#### Answer

${x = -2, 3}$

#### Work Step by Step

Original Equation: ${f(x) = 2x^{3} -3x^{2} -36x}$
The critical numbers of a function are found when the derivative of a function is set to zero.
Finding the derivative of the function using the power rule:
${f'(x) = 6x^{2} - 6x-36}$
Setting the derivative to zero and factoring out the 6:
${f'(x) = 6(x^{2} - x-6) = 0}$
Factoring out the quadratic equation:
${f'(x) = 6(x+2)(x-3) = 0}$
Solving for x:
${x = -2, 3}$