Answer
\[\begin{align}
& \left( \mathbf{a} \right)\text{increasing on the interval }\left[ 0,+\infty \right) \\
& \left( \mathbf{b} \right)\text{decreasing on the interval }\left( -\infty ,0 \right] \\
& \left( \mathbf{c} \right)\text{concave upward on the interval }\left( -\infty ,1 \right),\left( \frac{3}{2},+\infty \right) \\
& \left( \mathbf{d} \right)\text{concave downward on the interval }\left( 1,\frac{3}{2} \right) \\
& \left( \mathbf{e} \right)\text{inflection points at }x=0,\text{ }x=\frac{2}{3} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{x}^{4}}-5{{x}^{3}}+9{{x}^{2}} \\
& \text{The domain of the function is }\left( -\infty ,\infty \right) \\
& \text{Calculate the first and second derivatives} \\
& f'\left( x \right)=\frac{d}{dx}\left[ {{x}^{4}}-5{{x}^{3}}+9{{x}^{2}} \right] \\
& f'\left( x \right)=4{{x}^{3}}-15{{x}^{2}}+18x \\
& \text{Find the critical points, set }f'\left( x \right)=0 \\
& f'\left( x \right)=0 \\
& 4{{x}^{3}}-15{{x}^{2}}+18x=0 \\
& x=0 \\
& \\
& f''\left( x \right)=\frac{d}{dx}\left[ 4{{x}^{3}}-15{{x}^{2}}+18x \right] \\
& f''\left( x \right)=12{{x}^{2}}-30x+18 \\
& 12{{x}^{2}}-30x+18=0 \\
& {{x}_{1}}=\frac{3}{2},\text{ }{{x}_{2}}=1 \\
& \text{We obtain the sign analysis shown in the following tables} \\
& \begin{matrix}
\text{Interval} & \left( -\infty ,0 \right) & \left( 0,\infty \right) \\
\text{Test Value} & x=-1 & x=1 \\
\text{Sign of }f'\left( x \right) & - & + \\
\text{Conclusion} & \text{Decreasing} & \text{Increasing} \\
\end{matrix} \\
& \\
& \begin{matrix}
\text{Interval} & \left( -\infty ,1 \right) & \left( 1,\frac{3}{2} \right) & \left( \frac{3}{2},+\infty \right) \\
\text{Test Value} & 0 & 5/4 & 2 \\
\text{Sign of }f''\left( x \right) & + & - & + \\
\text{Conclusion} & \text{C}\text{. upward} & \text{C}\text{. downward} & \text{C}\text{. upward} \\
\end{matrix} \\
& \\
& \text{Summary:} \\
& \left( \mathbf{a} \right)\text{ }f\left( x \right)\text{ is increasing on the interval }\left[ 0,+\infty \right) \\
& \left( \mathbf{b} \right)\text{ }f\left( x \right)\text{ is decreasing on the interval }\left( -\infty ,0 \right] \\
& \left( \mathbf{c} \right)\text{ }f\left( x \right)\text{ is concave upward on the interval }\left( -\infty ,1 \right),\left( \frac{3}{2},+\infty \right) \\
& \left( \mathbf{d} \right)\text{ }f\left( x \right)\text{ is concave downward on the interval }\left( 1,\frac{3}{2} \right) \\
& \left( \mathbf{e} \right)\text{ Inflection points at }x=0,\text{ }x=\frac{2}{3} \\
\end{align}\]