Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14 point 1 question 8 sub question (iii)

$\left ( \frac{1}{2}, -27 \right )$

Hint:

$f' (x) = 124x -12$ exist$(-1,2)$

Given:

$12(x+1)(x-2), x \in[-1,2]$

Explanation:

$f(x) = 12(x+1)(x-2), x \in[-1,2]$

1. Being polynomial f(x) is continuous for all x and hence continuous in$[-1,2]$.

2.

$\\ f^{\prime}(x)=12[(x+1)(1)+(x-2)(1)] \\\\ f^{\prime}(x)=12[x+1+x-2] \\\\ f^{\prime}(x)=12[2 x-1]=24 x-12$

$\therefore f(x)$ is derivable in

3.

$\\ f(-1)=12(-1+1)(-1-2)=0\\\\ f(2)=12(2+1)(2-2)=0\\\\ \therefore f(-1)=f(2)$

Thus all conditions of Rolle’s Theorem are satisfied.

Therefore, there exists at least one$c \in(-1,2)$ such that$f'(c) = 0$

Now $f'(c) = 0$

$\Rightarrow 24 c-12=0 \\\\ \Rightarrow 24 c=12 \\\\ \Rightarrow \quad c=\frac{12}{24}=\frac{1}{2} \\\\ \Rightarrow f\left(\frac{1}{2}\right)=12\left(\frac{1}{2}+1\right)\left(\frac{1}{2}-2\right) \\\\ \Rightarrow f\left(\frac{1}{2}\right)=12\left(\frac{3}{2}\right)\left(\frac{-3}{2}\right) \\\\ f\left (\frac{1}{2} \right ) = -27$

Therefore, the tangent is parallel to x-axis and the point will be$\left ( \frac{1}{2}, -27 \right )$