Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 60

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

$$\eqalign{ & f\left( x \right) = \frac{1}{{1 + {x^2}}} \cr & \cr & {\text{Calculate the second derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{1 + {x^2}}}} \right] \cr & f'\left( x \right) = - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \right] \cr & f''\left( x \right) = - \frac{{2{{\left( {1 + {x^2}} \right)}^2} - 2x\left( 2 \right)\left( {1 + {x^2}} \right)\left( {2x} \right)}}{{{{\left( {1 + {x^2}} \right)}^4}}} \cr & f''\left( x \right) = - \frac{{2\left( {1 + {x^2}} \right) - 2x\left( 2 \right)\left( {2x} \right)}}{{{{\left( {1 + {x^2}} \right)}^3}}} \cr & f''\left( x \right) = - \frac{{2 + 2{x^2} - 8{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}} \cr & f''\left( x \right) = \frac{{6{x^2} - 2}}{{{{\left( {1 + {x^2}} \right)}^3}}} \cr & \cr & {\text{Set the second derivative to 0}} \cr & \frac{{6{x^2} - 2}}{{{{\left( {1 + {x^2}} \right)}^3}}} = 0 \cr & 6{x^2} - 2 = 0 \cr & 3{x^2} - 1 = 0 \cr & {\text{We see that }}\,f''\left( x \right) = 0{\text{ at }}x = - \frac{1}{{\sqrt 3 }}{\text{ and}}\,\,\,x = \frac{1}{{\sqrt 3 }} \cr & {\text{These points are candidates for inflection points}} \cr & {\text{We need evaluate the intervals }} \cr & \left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right),\,\,\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right){\text{ and }}\left( {\frac{1}{{\sqrt 3 }},\infty } \right) \cr & \cr & {\text{Now}}{\text{, we will evaluate test values and resume in a table}} \cr & {\text{to determinate whether the concavity changes at these points}} \cr} $$ \[\begin{array}{*{20}{c}} {{\text{Interval}}}&{{\text{Test value }}\left( x \right)}&{{\text{Sign of }}f''\left( x \right)}&{{\text{Behavior of }}f\left( x \right)} \\ {\left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right)}&{ - 1}&{f''\left( { - 1} \right) > 0}&{{\text{Concave up}}} \\ {\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)}&0&{f''\left( 0 \right) < 0}&{{\text{Concave down}}} \\ {\left( {\frac{1}{{\sqrt 3 }},\infty } \right)}&1&{f''\left( { - 1} \right) > 0}&{{\text{Concave up}}} \end{array}\] $$\eqalign{ & {\text{From the table we can conclude that the function is:}} \cr & {\text{Concave up on }}\left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right){\text{ and }}\left( {\frac{1}{{\sqrt 3 }},\infty } \right) \cr & {\text{Concave down on }}\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right) \cr & {\text{The inflection points are at: }}x = - \frac{1}{{\sqrt 3 }}{\text{ and}}\,\,\,x = \frac{1}{{\sqrt 3 }} \cr} $$