Medium
If $f:[-5,5] \rightarrow R$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere,then prove that $f(-5) \neq f(5).$
(N/A) 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 differentiable 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,which means $f^{\prime}(x) \neq 0$ for all $x \in [-5,5].$
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,it is proved that $f(-5) \neq f(5).$
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 $c \in (-5,5)$ 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,which means $f^{\prime}(x) \neq 0$ for all $x \in [-5,5].$
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,it is proved that $f(-5) \neq f(5).$
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