Question

If $f:[-5,5]\to R$ is a differentiable function and if $f^{\prime }\left(x\right)$ does not vanish anywhere, prove that $f(-5)\ne f\left(5\right)$.

Answer 1

Given: $f:[-5,5]\to R$ is a differentiable function and if $f^{\prime }\left(x\right)$ does not vanish anywhere
To prove: $f(-5)\ne f\left(5\right)$.

We know that every differentiable function is continuous. Therefore, $f$ is continuous & differentiable in $(-5,5)$.

By Mean Value Theorem there exist some $c$ in $(-5,5)$ such that

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
It is given that $f^{\prime }\left(x\right)$ does not vanish anywhere.

$\Rightarrow f^{\prime }\left(x\right)\ne 0$ for any value of $x$
Thus, $f^{\prime }\left(c\right)\ne 0$

$\Rightarrow \frac{f\left(5\right)-f(-5)}{5-(-5)}\ne 0$

$\Rightarrow \frac{f\left(5\right)-f(-5)}{5+5}\ne 0$

$\Rightarrow f\left(5\right)\ne f(-5)$

Hence proved