Question

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

Answer 1

The given function is

\[f\left( x \right) = \sin2x\] .

Since \[\sin2x\] is everywhere continuous and differentiable.

Therefore, \[\sin2x\]  is continuous on  \[\left[ 0, \frac{\pi}{2} \right]\] and differentiable on \[\left( 0, \frac{\pi}{2} \right)\] .

Also,

\[f\left( \frac{\pi}{2} \right) = f\left( 0 \right) = 0\]

Thus, 

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

Now, we have to show that there exists

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

\[f'\left( c \right) = 0\] .

We have 

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

\[\therefore f'\left( x \right) = 0\]
\[ \Rightarrow 2\cos2x = 0\]
\[ \Rightarrow \cos2x = 0\]
\[ \Rightarrow x = \frac{\pi}{4}\]

Thus,

\[c = \frac{\pi}{4} \in \left( 0, \frac{\pi}{2} \right)\] such that \[f'\left( c \right) = 0\] .

​Hence, Rolle's theorem is verified.