#### Answer

$x=-2$ and $x=3$ are the critical numbers.

#### Work Step by Step

Find the critical numbers of $f(x) = 2x^3-3x^2-36x$
Differentiate and set the derivative $=0$
$f'(x) = 6x^2-6x-36$
$0=6(x^2-x-6)$
$0=6(x+2)(x-3)$
$x=-2$ and $x=3$ are the critical numbers.