(adsbygoogle = window.adsbygoogle || []).push({}); 1. Find the local max and min of x/(1+x^2)

3. The attempt at a solution

Take the derivative of x/(1+x^2) and set it equal to zero.

f'(x)= (1-x^2)/(1+x^2)^2 = 0

This is where I am stuck. My first guess was to factor the top, resulting in (1-x)(1+x)/(1+x^2)^2 = 0, and then cancel the 1+x. But after that I'm still left with (1-x)/(1+x)^2 = 0. I know if you let x=0 you will get f'(x) = 0 because 0/anything = 0.

Really, I'm not sure what I can do after I take the derivative.

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# Homework Help: Local Max & Min

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