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: 26

Answer

$$ f(x)=\frac{\ln x}{ x^{2}} ; \quad\quad[1,4] $$ The critical number in the given interval is: $e ^{\frac{1}{2}}$. The absolute maximum, $0.18393 $ occurs when $x=e ^{\frac{1}{2}} $ and the absolute minimum, $0 $ occurs when $x=1.$
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Work Step by Step

$$ f(x)=\frac{\ln x}{ x^{2}} ; \quad\quad[1,4] $$ First look for critical numbers in the interval $(1,4)$. find the points where the derivative is $0$ $$ \begin{aligned} f^{\prime}(x) &=\frac{x^{2}.(\frac{1}{x}) -\ln x . (2x)}{ x^{4}} \\ &=\frac{x - 2x\ln x }{ x^{4}} \\ &=\frac{x(1- 2\ln x) }{ x^{4}}\\ &=\frac{(1- 2\ln x) }{ x^{3}}\\ \end{aligned} $$ Solve the equation $f^{\prime}(x)=0 $ to get $$ \begin{aligned} f^{\prime}(x)& =\frac{(1- 2\ln x) }{ x^{3}}=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & (1- 2\ln x)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & 2\ln x=1\\ \Rightarrow\quad\quad\quad\quad\quad\\ & \ln x=\frac{1}{2}\\ \Rightarrow\quad\quad\quad\quad\quad\\ & x=e ^{\frac{1}{2}} \approx 1.64872\\ \end{aligned} $$ So, the critical number in the given interval is: $e ^{\frac{1}{2}}$, also $ f^{\prime}(x) $ is not exist when $x=0 $ but $0 \notin [1,4] $, so $x=0$ is not critical number in the given interval. Now, evaluate the function at the critical numbers and the endpoints: $$ \begin{aligned} \text{when } x=1: \\ f(1) &=\frac{\ln (1)}{ (1)^{2}} \\ &=\frac{0}{ 1} \\ &=0 \end{aligned} $$ $$ \begin{aligned} \text{when } x=e ^{\frac{1}{2}}: \\ f(e ^{\frac{1}{2}}) &=\frac{\ln (e ^{\frac{1}{2}})}{ (e ^{\frac{1}{2}})^{2}} \\ &=\frac{1}{2e}\\ &\approx 0.18393 \end{aligned} $$ and $$ \begin{aligned} \text{when } x=4: \\ f(4) &=\frac{\ln (4)}{ (4)^{2}} \\ &=\frac{\ln \left(2\right)}{8} \\ &\approx 0.08664 \end{aligned} $$ We conclude that, the absolute maximum, $0.18393 $ occurs when $x=e ^{\frac{1}{2}} $ and the absolute minimum, $0 $ occurs when $x=1.$
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