Chapter 3 - Applications of Differentiation - 3.4 Exercises - Page 192: 23

$$\eqalign{ & {\text{Inflection points: }}\left( { - \frac{1}{{\sqrt 3 }},3} \right){\text{ and }}\left( {\frac{1}{{\sqrt 3 }},3} \right) \cr & {\text{Concave downward}}:{\text{ }}\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right) \cr & {\text{Concave upward}}:{\text{ }}\left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right){\text{ and }}\left( { \frac{1}{{\sqrt 3 }},\infty } \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{4}{{{x^2} + 1}} \cr & {\text{Calculate the second derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{4}{{{x^2} + 1}}} \right] \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {4{{\left( {{x^2} + 1} \right)}^{ - 1}}} \right] \cr & f'\left( x \right) = - 4{\left( {{x^2} + 1} \right)^{ - 2}}\left( {2x} \right) \cr & f'\left( x \right) = \frac{{ - 8x}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr & f''\left( x \right) = \underbrace {\frac{d}{{dx}}\left[ {\frac{{ - 8x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right]}_{{\text{Quotient rule}}} \cr & f''\left( x \right) = \frac{{{{\left( {{x^2} + 1} \right)}^2}\left( { - 8} \right) + 8x\left( 2 \right)\left( {{x^2} + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}} \cr & f''\left( x \right) = \frac{{ - 8{{\left( {{x^2} + 1} \right)}^2} + 32{x^2}\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}} \cr & f''\left( x \right) = \frac{{ - 8\left( {{x^2} + 1} \right) + 32{x^2}}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & f''\left( x \right) = \frac{{ - 8{x^2} - 8 + 32{x^2}}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & f''\left( x \right) = \frac{{24{x^2} - 8}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & f''\left( x \right) = \frac{{8\left( {3{x^2} - 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & {\text{Set }}f''\left( x \right) = 0 \cr & \frac{{8\left( {3{x^2} - 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} = 0 \cr & 3{x^2} - 1 = 0 \cr & x = \pm \frac{1}{{\sqrt 3 }} \cr & {\text{Set the intervals }}\left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right),\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right),\left( { \frac{1}{{\sqrt 3 }},\infty } \right) \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$ \[\boxed{\begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right)}&{\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)}&{\left( { \frac{1}{{\sqrt 3 }},\infty } \right)} \\ {{\text{Test Value}}}&{x = - 1}&{x = 0}&{x = 1} \\ {{\text{Sign of }}f''\left( x \right)}&{2 > 0}&{ - 8 < 0}&{2 > 0} \\ {{\text{Conclusion}}}&{{\text{C}}{\text{. upward}}}&{{\text{C}}{\text{. downward}}}&{{\text{C}}{\text{. upward}}} \end{array}}\] $$\eqalign{ & {\text{The inflection points occur at }}x = - \frac{1}{{\sqrt 3 }}{\text{ and }}x = \frac{1}{{\sqrt 3 }} \cr & f\left( { - \frac{1}{{\sqrt 3 }}} \right) = \frac{4}{{{{\left( { - 1/\sqrt 3 } \right)}^2} + 1}} = 3 \to \left( { - \frac{1}{{\sqrt 3 }},3} \right) \cr & f\left( {\frac{1}{{\sqrt 3 }}} \right) = \frac{4}{{{{\left( {1/\sqrt 3 } \right)}^2} + 1}} = 3 \to \left( {\frac{1}{{\sqrt 3 }},3} \right) \cr & {\text{Inflection points: }}\left( { - \frac{1}{{\sqrt 3 }},3} \right){\text{ and }}\left( {\frac{1}{{\sqrt 3 }},3} \right) \cr & {\text{Concave downward}}:{\text{ }}\left( { - \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right) \cr & {\text{Concave upward}}:{\text{ }}\left( { - \infty , - \frac{1}{{\sqrt 3 }}} \right){\text{ and }}\left( { \frac{1}{{\sqrt 3 }},\infty } \right) \cr} $$