Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 62

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

$$\eqalign{ & f\left( x \right) = 2{x^2}\ln x - 5{x^2} \cr & \cr & {\text{Calculate the second derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^2}\ln x - 5{x^2}} \right] \cr & f'\left( x \right) = 2{x^2}\left( {\frac{1}{x}} \right) + 4x\ln x - 10x \cr & f'\left( x \right) = 4x\ln x - 8x \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {4x\ln x - 8x} \right] \cr & f''\left( x \right) = 4x\left( {\frac{1}{x}} \right) + 4\ln x - 8 \cr & f''\left( x \right) = 4\ln x - 4 \cr & \cr & {\text{Set the second derivative to 0}} \cr & f''\left( x \right) = 4\ln x - 4 \cr & 4\ln x - 4 = 0 \cr & \ln x - 1 = 0 \cr & {\text{We see that }}\,f''\left( x \right) = 0{\text{ at }}x = e \cr & {\text{That points is candidate for inflection points}} \cr & {\text{We need evaluate the intervals }}\left( { - \infty ,e} \right){\text{ and }}\left( {e,\infty } \right) \cr & {\text{Now}}{\text{, we will evaluate test values and resume in a table}} \cr & {\text{to determinate whether the concavity changes at these points}} \cr} $$ \[\begin{array}{*{20}{c}} {{\text{Interval}}}&{{\text{Test value }}\left( x \right)}&{{\text{Sign of }}f''\left( x \right)}&{{\text{Behavior of }}f\left( x \right)} \\ {\left( { - \infty ,e} \right)}&2&{f''\left( 2 \right) < 0}&{{\text{Concave down}}} \\ {\left( {e,\infty } \right)}&3&{f''\left( 3 \right) > 0}&{{\text{Concave up}}} \end{array}\] $$\eqalign{ & {\text{From the table we can conclude that the function is:}} \cr & {\text{Concave up on }}\left( { - \infty ,e} \right) \cr & {\text{Concave down on }}\left( {e,\infty } \right) \cr & {\text{The inflection points is at: }}x = e \cr} $$