(i) Given $f(x)=(x-1)(x+2)^{2}$
$
\begin{array}{l}
\therefore \mathrm{f}(\mathrm{x})=(\mathrm{x}+2)^{2}+2(\mathrm{x}-1)(\mathrm{x}+2) \\
=(\mathrm{x}+2)(\mathrm{x}+2+2 \mathrm{x}-2) \\
=(\mathrm{x}+2)(3 \mathrm{x})
\end{array}
$
And $f^{\prime}(x)=3(x+2)+3 x$
$
=6 x+6
$
For maxima and minima, $f(x)=0$
$
(x+2)(3 x)=0
$
So roots will be $\mathrm{x}=0,-2$
Now, $f^{\prime}(0)=6>0$
$x=0$ is point of local minima
$
f(-2)=-6<0
$
$x=-2$ is point of local maxima
Local maxima value $=f(-2)=0$ and local minima value $=f(0)=-4$
(ii)
\[-2/{{x}^{2}}~+\text{ }4/{{x}^{3}}~=\text{ }0\]
\[-2\left( x\text{ }-\text{ }2 \right)/{{x}^{3}}~=\text{ }0\]
\[\Rightarrow \text{ }\left( x\text{ }-\text{ }2 \right)\text{ }=\text{ }0\]
\[\Rightarrow \text{ }x\text{ }=\text{ }2\]
Now,
\[f\left( 2 \right)\text{ }=\text{ }4/8\text{ }-\text{ }12/16\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }-\text{ }{\scriptscriptstyle 3\!/\!{ }_4}\text{ }=\text{ }-\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\text{ }<\text{ }0\]
