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 - Chapter 3 Review Exercises - Page 261: 37

Answer

$${\text{relative minimum of 0 at }}x = 0$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{{{x^2}}}{{{x^2} + 1}} \cr & {\text{Calculate the first derivative }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{{x^2} + 1}}} \right] \cr & f'\left( x \right) = \frac{{2x\left( {{x^2} + 1} \right) - {x^2}\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2{x^3} + 2x - 2{x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & {\text{Find the critical point}}{\text{, set }}f'\left( x \right) = 0 \cr & \frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0 \cr & {\text{The critical points is }}x = 0 \cr & \cr & {\text{Calculate the second derivative }} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right] \cr & f''\left( x \right) = \frac{{2{{\left( {{x^2} + 1} \right)}^2} - 4x\left( {{x^2} + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}} \cr & f''\left( x \right) = \frac{{2\left( {{x^2} + 1} \right) - 4x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & f''\left( x \right) = \frac{{2{x^2} + 2 - 8{x^2}}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & f''\left( x \right) = \frac{{2 - 6{x^2}}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & {\text{Using the second test derivative into the critical points}} \cr & {\text{Evaluate }}f''\left( 0 \right) \cr & f''\left( 0 \right) = \frac{{2 - 6{{\left( 0 \right)}^2}}}{{{{\left( {{{\left( 0 \right)}^2} + 1} \right)}^3}}} \cr & f''\left( 0 \right) = 2 \cr & f''\left( 0 \right) > 0,{\text{ then }}f\left( x \right){\text{ has a relative minimum at }}x = 0 \cr} $$
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