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: 18

Answer

$$ f(x)=\frac{8+x}{8-x}; \quad\quad[4,6] $$ The graph has no critical numbers. The absolute maximum, $7 $ occurs when $x=6 $ and the absolute minimum, $3 $ occurs when $x=4.$
1574900248

Work Step by Step

$$ f(x)=\frac{8+x}{8-x}; \quad\quad[4,6] $$ First look for critical numbers in the interval $(4,6)$. find the points where the derivative is $0$ $$ \begin{aligned} f^{\prime}(x) &=\frac{8+x}{8-x};\\ &=\frac{(8-x) (1)-(8+x)(-1)}{(8-x)^{2}}\\ &=\frac{16}{(8-x)^{2}} \end{aligned} $$ Solve the equation $f^{\prime}(x)=0 $ to get $$ \begin{aligned} f^{\prime}(x) &=\frac{16}{(8-x)^{2}}=0 \end{aligned} $$ Since this equation has no solution, also $ f^{\prime}(x) $is not exist when $x=8 $ but $8 \notin [4,6] $, so the graph has no critical numbers. Now, evaluate the function at the endpoints: $$ \text{when } x=4: f(4)=\frac{8+(4)}{8-(4)} =3\\ \text{when } x=6: f(6)=\frac{8+(6)}{8-(6)} =7\\ $$ We conclude that, the absolute maximum, $7 $ occurs when $x=6 $ and the absolute minimum, $3 $ occurs when $x=4.$
Small 1574900248
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.