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: 15

Answer

$$\eqalign{ & {\text{Inflection point }}\left( {2,8} \right) \cr & {\text{Concave upward}}:{\text{ }}\left( {2,\infty } \right) \cr & {\text{Concave downward}}:{\text{ }}\left( { - \infty ,2} \right) \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {x^3} - 6{x^2} + 12x \cr & {\text{Calculate the second derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3} - 6{x^2} + 12x} \right] \cr & f'\left( x \right) = 3{x^2} - 12x + 12 \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {3{x^2} - 12x + 12} \right] \cr & f''\left( x \right) = 6x - 12 \cr & {\text{Set }}f''\left( x \right) = 0 \cr & 6x - 12 = 0 \cr & x = 2 \cr & {\text{Set the intervals }}\left( { - \infty ,2} \right),\left( {2,\infty } \right) \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$ \[\boxed{\begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( { - \infty ,2} \right)}&{\left( {2,\infty } \right)} \\ {{\text{Test Value}}}&{x = - 1}&{x = 5} \\ {{\text{Sign of }}f''\left( x \right)}&{f''\left( { - 1} \right) = - 18 < 0}&{f''\left( 5 \right) = 18 > 0} \\ {{\text{Conclusion}}}&{{\text{Concave downward}}}&{{\text{Concave upward}}} \end{array}}\] $$\eqalign{ & {\text{Inflection point }}\left( {2,8} \right) \cr & {\text{Concave upward}}:{\text{ }}\left( {2,\infty } \right) \cr & {\text{Concave downward}}:{\text{ }}\left( { - \infty ,2} \right) \cr} $$
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