2.9 Practice Prelim Answers

2.9 Practice Prelim Answers - Solutions to Review Problems...

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Solutions to Review Problems for Prelim 2 Solutions to Maxima, minima, convexity, concavity 1. f ( x ) = x 3 + 3 x 2 - 9 x + 6, f ( x ) = 3 x 2 + 6 x - 9, f ( x ) = 6 x + 6. (a) 0 = 3( x 2 + 2 x - 3) = 3( x + 3)( x - 1) when x = 1, x = - 3. (b) f (1) = 12 > 0, local min., f ( - 3) = - 12 < 0, local max. (c) increasing when x < - 3, x > 1, (d) convex when x > - 1. 2. f ( x ) = x 3 / 3 - 5 x 2 / 2 + 4 x , f ( x ) = x 2 - 5 x + 4, f ( x ) = 2 x - 5. (a) 0 = x 2 - 5 x + 4 = ( x - 1)( x - 4) when x = 1, x = 4. (b) f (4) = 3 > 0, local min., f (1) = - 3 < 0, local max. (c) increasing when x < 1, x > 4, (d) convex when x > 5 / 2. 3. f ( x ) = x - 6 x + 2 ln x , f ( x ) = 1 - 3 x - 1 / 2 + 2 x - 1 , f ( x ) = (3 / 2) x - 3 / 2 - 2 x - 2 . (a) 0 = 1 - 3 x - 1 / 2 - 2 x - 1 = (1 - 2 x - 1 / 2 )(1 - x - 1 / 2 ) when x = 1, x = 2. (b) f (1) = 3 / 2 - 2 < 0, local max., f (4) = 3 / 16 - 2 / 16 > 0, local min. (c) increasing when 0 < x < 1, x > 2 (d) convex when x 1 / 2 > 4 / 3. 4. f ( x ) = x 8 / 5 + x - 2 / 5 , f ( x ) = (8 / 5) x 3 / 5 - (2 / 5) x - 7 / 5 , f ( x ) = (24 / 25) x - 2 / 5 +(14 / 25) x - 12 / 5 (a) 0 = (8 / 5) x 3 / 5 - (2 / 5) x - 7 / 5 when x 2 = 1 / 4 or x = 1 / 2 (b) f ( x ) > 0 for all x > 0 so 1/2 is a minimum. (c) increasing when x > 1 / 2, (d) convex for all x > 0. 5. f ( x ) = x ln x , f ( x ) = ln x + 1, f ( x ) = 1 /x . (a) 0 = ln x + 1 when x = 1 /e . (b) f (1 /e ) = e > 0, local min. (c) increasing when x > 1 /e , (d) convex for all x > 0.
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