Given, $f(x)=x^{3}-6 x^{2}+9 x+15$
Differentiate with respect to $x$, we get, $f^{\prime}(x)=3 x^{2}-12 x+9=3\left(x^{2}-4 x+3\right)$ $=3(x-3)(x-1)$
For all maxima and minima,
$
\begin{array}{l}
f^{\prime}(x)=0 \\
=3(x-3)(x-1)=0 \\
=x=3,1
\end{array}
$
At $x=1, f^{\prime}(x)$ changes from positive to negative
Since, $x=1$ is a point of Maxima
At $x=3, f^{\prime}(x)$ changes from negative to positive
Since, $x=3$ is point of Minima.
Hence, local maxima value $f(1)=(1)^{3}-6(1)^{2}+9(1)+15=19$
Local minima value $f(3)=(3)^{3}-6(3)^{2}+9(3)+15=15$