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If $f:[-5,5] \rightarrow R$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere then prove that $f(-5) \neq f(5)$.
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For Rolle's theorem
If $(\mathrm{i}) \mathrm{f}$ is continous in [ $\mathrm{a}, \mathrm{b}]$
(ii) $\mathrm{f}$ is derivable in [ $\mathrm{a}, \mathrm{b}]$
(iii) $f(a)=f(b)$
then $\mathrm{f}^{\prime}(\mathrm{c})=0, \mathrm{c} \in(\mathrm{a}, \mathrm{b})$
$\therefore \mathrm{f}$ is continuous and derivable
but $\mathrm{f}^{\prime}(\mathrm{c}) \neq 0 \Rightarrow \mathrm{f}(\mathrm{a}) \neq \mathrm{f}(\mathrm{b}) \quad$ i.e., $\quad \mathrm{f}(-5) \neq \mathrm{f}(5)$
If $(\mathrm{i}) \mathrm{f}$ is continous in [ $\mathrm{a}, \mathrm{b}]$
(ii) $\mathrm{f}$ is derivable in [ $\mathrm{a}, \mathrm{b}]$
(iii) $f(a)=f(b)$
then $\mathrm{f}^{\prime}(\mathrm{c})=0, \mathrm{c} \in(\mathrm{a}, \mathrm{b})$
$\therefore \mathrm{f}$ is continuous and derivable
but $\mathrm{f}^{\prime}(\mathrm{c}) \neq 0 \Rightarrow \mathrm{f}(\mathrm{a}) \neq \mathrm{f}(\mathrm{b}) \quad$ i.e., $\quad \mathrm{f}(-5) \neq \mathrm{f}(5)$
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