The maximum and minimum values of $x^3-18x^2+96$ in interval $(0,9)$ are:

Let $y=x^3-18x^2+96x$

$\begin{array}{l}\Rightarrow \frac{dy}{dx}=3x^2-36x+96=0\\ \therefore x^2-12x+32=0\\ \Rightarrow (x-4)(x-8)=0,x=4,8\end{array}$

Now, $\frac{d^{2}y}{d{x}^2}=6x-36\text{ at }x=4,\frac{d^{2}y}{d{x}^2}=24-36=-12<0$

$\therefore$ at $x=4$ function will be maximum and

$\begin{array}{l}\left[f\right(x){]}_{max.}=64-288+384=160\text{ at }\\ x=8\frac{d^{2}y}{d{x}^2}=48-36=12>0\end{array}$

$\therefore$ at $x=8$ function will be minimum and $\left[f\right(x){]}_{min}=128.$