# The sum of all the local minimum

Question:

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

1. -22

2. 5

3. -27

4. 0

Correct Option: , 3

Solution:

$f(x)=x^{3}-3 x^{2}-\frac{3}{2} f^{\prime \prime}(2) x+f^{\prime \prime}(1) \ldots(1)$

$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2} f^{\prime \prime}(2) \ldots(2)$

$f^{\prime \prime}(x)=6 x-6$..(3)

Now is $3^{\text {rd }}$ equation

$f^{\prime \prime}(2)=12-6=6$

$f^{\prime \prime}(11=0)$

Use (ii)

$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2} f^{\prime \prime}(2)$

$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2} \times 6$

$f^{\prime}(x)=3 x^{2}-6 x-9$

$f^{\prime}(x)=0$

$3 x^{2}-6 x-9=0$

$\Rightarrow x=-1 \& 3$

Use (iii)

$f^{\prime \prime}(x)=6 x-6$

$f^{\prime \prime}(-1)=-12<0$ maxima

$f^{\prime}(3)=12>0$ minima.

Use (i)

$f(x)=x^{3}-3 x^{2}-\frac{3}{2} f^{\prime \prime}(2) x+f^{\prime \prime}(1)$

$f(x)=x^{3}-3 x^{2}-\frac{3}{2} \times 6 \times x+0$

$f(x)=x^{3}-3 x^{2}-9 x$

$f(3)=27-27-9 \times 3=-27$