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: 57


No maximum. Local minimum: $f(x)=0$ and $x=1$ or $x=-1$

Work Step by Step

Given $y=({(x^2-1})^{(\frac{1}{2})}$ The critical point is where the first derivative is zero: so $\frac{dy}{dx}=\frac{d{({x^2}-1)^{(\frac{1}{2})})}}{dx}$ ${\frac{dy}{dx}}=\frac{1}{2({({x^2}-1)}^{(\frac{1}{2})}}{({{2x}})}$ so the critical point is: ${\frac{dy}{dx}}=0$ $f'(x)=\frac{1}{2({({x^2}-1)}^{(\frac{1}{2})}}{({{2x}})}=0$ $x=0$ and ${({x^2}-1)}^{(\frac{1}{2})}=0$ $x=0,x=1,x=-1$, but note that zero is not in the domain of the function the critical value of y can be obtained by putting in the value of x: $f(-1)=f(1)=0$ Thus we have: No maximum. Local minimum: $f(x)=0$ and $x=1$ or $x=-1$
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