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critical point of 2x^4 - 4x^2+3
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Take the derivative of 2x^4-4x^2+3, and set it equal to zero. You will get 2 values of x. x=0, x=1 These are the critical points of 2x^4-4x^2+3
Don't forget to fan :$
i get the derivative: 8x^3 - 8x is this correct?
Yes, this is correct.
then: 8x(x+1)(x-1) = 0 is the correct? How do you get to the point x=0, x=1
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8x=0 ;x=0 x+1=0 ;x=-1 x-1=0 ;x=1 So you actually get three values for x!
thanks alot!
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