# What are the critical points of f(x) = 2x^3-3x^2-12x?

Oct 28, 2015

The critical points are $- 1$ and $2$.
$f ' \left(x\right) = 6 {x}^{2} - 6 x - 12 = 6 \left({x}^{2} - x - 2\right)$
So. to find its zeros, we must look for the zeroes of $\left({x}^{2} - x - 2\right)$, which are $- 1$ and $2$. So, these are the critical points of $f \left(x\right)$.