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 November 11th, 2012, 05:29 PM #1 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Critical Points f(x) = x^4 -2x^2 +4 f ' (x) = 4x^3 -4x set to zero to obtain critcal points 4x^3 - 4x = 0 factor out 4x and get 4x(x^2-1) = 0 I am seeing 3 critical points x=0, x = -1 and x =1 but... I think that at -1 and 1 the function DNE therefore does that mean I only have 1 critical number 0 ? or do I have 3 critical #'s ? ty
 November 11th, 2012, 06:34 PM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Re: Critical Points I think you have 3 critical points as the function does exist at x=-1 and x=1. In each case, the function value is 3.

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