Question

Verify Rolle's theorem for the following function on the indicated interval \[f\left( x \right) = \frac{6x}{\pi} - 4 \sin^2 x \text { on } [0, \pi/6]\] ?

Answer 1

The given function is \[f\left( x \right) = \frac{6x}{\pi} - 4 \sin^2 x\] .

Since \[\sin^2 x \text { & }x\] are everywhere continuous and differentiable, \[f\left( x \right)\] is continuous on \[\left[ 0, \frac{\pi}{6} \right]\] and differentiable on \[\left( 0, \frac{\pi}{6} \right)\] .

Also,

\[f\left( \frac{\pi}{6} \right) = f\left( 0 \right) = 0\]

Thus, 

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

Now, we have to show that there exists\[c \in \left( 0, \frac{\pi}{6} \right)\] such that \[f'\left( c \right) = 0\] .

We have

\[f\left( x \right) = \frac{6x}{\pi} - 4 \sin^2 x\]

\[ \Rightarrow f'\left( x \right) = \frac{6}{\pi} - 8 \sin x \cos x\]

\[\therefore f'\left( x \right) = 0\]

\[ \Rightarrow \frac{6}{\pi} - 8\sin x\cos x = 0\]

\[ \Rightarrow \sin2x = \frac{3}{2\pi}\]

\[ \Rightarrow x = \frac{1}{2} \sin^{- 1} \left( \frac{3}{2\pi} \right)\]

Thus, 

\[c = \frac{1}{2} \sin^{- 1} \left( \frac{3}{2\pi} \right) \in \left( 0, \frac{\pi}{6} \right)\] such that \[f'\left( c \right) = 0\] .

​Hence, Rolle's theorem is verified.