Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 67

Answer

$$\eqalign{ & {\text{Concave up on }}\left( {0,1} \right) \cr & {\text{Concave down on }}\left( {1,\infty } \right) \cr & {\text{The inflection points is at: }}x = 1 \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \sqrt x \ln x \cr & {\text{Calculate the second derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\sqrt x \ln x} \right] \cr & f'\left( x \right) = \sqrt x \left( {\frac{1}{x}} \right) + \frac{{\ln x}}{{2\sqrt x }} \cr & f'\left( x \right) = \frac{1}{{\sqrt x }} + \frac{{\ln x}}{{2\sqrt x }} \cr & f'\left( x \right) = \frac{1}{{\sqrt x }}\left( {1 + \frac{1}{2}\ln x} \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{\sqrt x }}\left( {1 + \frac{1}{2}\ln x} \right)} \right] \cr & f''\left( x \right) = \frac{1}{{\sqrt x }}\left( {\frac{1}{{2x}}} \right) - \frac{1}{{2{x^{3/2}}}}\left( {1 + \frac{1}{2}\ln x} \right) \cr & f''\left( x \right) = \frac{1}{{2{x^{3/2}}}} - \frac{1}{{2{x^{3/2}}}}\left( {1 + \frac{1}{2}\ln x} \right) \cr & \cr & {\text{Set the second derivative to 0}} \cr & \frac{1}{{2{x^{3/2}}}} - \frac{1}{{2{x^{3/2}}}}\left( {1 + \frac{1}{2}\ln x} \right) = 0 \cr & \frac{1}{{2{x^{3/2}}}}\left( {1 - 1 + \frac{1}{2}\ln x} \right) = 0 \cr & \frac{{\ln x}}{{4{x^{3/2}}}} = 0 \cr & {\text{The function is not defined for }}x = 0,{\text{ but the domain of}} \cr & f\left( x \right) = \sqrt x \ln x{\text{ is }}\left\{ {x/x > 0} \right\}.{\text{ Then }} \cr & {\text{We see that }}\,f''\left( x \right) = 0{\text{ at }}x = 1 \cr & {\text{That points is candidate for inflection points}} \cr & {\text{We need evaluate the intervals }}\left( {0,1} \right){\text{ and }}\left( {1,\infty } \right) \cr & {\text{Now}}{\text{, we will evaluate test values and resume in a table}} \cr & {\text{to determinate whether the concavity changes at these points}} \cr} $$ \[\begin{array}{*{20}{c}} {{\text{Interval}}}&{{\text{Test value }}\left( x \right)}&{{\text{Sign of }}f''\left( x \right)}&{{\text{Behavior of }}f\left( x \right)} \\ {\left( {0,1} \right)}&{1/2}&{f''\left( {1/2} \right) > 0}&{{\text{Concave up}}} \\ {\left( {1,\infty } \right)}&2&{f''\left( 2 \right) < 0}&{{\text{Concave down}}} \end{array}\] $$\eqalign{ & {\text{From the table we can conclude that the function is:}} \cr & {\text{Concave up on }}\left( {0,1} \right) \cr & {\text{Concave down on }}\left( {1,\infty } \right) \cr & {\text{The inflection points is at: }}x = 1 \cr} $$
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.