Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 63

$$\eqalign{ & {\text{Concave down on }}\left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right){\text{ and }}\left( {\frac{1}{{\sqrt 3 }},\infty } \right) \cr & {\text{Concave up on }}\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right) \cr & {\text{The inflection points are at: }}t = - \frac{1}{{\sqrt 3 }}{\text{ and}}\,\,\,t = \frac{1}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & g\left( t \right) = \ln \left( {3{t^2} + 1} \right) \cr & \cr & {\text{Calculate the second derivative}} \cr & g'\left( t \right) = \frac{d}{{dt}}\left[ {\ln \left( {3{t^2} + 1} \right)} \right] \cr & g'\left( t \right) = \frac{{6t}}{{3{t^2} + 1}} \cr & g''\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{6t}}{{3{t^2} + 1}}} \right] \cr & g''\left( t \right) = \frac{{6\left( {3{t^2} + 1} \right) - 6t\left( {6t} \right)}}{{{{\left( {3{t^2} + 1} \right)}^2}}} \cr & g''\left( t \right) = \frac{{18{t^2} + 6 - 36{t^2}}}{{{{\left( {3{t^2} + 1} \right)}^2}}} \cr & g''\left( t \right) = \frac{{6 - 18{t^2}}}{{{{\left( {3{t^2} + 1} \right)}^2}}} \cr & {\text{Set the second derivative to 0}} \cr & g''\left( t \right) = 0 \cr & 6 - 18{t^2} = 0 \cr & {t^2} = \frac{1}{3} \cr & {\text{We see that }}\,g''\left( t \right) = 0{\text{ at }}t = - \frac{1}{{\sqrt 3 }}{\text{ and}}\,\,\,t = \frac{1}{{\sqrt 3 }} \cr & {\text{These points are candidates for inflection points}} \cr & {\text{We need evaluate the intervals }} \cr & \left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right),\,\,\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right){\text{ and }}\left( {\frac{1}{{\sqrt 3 }},\infty } \right) \cr & \cr & {\text{Now}}{\text{, we will evaluate test values and resume in a table}} \cr & {\text{to determinate whether the concavity changes at these points}} \cr} $$ \[\begin{array}{*{20}{c}} {{\text{Interval}}}&{{\text{Test value }}\left( t \right)}&{{\text{Sign of }}g''\left( t \right)}&{{\text{Behavior of }}g\left( t \right)} \\ {\left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right)}&{ - 1}&{g''\left( { - 1} \right) < 0}&{{\text{Concave down}}} \\ {\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)}&0&{g''\left( 0 \right) > 0}&{{\text{Concave up}}} \\ {\left( {\frac{1}{{\sqrt 3 }},\infty } \right)}&1&{g''\left( { - 1} \right) < 0}&{{\text{Concave down}}} \end{array}\] $$\eqalign{ & {\text{From the table we can conclude that the function is:}} \cr & {\text{Concave down on }}\left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right){\text{ and }}\left( {\frac{1}{{\sqrt 3 }},\infty } \right) \cr & {\text{Concave up on }}\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right) \cr & {\text{The inflection points are at: }}t = - \frac{1}{{\sqrt 3 }}{\text{ and}}\,\,\,t = \frac{1}{{\sqrt 3 }} \cr} $$