If $f:[-5,5] \rightarrow \mathrm{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5).$
If $f:[-5,5] \rightarrow \mathrm{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5).$
It is given that $f:[-5,5] \rightarrow R$ is a differentiable function.
Since every differentiable function is a continuous function, we obtain
a) $f$ is continuous on $[-5,5].$
b) $f$ is continuous on $(-5,5).$
Therefore, by the Mean Value Theorem, there exists $c \in(-5,5)$ 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.
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