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 215: 30

Answer

$g(x)$ has the absolute maximum value of $0$ at $x=-\sqrt5$ and the absolute minimum value of $-\sqrt5$ at $x=0$ in the defined domain.
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Work Step by Step

$$g(x)=-\sqrt{5-x^2}=-(5-x^2)^{1/2}\hspace{1cm}-\sqrt5\le x\le0$$ 1) Find the derivative $g'(x)$: $$g'(x)=-\frac{1}{2}(5-x^2)^{-1/2}(5-x^2)'$$ $$g'(x)=-\frac{1}{2}(5-x^2)^{-1/2}(-2x)$$ $$g'(x)=\frac{2x}{2\sqrt{5-x^2}}=\frac{x}{\sqrt{5-x^2}}$$ - At $x=0$, we have $g'(x)=0$ - Also, $g'(x)$ is undefined when $\sqrt{5-x^2}=0$ or $x=\pm\sqrt5$. All these values of $x=0$ and $x=\pm\sqrt5$ are critical points of $g(x)$. However, we would leave out $x=\sqrt5$ here since this point does not lie in the defined interval. 2) Evaluate $g'(x)$ at the critical points and endpoints: - For $x=-\sqrt5$: $$g(-\sqrt5)=-\sqrt{5-(\sqrt5)^2}=-\sqrt{5-5}=\sqrt0=0$$ - For $x=0$: $$g(0)=-\sqrt{5-0^2}=-\sqrt{5}$$ This means $g(x)$ has the absolute maximum value of $0$ at $x=-\sqrt5$ and the absolute minimum value of $-\sqrt5$ at $x=0$ in the defined domain.
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