Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 6 - Applications of the Derivative - 6.1 Absolute Extrema - 6.1 Exercises - Page 310: 22

Answer

$$ f(x)=(x^{2}-16)^{\frac{2}{3}}; \quad\quad[-5,8] $$ Critical numbers in the given interval is: -4, 0, and 4. The absolute maximum, $13.20770 $ occurs when $x=8 $ and the absolute minimum, $0$ occurs when $x=-4$ and $x=4$
1575237573

Work Step by Step

$$ f(x)=(x^{2}-16)^{\frac{2}{3}}; \quad\quad[-5,8] $$ First look for critical numbers in the interval $(-5,8)$. find the points where the derivative is $0$ $$ \begin{aligned} f^{\prime}(x) &=(\frac{2}{3})(2x)(x^{2}-16)^{\frac{-1}{3}}\\ &=(\frac{4x}{3})(x^{2}-16)^{\frac{-1}{3}}\\ &=\frac{4x}{3(x^{2}-16)^{\frac{1}{3}}}\\ \end{aligned} $$ Solve the equation $f^{\prime}(x)=0 $ to get $$ \begin{aligned} f^{\prime}(x)& =\frac{4x}{3(x^{2}-16)^{\frac{1}{3}}}=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & 4x=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & x=0 \end{aligned} $$ The derivative $f^{\prime}(x) $ is undefined when $$ \begin{aligned} 3(x^{2}-16)^{\frac{1}{3}}=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & x^{2}-16=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & (x-4)(x+4)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & x=4 \quad \text {or} \quad x=-4 \\ \end{aligned} $$ So, critical numbers in the given interval is: -4, 0, and 4. Now, evaluate the function at the critical numbers and the endpoints: $$ \begin{aligned} \text{when } x=-5: &\\ f(-5) &=((-5)^{2}-16)^{\frac{2}{3}}\\ &=(25-16)^{\frac{2}{3}}\\ &=(9)^{\frac{2}{3}}\\ &\approx \:4.32674 \end{aligned} $$ $$ \begin{aligned} \text{when } x=-4: &\\ f(-4) &=((-4)^{2}-16)^{\frac{2}{3}}\\ &=(0)^{\frac{1}{3}}\\ &=0 \end{aligned} $$ $$ \begin{aligned} \text{when } x=0: &\\ f(0) &=((0)^{2}-16)^{\frac{2}{3}}\\ &=(-16)^{\frac{2}{3}}\\ &\approx 6.34960 \end{aligned} $$ $$ \begin{aligned} \text{when } x=4: &\\ f(4) &=((4)^{2}-16)^{\frac{2}{3}}\\ &=(0)^{\frac{1}{3}}\\ &=0 \end{aligned} $$ and $$ \begin{aligned} \text{when } x=8: &\\ f(8)&=((8)^{2}-16)^{\frac{2}{3}}\\ &=(64-16)^{\frac{2}{3}}\\ &=(48)^{\frac{2}{3}}\\ &\approx 13.20770 \end{aligned} $$ We conclude that, the absolute maximum, $13.20770 $ occurs when $x=8 $ and the absolute minimum, $0$ occurs when $x=-4$ and $x=4$
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