#### Explain solution for RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 9 sub question (i) math

$f(-5)=f(5)$

Hint:

f is continuous in$[-5,5]$

Given:

$f:[-5,5]\rightarrow R$ is differentiable

$\therefore f$ is continuous in  $[-5,5]$ of possible

Let         $f(-5)=f(5)$

Thus f satisfies all the conditions of Rolle’s Theorem

Therefore, there exists$c \in(-5,5)$ such that$f^{\prime}(c)=0$

But this contradicts the hypothesis that$f^{\prime}(x)$does not vanish anywhere.

Hence $f(-5)=f(5)$