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Unformatted text preview: Problem 2 on review sheet. Graph f ( x ) = x 3 ln(1+ x 2 ) on (- , ) . First, we note that f ( x ) = 0 at x = 0 and that lim x f ( x ) = , lim x - f ( x ) =- . This is done by Lhopitals rule. This info will help draw the graph as x goes off to . Now, we need to know the critical points of f ( x ). Compute f ( x ): f ( x ) = 3 x 2 ln(1 + x 2 )- 2 x 4 1 1+ x 2 ln(1 + x 2 ) 2 . To find potential critical points we look for where the numerator ( n ( x ) from now on) is 0. That is, 0 = n ( x ) = 3 x 2 ln(1 + x 2 )- 2 x 4 1 1 + x 2 . Notice that n ( x ) is 0 at x = 0. This means that f ( x ) might be 0 at x = 0, but notice that we cant just plug in 0 for f ( x ) cause the denominator of f ( x ) is 0 at x = 0. So we take the limit at x 0 and see what we get. Dont use lHopital, it takes forever... Consider the easiest Taylor approximation to ln(1 + x 2 ) centered at x = 0: ln(1 + x 2 ) x 2 ....
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- Winter '09