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Unformatted text preview: Math 2250 Exam #3 Solutions 1. The function g ( x ) = x 3 2 x 2 4 x +9 has no absolute minimum, but what is the local minimum value of g ( x )? Answer: To find the local minimum, first we want to determine the critical points of g , so we take the derivative: g ( x ) = 3 x 2 4 x 4 = (3 x + 2)( x 2) Therefore, g ( x ) = 0 when (3 x +2)( x 2) = 0, meaning when x = 2 / 3 or x = 2. To determine which is the local minimum, we’ll use the Second Derivative Test. g 00 ( x ) = 6 x 4 , so g 00 ( 2 / 3) = 6( 2 / 3) 4 = 8 g 00 (2) = 6(2) 4 = 8 . Therefore, by the Second Derivative Test, since x = 2 is a critical point and g 00 (2) > 0, we know g has a local minimum at x = 2. Therefore, the local minimum value of g is g (2) = 2 3 2(2) 2 4(2) + 9 = 8 8 8 + 9 = 1 . 2. Evaluate the limit lim x → + sin( x )tan( x + π/ 2) . Answer: Notice that lim x → + sin( x ) = sin(0) = 0 since the sine function is continuous, while lim x → + tan( x + π/ 2) = lim x → + sin( x + π/ 2) cos( x + π/ 2) =∞ since the numerator goes to 1 and the denominator is negative but approaching zero. Therefore, the limit we’re computing is of the form 0 · ∞ , so before we can use L’Hˆ opital’s Rule, we need to rewrite it in the following form: lim x → + sin( x )tan( x + π/ 2) = lim x → + sin( x ) 1 tan( x + π/ 2) = lim x → + sin( x ) cot( x + π/ 2) . Now both numerator and denominator are going to zero, so, by L’Hˆ opital’s rule, the above limit is equal to lim x → + d dx (sin( x )) d dx (cot( x + π/ 2)) = lim x → + cos( x ) csc 2 ( x + π/ 2) ....
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This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at UGA.
 Fall '08
 CHESTKOFSKY
 Calculus, Critical Point, Derivative

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