finalreview1_soln2 - Problem 2 on review sheet Graph f(x =...

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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 L’hopital’s 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 0 ( x ): f 0 ( 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 0 ( x ) might be 0 at x = 0, but notice that we can’t just plug in 0 for f 0 ( x ) cause the denominator of f 0 ( x ) is 0 at x = 0. So we take the limit at x 0 and see what we get. Don’t use l’Hopital, it takes forever... Consider the easiest Taylor approximation to ln(1 + x 2 ) centered at x = 0: ln(1 + x 2 ) x 2 . Compute the limit: lim x 0 f 0 ( x ) = lim x 0 3 x 2 ( x 2 ) - 2 x 4 1 1+ x 2 x 4 = 1 . Note that in the limit we are free to replace the ln(1 + x 2 ) by it’s Taylor series
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