Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 26

Answer

$$\eqalign{ & {\text{Domain }}\left( { - \infty ,\infty } \right) \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{No }}x{\text{ - intercepts}} \cr & {\text{Local minimum at }}\left( { \pm \frac{1}{{2\sqrt 2 }},\frac{7}{4}} \right) \cr & {\text{Local maximum at }}\left( {0,2} \right) \cr & {\text{No Inflection points}} \cr & {\text{No vertical asymptotes}} \cr & {\text{No horizontal asymptotes}} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = 2 - {x^{2/3}} + {x^{4/3}} \cr & {\text{Domain }}\left( { - \infty ,\infty } \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = 2 - {\left( 0 \right)^{2/3}} + {\left( 0 \right)^{4/3}} \cr & f\left( 0 \right) = 2 \cr & y{\text{ - intercept }}\left( {0,2} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 2 - {x^{2/3}} + {x^{4/3}} = 0,{\text{ No real solutions}} \cr & {\text{There are no }}x{\text{ - intercepts}} \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2 - {x^{2/3}} + {x^{4/3}}} \right] \cr & f'\left( x \right) = - \frac{2}{3}{x^{ - 1/3}} + \frac{4}{3}{x^{1/3}} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & - \frac{2}{3}{x^{ - 1/3}}\left( {1 - 2{x^{2/3}}} \right) = 0 \cr & - \frac{2}{3}{x^{ - 1/3}} = 0,{\text{ }}\left( {{\text{Undefined at }}x = 0} \right) \cr & 1 - 2{x^{2/3}} = 0 \to x = \pm \frac{1}{{2\sqrt 2 }} \cr & \cr & *{\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - \frac{2}{3}{x^{ - 1/3}} + \frac{4}{3}{x^{1/3}}} \right] \cr & f''\left( x \right) = \frac{2}{9}{x^{ - 4/3}} + \frac{4}{9}{x^{ - 2/3}} \cr & {\text{*Evaluate }}f''\left( x \right){\text{ at the critical points }}x = \pm \frac{1}{{2\sqrt 2 }} \cr & f''\left( { - \frac{1}{{2\sqrt 2 }}} \right) = \frac{2}{9}{\left( { - \frac{1}{{2\sqrt 2 }}} \right)^{ - 4/3}} + \frac{4}{9}{\left( { - \frac{1}{{2\sqrt 2 }}} \right)^{ - 2/3}} = \frac{{16}}{9} > 0 \cr & {\text{There is a local minimum at }}\left( { - \frac{1}{{2\sqrt 2 }},f\left( { - \frac{1}{{2\sqrt 2 }}} \right)} \right) \cr & f\left( { - \frac{1}{{2\sqrt 2 }}} \right) = 2 - {\left( { - \frac{1}{{2\sqrt 2 }}} \right)^{2/3}} + {\left( { - \frac{1}{{2\sqrt 2 }}} \right)^{4/3}} = \frac{7}{4} \cr & {\text{Local minimum at }}\left( { - \frac{1}{{2\sqrt 2 }},\frac{7}{4}} \right) \cr & and \cr & f''\left( {\frac{1}{{2\sqrt 2 }}} \right) = \frac{2}{9}{\left( {\frac{1}{{2\sqrt 2 }}} \right)^{ - 4/3}} + \frac{4}{9}{\left( {\frac{1}{{2\sqrt 2 }}} \right)^{ - 2/3}} = \frac{{16}}{9} > 0 \cr & {\text{There is a local minimum at }}\left( {\frac{1}{{2\sqrt 2 }},f\left( {\frac{1}{{2\sqrt 2 }}} \right)} \right) \cr & f\left( {\frac{1}{{2\sqrt 2 }}} \right) = 2 - {\left( {\frac{1}{{2\sqrt 2 }}} \right)^{2/3}} + {\left( {\frac{1}{{2\sqrt 2 }}} \right)^{4/3}} = \frac{7}{4} \cr & {\text{Local minimum at }}\left( {\frac{1}{{2\sqrt 2 }},\frac{7}{4}} \right) \cr & {\text{Use the first derivative test at }}x = 0 \cr & f'\left( { - 0.3} \right) = - \frac{2}{3}{\left( { - 0.3} \right)^{ - 1/3}} + \frac{4}{3}{\left( { - 0.3} \right)^{1/3}} > 0{\text{ Increasing}} \cr & f'\left( 0 \right) = - \frac{2}{3}{\left( 0 \right)^{ - 1/3}} + \frac{4}{3}{\left( 0 \right)^{1/3}}{\text{ Undefined}} \cr & f'\left( {0.3} \right) = - \frac{2}{3}{\left( {0.3} \right)^{ - 1/3}} + \frac{4}{3}{\left( {0.3} \right)^{1/3}} < 0{\text{ Decreasing}} \cr & x = 0{\text{ is in the domain of the function}}{\text{, then}} \cr & {\text{There is a local maximum at }}x = 0 \cr & f\left( 0 \right) = 2 \cr & {\text{Local maximum at }}\left( {0,2} \right) \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{2}{9}{x^{ - 4/3}} + \frac{4}{9}{x^{ - 2/3}} = 0 \cr & \frac{2}{9}{x^{ - 4/3}}\left( {1 + 2{x^{2/3}}} \right) = 0 \cr & \frac{2}{9}{x^{ - 4/3}} = 0,{\text{ undefined at }}x = 0 \cr & 1 + 2{x^{2/3}} = 0,{\text{ No real solutions}} \cr & {\text{The second derivative is not defined at }}x = 0 \cr & f''\left( { - 0.3} \right) \approx 4.311 > 0,{\text{ }}\left( {{\text{Concave up}}} \right) \cr & f''\left( {0.3} \right) \approx 4.311 > 0,{\text{ }}\left( {{\text{Concave up}}} \right) \cr & {\text{Then}}{\text{, no inflection points}}{\text{.}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {2 - {x^{2/3}} + {x^{4/3}}} \right) = + \infty \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {2 - {x^{2/3}} + {x^{4/3}}} \right) = + \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr & {\text{Graph}} \cr} $$
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