2.9 Prelim Answers - g ( x ) < 0 when 1...

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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 . ² 3( x 2 - 2 x +1) e x 3 - 3 x 2 +3 x dx = ² e u du = e u = e x 3 - 3 x 2 +3 x c. Let u = x 5 + 27, du =5 x 4 dx ² 5 x 13 +27 x 8 dx = ² 5 x 4 x 5 +27 dx = ² udu = 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 +4ln | 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 inFection point is at 2. 0= f (2) = 8 - 6 · 4+9 · 2+ c =26 - 24 + c ,so c = - 2. 3. g ( x )= x - 5ln 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) - (8 / 64) >
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Unformatted text preview: g ( x ) &lt; 0 when 1 &lt; x &lt; 4 (d) g ( x ) &gt; when 5 /x 2 &gt; 8 /x 3 or x &gt; 8 / 5. 4. Minimize P = x + y + 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 x 2 + 16 x 2 -1 / 2 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 &lt; 2, P ( x ) &lt; 0 since (1-4 /x 2 ) &lt; 0 and 2 x-32 x 3 &lt; 0. When x &gt; 2, P ( x ) &gt; 0 since (1-4 /x 2 ) &gt; 0 and 2 x-32 x 3 &gt; 0....
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This note was uploaded on 11/14/2007 for the course MATH 1106 taught by Professor Durrett during the Spring '07 term at Cornell University (Engineering School).

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