## Calculus: Early Transcendentals 8th Edition

${x = -5, 1}$
Original equation: ${f(x) = x^{3} + 6x^{2} - 15x}$ The critical numbers of a function are found when the derivative of a function is set to 0. Finding the derivative of the function using the power rule: ${f'(x) = 3x^{2} + 12x - 15}$ Setting the derivative to equal 0: ${f'(x) = 3x^{2} + 12x - 15 = 0}$ Factoring out the 3: ${3x^{2} + 12x - 15 = 0}$ ${3(x^{2} + 4x - 5) = 0}$ Solving the quadratic equation through factoring: ${3(x^{2} + 4x - 5) = 0}$ ${3(x+5)(x-1) = 0}$ Solve for x: ${x = -5, 1}$