Solutions Test 3A

# Solutions Test 3A - Solutions to Test 3 Version A Math 1501...

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Solutions to Test 3, Version A Math 1501, Fall 02. WG * November 8, 2002 Exercise 1. (a) We have that f 0 ( x )= - 2 - 2 x if - 2 <x< 0 - 1i f 0 2 f 2 3 ( x - 2) 2 if 3 4 . Since f 0 - (0) = - 2and f 0 + (0) = - 1 we deduce that f 0 (0) does not exist. Similarly, f 0 - (2) = - 1and f 0 + (2) = 1 and so f 0 (2) does not exist. We have that f 0 - (3) = 1 = f 0 + (3) and so, f 0 (3) = 1 . The only point where f 0 vanishes is - 1 . Therefore, the critical points of f are - 1 , 0 , 2 . (b) and (c). Since f 0 is positive on ( - 2 , - 1), negative on ( - 1 , 0) we conclude that f has a local maximum at - 1 . Also, f is negative on (0 , 2) and positive on (2 , 3) . This proves that f has a local minimum at 2 . Exercise 2. (a) We have that f 0 ( x - 4sin x cos x - 2 x and so, f ”( x - 4[cos 2 x - sin 2 x ] - 2= - 6+8sin 2 x =8(s in x - 3 2 )(sin x + 3 2 ) . Note that sin x + 3 2 3 2 on [0 ] and so, the sign of f ” depends only on the sign of sin x - 3 2 . (b) f ”( x ) = 0 if and only if sin x = 3 2 . This means x = π 3 or x = 2 π 3 . Also f ”( x ) < 0 if and only if sin x< 3 2 , which means that x (0 , π 3 ) ( 2 π 3 ) . Hence f ” changes sign at * School of Mathematics, Georgia Institute of technology. 1

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x = π 3 and x = 2 π 3 . These two points are points of inﬂection. Exercise 3. We have that f 0 ( x )=6 / 5( x - 2 5 - x 1 5 )andso , lim x 0 f 0 ( x / 5 lim x 0 x - 2 5 =+ . This proves that the graph of
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## This note was uploaded on 02/07/2011 for the course MATH 1501 taught by Professor N/a during the Fall '08 term at Georgia Tech.

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Solutions Test 3A - Solutions to Test 3 Version A Math 1501...

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