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November 11th, 2012, 06:29 PM   #1
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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
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November 11th, 2012, 07:34 PM   #2
jks
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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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