University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.1 - Extreme Values of Functions - Exercises - Page 216: 47


critical points are (1,27) and (4,0)

Work Step by Step

Given $y=x(4-x)^3$ The critical points occur where the first derivative is zero: so$\frac{dy}{dx}=\frac{d{(x(4-x)^3)}}{dx}$ ${\frac{dy}{dx}}=(4-x)^3-x(4-x)^2$ so the critical point is ${\frac{dy}{dx}}=0$ $f'(x)=(4-x)^3-x(4-x)^2=0$ $x=1,x=4$ the critical value of y can be obtained by putting in the value of x: $y=1(4-1)^3=27$ $y=4(4-4)^3=0$ thus the critical points are (1,27) and (4,0)
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