Question

Verify Rolle's theorem for the following function on the indicated interval  f(x) = cos 2x on [0, π] ?

Answer 1

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

Since \[\cos2 \ x\] is everywhere continuous and differentiable. 

Therefore, \[f\left( x \right)\]  is continuous on  \[\left[ 0, \pi \right]\] and differentiable on \[\left( 0, \pi \right)\] .

Also,\[f\left( \pi \right) = f\left( 0 \right) = 1\]

Thus, 

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

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

We have

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

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

Thus,

\[c = \frac{\pi}{2} \in \left( 0, \pi \right)\] such that  \[f'\left( c \right) = 0\] .

​Hence, Rolle's theorem is verified.