If \(f(x)\) is cubic polynomial which has local maximum at \(x=-1\). If \(f(2)=18, f(1)=-1\) and \(f^{\prime}(x)\) has local minimum at \(x=0\), then

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,
\[
f(0)=\frac{34}{4} \text {. }
\]