The sum of all the local minimum values of the twice differentiable function $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ defined by $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3{\mathrm{x}}^2-\frac{3\mathrm{f}\text{'}\text{'}\left(2\right)}{2}\mathrm{x}+\mathrm{f}\text{'}\text{'}\left(1\right)$ is

Given,

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3{\mathrm{x}}^2-\frac{3}{2}\mathrm{f}\text{'}\text{'}\left(2\right)\cdot \mathrm{x}+\mathrm{f}\text{'}\text{'}\left(1\right)$

Then, $\mathrm{f}\text{'}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-6\mathrm{x}-\frac{3}{2}\mathrm{f}\text{'}\text{'}\left(2\right)$

$\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)=6\mathrm{x}-6$

Put x = 1and x= 2,

$\mathrm{f}\text{'}\text{'}\left(2\right)=12-6=6\text{ and }\mathrm{f}\text{'}\text{'}\left(1\right)=0$

Therefore, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3{\mathrm{x}}^2-\frac{3}{2}\times 6\mathrm{x}+0$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3{\mathrm{x}}^2-9\mathrm{x}$           ...(i)

and $\mathrm{f}\text{'}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-6\mathrm{x}-9\text{ and }\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)=6\mathrm{x}-6$

Equate $\mathrm{f}\text{'}\left(\mathrm{x}\right)=0$ gives,

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3{\mathrm{x}}^2-6\mathrm{x}-9=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\left({\mathrm{x}}^2-2\mathrm{x}-3\right)=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3(\mathrm{x}+1)(\mathrm{x}-3)=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}=-1,\mathrm{x}=3\end{array}$

Now, $\mathrm{f}\text{'}\text{'}(-1)=6(-1)-6=-12<0$

$\mathrm{f}\text{'}\text{'}\left(3\right)=6\left(3\right)-6=12>0$

$\therefore (-1)$ is maxima. 3 is minima.

Local minimum value = f(3)

$\begin{array}{l}\mathrm{f}\left(3\right)=(3{)}^3-3(3{)}^2-9\left(3\right) [\mathrm{using} \mathrm{Eq}. (\mathrm{i}\left)\right]\\ =27-27-27\\ \mathrm{f}\left(3\right)=-27\end{array}$