Since $f(x)$ has local maxima at $x=-1$ and $f^{\prime}(x)$ has local minima at $x=0$.
$$
\begin{aligned}
f \wedge(x) & =\lambda x \\
f^{\prime}(x) & =\lambda \frac{x^2}{2}+c \\
\frac{\lambda}{2}+c & =0 \Rightarrow \lambda=-2 c
\end{aligned}
$$
$$
\left[f^{\prime}(-1)=0\right]
$$
Again, On integrating both sides, we get
$$
\begin{aligned}
& f(x)=\lambda \frac{x^3}{6}+c x+d \\
& f(2)=\lambda\left(\frac{8}{6}\right)+2 c+d=18 \\
& f(1)=\frac{\lambda}{6}+c+d=-1
\end{aligned}
$$
and
$\therefore$ Using Eqs. (i), (ii) and (iii), we get
$$
\begin{aligned}
f(x) & =\frac{1}{4}\left(19 x^3-57 x+34\right) \\
\therefore \quad f^{\prime}(x) & =\frac{1}{4}\left(57 x^2-57\right) \\
& =\frac{57}{4}(x-1)(x+1), \text { using number line rule }
\end{aligned}
$$
$\therefore f(x)$ is increasing for $[1,2 \sqrt{5}]$ and $f(x)$ is increasing for $[1,2 \sqrt{5}]$ and $f(x)$ has local maximum at $x=-1$ and $f(x)$ has local minimum at $C x=1$.
Also,
$$
f(0)=\frac{34}{4} \text {. }
$$