Question

critical point of 2x^4 - 4x^2+3

Answer 1

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

Answer 2

Don't forget to fan :$

Answer 3

i get the derivative: 8x^3 - 8x is this correct?

Answer 4

Yes, this is correct.

Answer 5

then: 8x(x+1)(x-1) = 0 is the correct?

How do you get to the point x=0, x=1

Answer 6

8x=0 ;x=0
x+1=0 ;x=-1
x-1=0 ;x=1

So you actually get three values for x!

Answer 7

thanks alot!