Question

Verify Rolle's theorem for the following function on the indicated interval f(x) = cos 2x on [−π/4, π/4] ?

Answer 1

The given function is \[f\left( x \right) = \cos2x\].

Since 

\[\cos2x\] is everywhere continuous and differentiable, \[\cos2x\] is continuous on 

\[\left[ \frac{- \pi}{4}, \frac{\pi}{4} \right]\] and differentiable on \[\left( \frac{- \pi}{4}, \frac{\pi}{4} \right)\] .

Also,

\[f\left( \frac{\pi}{4} \right) = f\left( \frac{- \pi}{4} \right) = 0\]

Thus,\[f\left( x \right)\] satisfies all the conditions of Rolle's theorem. 

Now, we have to show that there exists \[c \in \left( \frac{- \pi}{4}, \frac{\pi}{4} \right)\] such that \[f'\left( c \right) = 0\] .

We have

\[f\left( x \right) = \cos2x\]
\[ \Rightarrow f'\left( x \right) = - 2\sin2x\]

\[\therefore f'\left( x \right) = 0\]
\[ \Rightarrow - 2\sin2x = 0\]
\[ \Rightarrow \sin2x = 0\]
\[ \Rightarrow \sin2x = 0\]
\[ \Rightarrow x = 0\]

Since \[c = 0 \in \left( \frac{- \pi}{4}, \frac{\pi}{4} \right)\] such that \[f'\left( c \right) = 0\] .

​Hence, Rolle's theorem is verified.