If the solve the problem


If $f:[-5,5] \rightarrow R$ is differentiable and if $f(x)$ doesnot vanish anywhere, then prove that $f(-5) \pm f(5)$


It is given that $f:[-5,5] \rightarrow \mathbf{R}$ is a differentiable function.

Every differentiable function is a continuous function. Thus,

(a) $f$ is continuous in $[-5,5]$.

(b) $f$ is differentiable in $(-5,5)$.

Therefore, by the Mean Value Theorem, there exists $c \in(-5,5)$ such that


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

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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