Verify Rolle's theorem for $f (x) = | cos x |$ in $[0, π]$ .

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Solution

$\begin{matrix} V e r i f y r o l l e^{'} s t h e o r e m f o r w e m a y r e s s i t f (x) = {\begin{matrix} cos x 0 x \frac{π}{2} - cos x \frac{π}{2} < x π \end{matrix} c l e a r l y, f (0) = cos 0^{^{\circ}} = 1 \Rightarrow f (π) = - cos π = 1 ∴ f (0) = f (π) = 1 \Rightarrow R f^{'} (\frac{π}{2} + h) = {lim}_{h \to 0} f \frac{(\frac{π}{2} + h) - f (\frac{π}{2})}{h} = {lim}_{h \to 0} \frac{| - sin 1 m u h | - 0}{h} = 1 \Rightarrow L f^{'} (\frac{π}{2}) = {lim}_{h \to 0} = \frac{f (\frac{π}{2} - h) - f (\frac{π}{2})}{- h} = {lim}_{h \to 0} = \frac{| sin 1 m u 1 m u h | - 0}{- h} = {lim}_{h \to 0} = \frac{- sin 1 m u 1 m u h}{h} = - 1 ∴ R f^{'} (o) \neq L f^{'} (o) \end{matrix}$

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