Calculus Calculus Math Forum

 November 11th, 2015, 12:57 PM #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, 01:24 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, 01:32 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. Tags critical, points Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mathkid Calculus 1 November 11th, 2012 07:34 PM Timk Calculus 3 November 29th, 2011 11:59 AM summerset353 Calculus 1 March 5th, 2010 02:50 AM SSmokinCamaro Calculus 2 April 3rd, 2009 08:04 PM stainsoftime Calculus 3 November 24th, 2008 05:24 AM

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