What you are doing will find the zeros if any.
For the critical numbers take the derivative equal to zero and solve for x, these are the critical numbers I think you are after.
Hello,
I am trying to find the critical numbers for
f'(x) = 36x^3 - 72x^2 - 72
I got it to:
36(x^3 - 2x^2 - 2)
However, I am confused to as if I am supposed to factor it more?
Or how would I solve for
36(x^3 - 2x^2 - 2) = 0
Thanks very much in advance.
If you get f'(x) = 36x^3 - 72x^2 - 72 as part of finding the critical values of f(x) then you're expected to solve x^3 - 2x^2 - 2 = 0. Since there is no simple solution I assume you're expected to use technology such as a CAS: solve x^3 - 2x^2 - 2 = 0 - Wolfram|Alpha
On the other hand, if you have to find the critical values of f'(x) = 36x^3 - 72x^2 - 72, then you need to solve f''(x) = 0.