Question

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

Answer 1

It is given that $f\left(x\right)$ is a differentiable function, so it is clear that it is also a continuous function.

Now let's apply mean value theorem.
So as per the theorem, there exist a $c$ $ϵ$ $(-5,5)$ such that

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

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

Given: $f^{{}^{\prime }}\left(c\right)\ne 0$

$\Rightarrow 10f^{{}^{\prime }}\left(c\right)\ne 0$

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

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

Hence proved.