < 0. When x > 2, P ± ( x ) > 0 since (1-4 /x 2 ) > 0 and 2 x-32 x 3 > 0....
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