Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.6 Exercises - Page 212: 6

Answer

Graph

Work Step by Step

$$\eqalign{ & y = \frac{x}{{{x^2} + 1}} \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & y = \frac{0}{{{0^2} + 1}} \cr & y = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & 0 = \frac{x}{{{x^2} + 1}} \cr & x = 0 \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {\frac{x}{{{x^2} + 1}}} \right] \cr & y' = \frac{{\left( {{x^2} + 1} \right)\left( 1 \right) - x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & y' = \frac{{{x^2} + 1 - 2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & y' = \frac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & {\text{Let }}y' = 0 \cr & 1 - {x^2} = 0 \cr & {x^2} = 1 \cr & x = \pm 1 \cr & \cr & {\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ {\frac{{1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right] \cr & y'' = \frac{{{{\left( {{x^2} + 1} \right)}^2}\left( { - 2x} \right) - 2\left( {1 - {x^2}} \right)\left( {{x^2} + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}} \cr & y'' = \frac{{{{\left( {{x^2} + 1} \right)}^2}\left( { - 2x} \right) - 2\left( {1 - {x^2}} \right)\left( {{x^2} + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}} \cr & y'' = \frac{{2x\left( {{x^2} + 1} \right)\left[ {\left( {{x^2} + 1} \right)\left( { - 1} \right) - 2\left( {1 - {x^2}} \right)} \right]}}{{{{\left( {{x^2} + 1} \right)}^4}}} \cr & y'' = \frac{{2x\left[ {\left( {{x^2} + 1} \right)\left( { - 1} \right) - 2\left( {1 - {x^2}} \right)} \right]}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & y'' = \frac{{2x\left( { - {x^2} - 1 - 2 + 2{x^2}} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & y'' = \frac{{2x\left( {{x^2} - 3} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & {\text{Evaluate }}y''\left( { - 1} \right) \cr & y''\left( { - 1} \right) = \frac{{2\left( { - 1} \right)\left( {{{\left( { - 1} \right)}^2} - 3} \right)}}{{{{\left( {{{\left( { - 1} \right)}^2} + 1} \right)}^3}}} = \frac{1}{2} > 0 \cr & {\text{Relative minimum at }}\left( { - 1,f\left( { - 1} \right)} \right) \to \left( { - 1, - \frac{1}{2}} \right) \cr & y''\left( 1 \right) = \frac{{2\left( 1 \right)\left( {{{\left( 1 \right)}^2} - 3} \right)}}{{{{\left( {{{\left( 1 \right)}^2} + 1} \right)}^3}}} = - \frac{1}{2} < 0 \cr & {\text{Relative maximum at }}\left( {1,f\left( 1 \right)} \right) \to \left( {1,\frac{1}{2}} \right) \cr & \cr & *{\text{Find the Inflection points}} \cr & y'' = \frac{{2x\left( {{x^2} - 3} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & \frac{{2x\left( {{x^2} - 3} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} = 0 \cr & x = 0,{\text{ }}x = - \sqrt 3 ,{\text{ }}x = \sqrt 3 \cr & \cr & {\text{*Calculate the asymptotes}} \cr & y = \frac{x}{{{x^2} + 1}} \cr & {x^2} + 1 = 0,{\text{ no real solutions}} \cr & {\text{No vertical asymptotes }} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {\frac{x}{{{x^2} + 1}}} \right) = 0 \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{x}{{{x^2} + 1}}} \right) = 0 \cr & {\text{Horizontal asymptote }}y = 0 \cr & \cr & {\text{Graph}} \cr} $$
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