the question is
on what interval is f'(x)>0 for f(x)=2x^3 - 3x^2 - 36x +2
i found f'(x) = 6x^2 - 6x -36
then what do i have to do next to find interval?
As the product is greater than zero we have two possibilities
1) $\displaystyle x-3 >0 \wedge x+2 > 0 $
We conclude that $\displaystyle x \in (3,\infty)$
2) $\displaystyle x-3 <0 \wedge x+2 < 0 $
We conclude that $\displaystyle x \in (-\infty, -2)$
Then the interval is $\displaystyle x \in (-\infty,-2)\cup (3,\infty)$