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 November 11th, 2015, 11:57 AM #1 Newbie   Joined: Nov 2015 From: Sweden Posts: 2 Thanks: 0 Critical Points Find the critical points for the function: $\displaystyle f(x)=(x^2 -2|x|)/(x-1)$ $\displaystyle f'(x) = ((2x*-2sgn(x)*(x-1))-(1*x^2-2|x|))/(x-1)^2$ If i use the derivative function above without simplifying it i seem to find critical points in x=2 and x=-2. I found these points by simply trying numbers just for fun, i did not simplify the function whatsoever. My professor simplified the derivative function to $\displaystyle (x^2-2x+2)/(x-1)^2 , x>0$ $\displaystyle (x^2-2x-2)/(x-1)^2 , x<0$ From this we find no critical points for $\displaystyle x>0$ and a critical point $\displaystyle x= 1-sqrt(3)$ Am i doing something wrong when i find the points x=2 and x=-2? I have tried these numbers multiple times and seem to find them correct.
 November 11th, 2015, 12:24 PM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 As an example, \begin{align*} f'(x) &= \dfrac{(2x - 2\,\text{sgn}\, x)(x - 1) - (x^2 - 2|x|)}{(x - 1)^2}\\\\ f'(2) &= \dfrac{(2(2) - 2(1))(1) - (2^2 - 2(2))}{(2 - 1)^2}\\ &= \dfrac{2 - 0}{1} = 2 \neq 0 \end{align*} Thanks from mathe
 November 11th, 2015, 12:32 PM #3 Newbie   Joined: Nov 2015 From: Sweden Posts: 2 Thanks: 0 Thank you, i have for some reason used sgn(2)=2 which it clearly isn't.

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