Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises: 30

Answer

${x = -5, 1}$

Work Step by Step

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: ${x = -5, 1}$
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