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Unformatted text preview: Solutions to Review Problems for the Final Exam Solutions to Maxima, minima, convexity, concavity 1. f ( x ) = 2 3 x 3 x 2 4 x + 1, f ( x ) = 2 x 2 2 x 4, f ( x ) = 4 x 2 (a) f ( x ) = 0 when 0 = x 2 x 2 = ( x + 1)( x 2), i.e., x = 1, x = 2. (b) f ( 1) = 6 < 0, maximum; f (2) = 6 > 0 minimum. (c) increasing for x < 1, x > 2, (d) convex for x > 1 / 2. 2. f ( x ) = 2 x + 18 /x , f ( x ) = 2 18 /x 2 , f ( x ) = 36 /x 3 (a) f ( x ) = 0 when 2 = 18 /x 2 , i.e., x = 3, x = 3 (b) f ( 3) = 36 / 27 < 0, maximum; f (3) = 36 / 27 > 0, minimum (c) increasing for x < 3, x > 3; (d) convex for x > 3. f ( x ) = x x 2 +1 , f ( x ) = ( x 2 +1) x 2 x ( x 2 +1) 2 = 1 x 2 ( x 2 +1) 2 (a) f ( x ) = 0 when x 2 = 1, i.e., x = 1, x = 1; (c) increasing for 1 < x < 1; (b) f ( x ) = 2 x ( x 2 +1) 2 (1 x 2 ) 2( x 2 +1)(2 x ) ( x 2 +1) 4 f ( 1) = 2 / 2 2 > 0, minimum; f (1) = 2 / 2 2 < 0, maximum (d) f ( x ) = 2 x ( x 2 +1)(3 x 2 ) ( x 2 +1) 4 , convex for 3 < x < 0, x > 3 4. f ( x ) = 2 x 1 / 2 3 ln x + 4 x , f ( x ) = x 3 / 2 3 /x + 2 x 1 / 2 , (a) f ( x ) = x 1 / 2 ( x 1 3 x 1 / 2 + 2) = 0 when x 1 / 2 = 1 , 2 or x = 1 , x = 1 / 4. (b) f ( x ) = (3 / 2) x 5 / 2 + 3 x 2 x 3 / 2 , f (1) = (3 / 2) + 3 1 > 0 minimum f (1 / 4) = (3 / 2)(32) + 3 16 8 = 8 < 0 maximum (c) increasing for x < 1 / 4 and x > 1; (d) convex when x 3 x 1 / 2 + (3 / 2) < 0, that is when r 1 < x < r 2 where r i = (3 9 6) / 2 5. f ( x ) = x ( x 1) 4 , f ( x ) = ( x 1) 4 + 4 x ( x 1) 3 = (5 x 1)( x 1) 3 f ( x ) = 0 when x = 1 / 5, x = 1. f is increasing for x < 1 / 5, decreasing on (1 / 5...
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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).
 Spring '07
 DURRETT
 Calculus

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