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 - Review Exercises - Page 238: 36

Answer

$$\eqalign{ & {\text{Inflection points: }}\left( { - \frac{1}{6}, - \frac{5}{{216}}} \right){\text{ and }}\left( {\frac{1}{6}, - \frac{5}{{216}}} \right) \cr & {\text{Concave downward}}:{\text{ }}\left( { - \frac{1}{6},\frac{1}{6}} \right) \cr & {\text{Concave upward}}:{\text{ }}\left( { - \infty , - \frac{1}{6}} \right){\text{ and }}\left( {\frac{1}{6},\infty } \right) \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = 6{x^4} - {x^2} \cr & {\text{Calculate the second derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {6{x^4} - {x^2}} \right] \cr & f'\left( x \right) = 24{x^3} - 2x \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {24{x^3} - 2x} \right] \cr & f''\left( x \right) = 72{x^2} - 2 \cr & {\text{Set }}f''\left( x \right) = 0 \cr & 72{x^2} - 2 = 0 \cr & {x^2} = \frac{2}{{72}} = \frac{1}{{36}} \cr & x = \pm \frac{1}{6} \cr & {\text{Set the intervals }}\left( { - \infty , - \frac{1}{6}} \right),\left( { - \frac{1}{6},\frac{1}{6}} \right),\left( {\frac{1}{6},\infty } \right) \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$ \[\boxed{\begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( { - \infty , - \frac{1}{6}} \right)}&{\left( { - \frac{1}{6},\frac{1}{6}} \right)}&{\left( {\frac{1}{6},\infty } \right)} \\ {{\text{Test Value}}}&{x = - 1}&{x = 0}&{x = 1} \\ {{\text{Sign of }}f''\left( x \right)}&{70 > 0}&{ - 2 < 0}&{70 > 0} \\ {{\text{Conclusion}}}&{{\text{C}}{\text{. upward}}}&{{\text{C}}{\text{. downward}}}&{{\text{C}}{\text{. upward}}} \end{array}}\] $$\eqalign{ & {\text{The inflection points occur at }}x = - \frac{1}{6}{\text{ and }}x = \frac{1}{6} \cr & f\left( { - \frac{1}{6}} \right) = 6{\left( { - \frac{1}{6}} \right)^4} - {\left( { - \frac{1}{6}} \right)^2} = - \frac{5}{{216}}, \to {\text{ }}\left( { - \frac{1}{6}, - \frac{5}{{216}}} \right) \cr & f\left( {\frac{1}{6}} \right) = 6{\left( {\frac{1}{6}} \right)^4} - {\left( {\frac{1}{6}} \right)^2} = - \frac{5}{{216}}, \to {\text{ }}\left( {\frac{1}{6}, - \frac{5}{{216}}} \right) \cr & {\text{Inflection points: }}\left( { - \frac{1}{6}, - \frac{5}{{216}}} \right){\text{ and }}\left( {\frac{1}{6}, - \frac{5}{{216}}} \right) \cr & {\text{Concave downward}}:{\text{ }}\left( { - \frac{1}{6},\frac{1}{6}} \right) \cr & {\text{Concave upward}}:{\text{ }}\left( { - \infty , - \frac{1}{6}} \right){\text{ and }}\left( {\frac{1}{6},\infty } \right) \cr} $$
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