Question

If $f:[-5,5]→R$ is a differentiable function and if $f'(x)$does not vanish anywhere, then prove that $f(-5)≠f(5)$

Answer 1

It is given that $f:[-5,5]→R$ 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
$f'(c)=\frac{f(5)-f(-5)}{5-(-5)}$
$⇒10f'(c)=f(5)-f(-5)$
It is also given that $f'(x)$ does not vanish anywhere
$∴f'(c)≠0$
$⇒10f'(c)≠0$
$⇒f(5)-f(-5)≠0$
$⇒f(-5)≠f(5)$
Hence, proved.