**Question:**

If $f:[-5,5] \rightarrow \mathbf{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5)$.

**Solution:**

It is given that $f:[-5,5] \rightarrow \mathbf{R}$ is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) *f* is continuous on [−5, 5].

(b) *f *is differentiable on (−5, 5).

Therefore, by the Mean Value Theorem, there exists *c* ∈ (−5, 5) such that

$f^{\prime}(c)=\frac{f(5)-f(-5)}{5-(-5)}$

$\Rightarrow 10 f^{\prime}(c)=f(5)-f(-5)$

It is also given that $f^{\prime}(x)$ does not vanish anywhere.

$\therefore f^{\prime}(c) \neq 0$

$\Rightarrow 10 f^{\prime}(c) \neq 0$

$\Rightarrow f(5)-f(-5) \neq 0$

$\Rightarrow f(5) \neq f(-5)$

Hence, proved.