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: 18

Answer

$$\eqalign{ & {\text{Inflection point }}\left( {0,4} \right) \cr & {\text{Concave downward}}:{\text{ }}\left( { - \infty ,\infty } \right) \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = 4 - x - 3{x^4} \cr & {\text{Calculate the second derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {4 - x - 3{x^4}} \right] \cr & f'\left( x \right) = - 1 - 12{x^3} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - 1 - 12{x^3}} \right] \cr & f''\left( x \right) = - 36{x^2} \cr & {\text{Set }}f''\left( x \right) = 0 \cr & - 36{x^2} = 0 \cr & x = 0 \cr & {\text{Set the intervals }}\left( { - \infty ,0} \right),\left( {0,\infty } \right) \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$ \[\boxed{\begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( { - \infty ,0} \right)}&{\left( {0,\infty } \right)} \\ {{\text{Test Value}}}&{x = - 5}&{x = 5} \\ {{\text{Sign of }}f''\left( x \right)}&{f''\left( { - 5} \right) = - 900 < 0}&{f''\left( 5 \right) = - 900 < 0} \\ {{\text{Conclusion}}}&{{\text{Concave downward}}}&{{\text{Concave downward}}} \end{array}}\] $$\eqalign{ & {\text{The inflection point occurs at }}x = 0 \cr & f\left( 0 \right) = 4 - \left( 0 \right) - 3{\left( 0 \right)^4} \cr & {\text{Inflection point }}\left( {0,4} \right) \cr & {\text{Concave downward}}:{\text{ }}\left( { - \infty ,\infty } \right) \cr} $$
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