Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 3 - The Derivative In Graphing And Applications - 3.2 Analysis Of Functions II: Relative Extrema; Graphing Polynomials - Exercises Set 3.2 - Page 205: 10

Answer

$$x = 0,{\text{ }}x = \root 3 \of {16} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{{{x^2}}}{{{x^3} + 8}} \cr & {\text{Calculate the derivative of the function}} \cr & f'\left( x \right) = \frac{{\left( {{x^3} + 8} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] - {x^2}\frac{d}{{dx}}\left[ {{x^3} + 8} \right]}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr & f'\left( x \right) = \frac{{\left( {{x^3} + 8} \right)\left( {2x} \right) - {x^2}\left( {3{x^2}} \right)}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2{x^4} + 16x - 3{x^4}}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr & f'\left( x \right) = \frac{{ - {x^4} + 16x}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr & {\text{Factor the numerator}} \cr & f'\left( x \right) = \frac{{ - x\left( {{x^3} - 16} \right)}}{{{{\left( {{x^3} + 8} \right)}^2}}} \cr & {\text{Set }}f'\left( x \right) = 0 \cr & - x\left( {{x^3} - 16} \right) = 0 \cr & - x\left( {{x^3} - 16} \right) = 0 \cr & - x = 0,{\text{ }}{x^3} - 16 = 0 \cr & x = 0,{\text{ }}{x^3} = 16 \cr & x = 0,{\text{ }}x = \root 3 \of {16} \cr & {\text{Therefore, the stationary points are:}} \cr & x = 0,{\text{ }}x = \root 3 \of {16} \cr} $$
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.