Chapter 3 - The Derivative In Graphing And Applications - 3.1 Analysis Of Functions I: Increase, Decrease, and Concavity - Exercises Set 3.1 - Page 195: 21

\[\begin{align} & \left( \mathbf{a} \right)\text{increasing on the interval }\left[ \frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2} \right] \\ & \left( \mathbf{b} \right)\text{decreasing on the intervals }\left( -\infty ,\frac{3-\sqrt{5}}{2} \right],\text{ } \\ & \left[ \frac{3+\sqrt{5}}{2},+\infty \right) \\ & \left( \mathbf{c} \right)\text{concave upward on the intervals }\left( 0,\frac{4-\sqrt{6}}{2} \right), \\ & \left( \frac{4+\sqrt{6}}{2},\infty \right) \\ & \left( \mathbf{d} \right)\text{concave downward on the intervals }\left( -\infty ,0 \right), \\ & \left( \frac{4-\sqrt{6}}{2},\frac{4+\sqrt{6}}{2} \right) \\ & \left( \mathbf{e} \right)\text{inflection points at }x=0,\text{ }x=\frac{4-\sqrt{6}}{2},\text{ }x=\frac{4+\sqrt{6}}{2} \\ \end{align}\]

\[\begin{align} & f\left( x \right)=\frac{x-2}{{{\left( {{x}^{2}}-x+1 \right)}^{2}}} \\ & {{x}^{2}}-x+1=0\text{ has complex solutions, then} \\ & \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-2}{{{\left( {{x}^{2}}-x+1 \right)}^{2}}} \right] \\ & f'\left( x \right)=\frac{{{\left( {{x}^{2}}-x+1 \right)}^{2}}-\left( x-2 \right)\left( 2 \right)\left( {{x}^{2}}-x+1 \right)\left( 2x-1 \right)}{{{\left( {{x}^{2}}-x+1 \right)}^{4}}} \\ & f'\left( x \right)=\frac{{{x}^{2}}-x+1-2\left( x-2 \right)\left( 2x-1 \right)}{{{\left( {{x}^{2}}-x-1 \right)}^{3}}} \\ & f'\left( x \right)=\frac{{{x}^{2}}-x+1-4{{x}^{2}}+10x-4}{{{\left( {{x}^{2}}-x-1 \right)}^{3}}} \\ & f'\left( x \right)=\frac{-3{{x}^{2}}-9x-3}{{{\left( {{x}^{2}}-x-1 \right)}^{3}}} \\ & \text{Find the critical points, set }f'\left( x \right)=0 \\ & f'\left( x \right)=0 \\ & -3{{x}^{2}}-9x-3=0 \\ & \text{By the quadratic formula we obtain} \\ & {{x}_{1}}=\frac{3-\sqrt{5}}{2},\text{ }{{x}_{2}}=\frac{3+\sqrt{5}}{2} \\ & \text{Interval analysis }\left( -\infty ,\frac{3-\sqrt{5}}{2} \right),\text{ }\left( \frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2} \right),\text{ }\left( \frac{3+\sqrt{5}}{2},\infty \right) \\ & f''\left( x \right)=\frac{d}{dx}\left[ \frac{-3{{x}^{2}}-9x-3}{{{\left( {{x}^{2}}-x-1 \right)}^{3}}} \right] \\ & \text{Differentiate using wolfram alpha website} \\ & f''\left( x \right)=\frac{6x\left( 2{{x}^{2}}-8x+5 \right)}{{{\left( {{x}^{2}}-x-1 \right)}^{4}}} \\ & f''\left( x \right)=0 \\ & 6x\left( 2{{x}^{2}}-8x+5 \right)=0 \\ & x=0 \\ & 2{{x}^{2}}-8x+5=0 \\ & \text{By the quadratic formula we obtain} \\ & x=\frac{4-\sqrt{6}}{2},\text{ }x=\frac{4+\sqrt{6}}{2} \\ & \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 ,\frac{3-\sqrt{5}}{2} \right) & 0 & - & \text{Decreasing} \\ \left( \frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2} \right) & 2 & + & \text{Increasing} \\ \left( \frac{3+\sqrt{5}}{2},+\infty \right) & 3 & - & \text{Decreasing} \\ \end{matrix} \\ & and \\ & \begin{matrix} \text{Interval} & \text{Test Value} & \text{Sign of }f''\left( x \right) & \text{Conclusion} \\ \left( -\infty ,0 \right) & -1 & - & \text{C}\text{. downward} \\ \left( 0,\frac{4-\sqrt{6}}{2} \right) & 1/2 & + & \text{C}\text{. upward} \\ \left( \frac{4-\sqrt{6}}{2},\frac{4+\sqrt{6}}{2} \right) & 3 & - & \text{C}\text{. downward} \\ \left( \frac{4+\sqrt{6}}{2},\infty \right) & 4 & + & \text{C}\text{. upward} \\ \end{matrix} \\ & \text{Summary:} \\ & \left( \mathbf{a} \right)\text{ }f\left( x \right)\text{ is increasing on the interval }\left[ \frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2} \right] \\ & \left( \mathbf{b} \right)\text{ }f\left( x \right)\text{ is decreasing on the intervals }\left( -\infty ,\frac{3-\sqrt{5}}{2} \right],\text{ } \\ & \left[ \frac{3+\sqrt{5}}{2},+\infty \right) \\ & \left( \mathbf{c} \right)\text{ }f\left(x\right)\text{ is concave upward on the intervals }\left( 0,\frac{4-\sqrt{6}}{2} \right), \\ & \left( \frac{4+\sqrt{6}}{2},\infty \right) \\ & \left( \mathbf{d} \right)\text{ }f\left( x \right)\text{ is concave downward on the intervals }\left( -\infty ,0 \right), \\ & \left( \frac{4-\sqrt{6}}{2},\frac{4+\sqrt{6}}{2} \right) \\ & \left( \mathbf{e} \right)\text{ Inflection points at }x=0,\text{ }x=\frac{4-\sqrt{6}}{2},\text{ }x=\frac{4+\sqrt{6}}{2} \\ \end{align}\]