Solutions Test 3B - Solutions to Test 3, Version B Math...

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Solutions to Test 3, Version B Math 1501, Fall 02. WG * November 8, 2002 Exercise 1. (a) We have that f 0 ( x )= - 1i f - 3 <x< - 1 f - 1 0 2 x - 4i f 0 3 2i f 3 4 . Since f 0 - ( - 1) = - 1and f 0 + ( - 1) = 1 we deduce that f 0 ( - 1) does not exist. Similarly, f 0 - (0) = 1 and f 0 + (0) = - 4andso f 0 (0) does not exist. We have that f 0 - (3) = 2 = f 0 + (3) and so, f 0 (3) = 2 . The only point where f 0 vanishes is 2 . Therefore, the critical points of f are - 1 , 0 , 2 . (b) and (c). Since f 0 is negative on ( - 3 , - 1) and positive on ( - 1 , 0) we conclude that f has a local minimum at - 1 . Since f 0 is positive on ( - 1 , 0) and negative on (0 , 2) we conclude that f has a local maximum at 0 . Since f 0 is negative on (0 , 2) and positive on (2 , 3) we conclude 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 . * School of Mathematics, Georgia Institute of technology. 1
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(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 x = π 3 and x = 2 π 3 . These two points are points of inflection. Exercise 3. We have that
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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 Institute of Technology.

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Solutions Test 3B - Solutions to Test 3, Version B Math...

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