Question

Verify Rolle's theorem for the following function on the indicated interval \[f\left( x \right) = \frac{x}{2} - \sin\frac{\pi x}{6} \text { on }[ - 1, 0]\]?

Answer 1

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

Since

\[\sin x \text { & } \frac{x}{2}\] are everywhere continuous and differentiable,\[f\left( x \right)\] is continuous on \[\left[ - 1, 0 \right]\] and differentiable on \[\left( - 1, 0 \right)\].

Also,

\[f\left( - 1 \right) = f\left( 0 \right) = 0\]

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

Now, we have to show that there exists\[c \in \left( - 1, 0 \right)\] such that \[f'\left( c \right) = 0\] .

We have

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

\[ \Rightarrow f'\left( x \right) = \frac{1}{2} - \frac{\pi}{6}\cos\frac{\pi x}{6}\]

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

\[ \Rightarrow \frac{1}{2} - \frac{\pi}{6}\cos\frac{\pi x}{6} = 0\]

\[ \Rightarrow \cos\frac{\pi x}{6} = \frac{3}{\pi}\]

\[ \Rightarrow x = \frac{- 6}{\pi} \cos^{- 1} \left( \frac{3}{\pi} \right)\]

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

​Hence, Rolle's theorem is verified.