(i)

Given $f(x)=x^{4}-62 x^{2+} 120 x+9$
$
\begin{array}{l}
\therefore \mathrm{f}(\mathrm{x})=4 \mathrm{x}^{\mathrm{a}}-124 x+120=4\left(\mathrm{x}^{a}-31 \mathrm{x}+30\right) \\
\mathrm{f}^{\prime}(\mathrm{x})=12 \mathrm{x}^{2}-124=4\left(3 x^{2}-31\right)
\end{array}
$
For maxima and minima, $f(x)=0$
$
4\left(x^{2}-31 x+30\right)=0
$
So roots will be $x=5,1,-6$
Now, $f^{\prime}(5)=176>0$
$\kappa=5$ is point of local minima
$
f^{\prime}(1)=-112<0
$
$x=1$ is point of local maxima
$
f^{\prime}(-6)=308>0
$
$x=-6$ is point of local minima
Local max value $=f(1)=68$
Local min value $=f(5)=-316$ and $f(-6)=-1647$

(ii)

(ii) Given $f(x)=x^{3}-6 x^{2}+9 x+15$
Differentiating $f$ with respect to $x$
$
\begin{array}{l}
f(x)=3 x^{2}-12 x+9=3\left(x^{2}-4 x+3\right) \\
f(x)=6 x-12=6(x-2)
\end{array}
$
For maxima and minima, $f(x)=0$
$
3\left(x^{2}-4 x+3\right)=0
$
Now, $f^{\prime}(3)=6>0$
$
f(1)=-6<0
$
$x=1$ is point of local maxima

Local maxima value $=f(1)=19$ and local minima value $=f(3)=15$