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^21) = 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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