Chapter 3 - The Derivative In Graphing And Applications - 3.1 Analysis Of Functions I: Increase, Decrease, and Concavity - Exercises Set 3.1 - Page 195: 22

\[\begin{align} & \left( \mathbf{a} \right)\text{increasing on the interval }\left[ -\sqrt{2},\sqrt{2} \right] \\ & \left( \mathbf{b} \right)\text{decreasing on the intervals }\left( -\infty ,-\sqrt{2} \right],\text{ } \\ & \left[ \sqrt{2},\infty \right) \\ & \left( \mathbf{c} \right)\text{concave upward on the intervals }\left( -\sqrt{6},0 \right), \\ & \left( -\sqrt{6},\infty \right) \\ & \left( \mathbf{d} \right)\text{concave downward on the intervals }\left( -\infty ,-\sqrt{6} \right), \\ & \left( 0,\sqrt{6} \right) \\ & \left( \mathbf{e} \right)\text{inflection points at }x=0,\text{ }x=\pm \sqrt{6} \\ \end{align}\]

\[\begin{align} & f\left( x \right)=\frac{x}{{{x}^{2}}+2} \\ & \text{The domain of the function is }\left( -\infty ,\infty \right) \\ & \text{Calculate the first and second derivatives} \\ & f'\left( x \right)=\frac{d}{dx}\left[ \frac{x}{{{x}^{2}}+2} \right] \\ & f'\left( x \right)=\frac{{{x}^{2}}+2-x\left( 2x \right)}{{{\left( {{x}^{2}}+2 \right)}^{2}}} \\ & f'\left( x \right)=\frac{{{x}^{2}}+2-2{{x}^{2}}}{{{\left( {{x}^{2}}+2 \right)}^{2}}} \\ & f'\left( x \right)=\frac{2-{{x}^{2}}}{{{\left( {{x}^{2}}+2 \right)}^{2}}} \\ & \text{Find the critical points, set }f'\left( x \right)=0 \\ & f'\left( x \right)=0 \\ & 2-{{x}^{2}}=0 \\ & {{x}_{1}}=-\sqrt{2},\text{ }{{x}_{2}}=\sqrt{2} \\ & \text{Interval analysis }\left( -\infty ,-\sqrt{2} \right),\text{ }\left( -\sqrt{2},\sqrt{2} \right),\text{ }\left( \sqrt{2},\infty \right) \\ & f''\left( x \right)=\frac{d}{dx}\left[ \frac{2-{{x}^{2}}}{{{\left( {{x}^{2}}+2 \right)}^{2}}} \right] \\ & f''\left( x \right)=\frac{{{\left( {{x}^{2}}+2 \right)}^{2}}\left( -2x \right)-\left( 2-{{x}^{2}} \right)\left( 2 \right)\left( {{x}^{2}}+2 \right)\left( 2x \right)}{{{\left( {{x}^{2}}+2 \right)}^{4}}} \\ & f''\left( x \right)=\frac{\left( {{x}^{2}}+2 \right)\left( -2x \right)-\left( 2-{{x}^{2}} \right)\left( 2 \right)\left( 2x \right)}{{{\left( {{x}^{2}}+2 \right)}^{3}}} \\ & f''\left( x \right)=\frac{-2{{x}^{3}}-4x-8x+4{{x}^{3}}}{{{\left( {{x}^{2}}+2 \right)}^{3}}} \\ & f''\left( x \right)=\frac{2{{x}^{3}}-12x}{{{\left( {{x}^{2}}+2 \right)}^{3}}} \\ & f''\left( x \right)=0 \\ & 2{{x}^{3}}-12x=0 \\ & 2x\left( {{x}^{2}}-6 \right)=0 \\ & x=0,\text{ }x=-\sqrt{6},\text{ }x=\sqrt{6} \\ & \text{We obtain the sign analysis shown in the following tables} \\ & \begin{matrix} \text{Interval} & \text{Test Value} & \text{Sign of }f'\left( x \right) & \text{Conclusion} \\ \left( -\infty ,-\sqrt{2} \right) & -2 & - & \text{Decreasing} \\ \left( -\sqrt{2},\sqrt{2} \right) & 0 & + & \text{Increasing} \\ \left( \sqrt{2},\infty \right) & 2 & - & \text{Decreasing} \\ \end{matrix} \\ & and \\ & \begin{matrix} \text{Interval} & \text{Test Value} & \text{Sign of }f''\left( x \right) & \text{Conclusion} \\ \left( -\infty ,-\sqrt{6} \right) & -3 & - & \text{C}\text{. downward} \\ \left( -\sqrt{6},0 \right) & -2 & + & \text{C}\text{. upward} \\ \left( 0,\sqrt{6} \right) & 2 & - & \text{C}\text{. downward} \\ \left( -\sqrt{6},\infty \right) & 3 & + & \text{C}\text{. upward} \\ \end{matrix} \\ & \text{Summary:} \\ & \left( \mathbf{a} \right)\text{ }f\left( x \right)\text{ is increasing on the interval }\left[ -\sqrt{2},\sqrt{2} \right] \\ & \left( \mathbf{b} \right)\text{ }f\left( x \right)\text{ is decreasing on the intervals }\left( -\infty ,-\sqrt{2} \right],\text{ } \\ & \left[ \sqrt{2},\infty \right) \\ & \left( \mathbf{c} \right)\text{ }f\left( x \right)\text{ is concave upward on the intervals }\left( -\sqrt{6},0 \right), \\ & \left( -\sqrt{6},\infty \right) \\ & \left( \mathbf{d} \right)\text{ }f\left( x \right)\text{ is concave downward on the intervals }\left( -\infty ,-\sqrt{6} \right), \\ & \left( 0,\sqrt{6} \right) \\ & \left( \mathbf{e} \right)\text{ Inflection points at }x=0,\text{ }x=\pm \sqrt{6} \\ \end{align}\]