Question
If $\text{f}:[-5,5]\rightarrow\text{R}$ is differentiable and if f'(x) does not vanish anywhere, then prove that $\text{f}(-5)\pm\text{f}(5).$

Answer

It is given that $\text{f}:[-5,5]\rightarrow\text{R}$ is a differentiable function.
Since every differentiable function is continuous function, we obtain
  1. f is continuous on [-5, 5].
  2. f is differentiable on (-5, 5).
Therefore, by the Mean Value Theorem, there exists $\text{c}\in(-5,5)$ such that

$\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-5)}{5-(-5)}$

$\Rightarrow10\text{f}'(\text{c})=\text{f}(5)-f(-5)$

It is also given that f'(x) does not vanish anywhere.

$\therefore\ \text{f}'(\text{c})\neq0$

$\Rightarrow10\text{f}'(\text{c})\neq0$

$\Rightarrow\text{f}(5)-\text{f}(-5)\neq0$

$\Rightarrow\text{f}(5)\neq\text{f}(-5)$

Hence, proved.

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