Given, $f(x)=x^{2}-3 x$
Differentiate with respect to $x$ then we get,
$
f(x)=3 x^{2}-3
$
$\mathrm{Now}, \mathrm{f}(x)=0$
$
3 x^{2}=3 \Rightarrow x=\pm 1
$
Again differentiate $f(x)=3 x^{2}-3$
$
\begin{array}{l}
f^{\prime}(x)=b x \\
f^{\prime}(1)=6>0 \\
f^{\prime}(-1)=-6<0
\end{array}
$
By second derivative test, $x=1$ is a point of local minima and local minimum value off at $x=1$ is $f(1)=1^{2}-3=1-3=-2$
However, $x=-1$ is a point of local maxima and local maxima value offf at $x=-1$ is
$
\begin{array}{l}
f(-1)=(-1)^{3}-3(-1) \\
=-1+3 \\
=2
\end{array}
$
Hence, the value of minima is $-2$ and maxima is 2 .