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: $\displaystyle f(x)=(x^2 2x)/(x1)$ $\displaystyle f'(x) = ((2x*2sgn(x)*(x1))(1*x^22x))/(x1)^2$ 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 $\displaystyle (x^22x+2)/(x1)^2 , x>0$ $\displaystyle (x^22x2)/(x1)^2 , x<0$ From this we find no critical points for $\displaystyle x>0$ and a critical point $\displaystyle x= 1sqrt(3)$ 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  2x)}{(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*} 
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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