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 - 3.4 Exercises - Page 192: 22

Answer

$$\eqalign{ & {\text{There are no inflection points}} \cr & {\text{Concave downward}}:\left( { - \infty ,9} \right) \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = x\sqrt {9 - x} \cr & {\text{The domain of the function is }}9 - x \geqslant 0 \cr & x \leqslant 9,{\text{ }}\left( { - \infty ,9} \right] \cr & {\text{Calculate the first and second derivatives}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {x\sqrt {9 - x} } \right] \cr & f'\left( x \right) = \sqrt {9 - x} \frac{d}{{dx}}\left[ x \right] + x\frac{d}{{dx}}\left[ {\sqrt {9 - x} } \right] \cr & f'\left( x \right) = \sqrt {9 - x} + x\left( { - \frac{1}{{2\sqrt {9 - x} }}} \right) \cr & f'\left( x \right) = \frac{{2\left( {9 - x} \right) - x}}{{2\sqrt {9 - x} }} \cr & f'\left( x \right) = \frac{{18 - 2x - x}}{{2\sqrt {9 - x} }} \cr & f'\left( x \right) = \frac{{18 - 3x}}{{2\sqrt {9 - x} }} \cr & f''\left( x \right) = \underbrace {\frac{d}{{dx}}\left[ {\frac{{18 - 3x}}{{2\sqrt {9 - x} }}} \right]}_{{\text{Use quotient rule}}} \cr & f''\left( x \right) = \frac{{2\sqrt {9 - x} \left( { - 3} \right) - \left( {18 - 3x} \right)\left( {\frac{{ - 1}}{{\sqrt {9 - x} }}} \right)}}{{{{\left( {2\sqrt {9 - x} } \right)}^2}}} \cr & f''\left( x \right) = \frac{{ - 6\sqrt {9 - x} + \frac{{18 - 3x}}{{\sqrt {9 - x} }}}}{{4\left( {9 - x} \right)}} \cr & f''\left( x \right) = \frac{{ - 6\left( {9 - x} \right) + 18 - 3x}}{{4{{\left( {9 - x} \right)}^{3/2}}}} \cr & f''\left( x \right) = \frac{{ - 54 + 6x + 18 - 3x}}{{4{{\left( {9 - x} \right)}^{3/2}}}} \cr & f''\left( x \right) = \frac{{3x - 36}}{{4{{\left( {9 - x} \right)}^{3/2}}}} \cr & {\text{Set the second derivative to }}0 \cr & 3x - 36 = 0 \cr & x = 12 \cr & {\text{This value is not in the domain of the function }}f\left( x \right) = x\sqrt {9 - x} \cr & {\text{So, there are no inflection points, and the domain is }}x \leqslant 9 \cr & {\text{Evaluating the second derivative at the interval }}\left( { - \infty ,9} \right) \cr} $$ \[\boxed{\begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( { - \infty ,9} \right)} \\ {{\text{Test Value}}}&{x = 0} \\ {{\text{Sign of }}f''\left( x \right)}&{ - \frac{1}{3} < 0} \\ {{\text{Conclusion}}}&{{\text{Concave downward}}} \end{array}}\] $$\eqalign{ & {\text{There are no inflection points}} \cr & {\text{Concave downward}}:\left( { - \infty ,9} \right) \cr} $$
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