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: 56


local minimum: $f(x)=0$ and $x=5$ and at $(0,0)$ local maximum: $f(x)=108$ and $x={3}$

Work Step by Step

Given $y=x^3(x-5)^2$ the critical point is where the first derivative is zero: so $\frac{dy}{dx}=\frac{d{(x^3(x-5)^2)}}{dx}$ $\frac{dy}{dx}=\frac{{(3x^2(x-5)^2)-2x^3(x-5)}}{(x-5)^2}$ ${\frac{dy}{dx}}={{5x^2(x-3)}}{(x-5)}$ so the critical point is: ${\frac{dy}{dx}}=0$ $f'(x)={{5x^2(x-3)}}{(x-5)}=0$ $x=0, x=3, x=5$ the critical value of y can be obtained by putting in the value of x: at $x=5$: $y=x^3(x-5)^2=0$ at $x=3$: $y=x^3(x-5)^2=3^3(-2)^2=108$ at $x=0$: $y=x^3(x-5)^2=0$ Thus we have: local minimum: $f(x)=0$ and $x=5$ and at $(0,0)$ local maximum: $f(x)=108$ and $x={3}$
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