## Calculus: Early Transcendentals 8th Edition

${x = -2, 3}$
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: ${x = -2, 3}$