Question

Verify Rolle's theorem for the following function on the indicated interval f(x) = sin 3x on [0, π] ?

Answer 1

The given function is \[f\left( x \right) = \sin3x\] .

Since \[\sin3x\] is everywhere continuous and differentiable, \[\sin3x\]  is continuous on \[\left[ 0, \pi \right]\] and differentiable on \[\left( 0, \pi \right)\] .

Also,

\[f\left( \pi \right) = f\left( 0 \right) = 0\]

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

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

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

We have

\[f\left( x \right) = \sin3x\]
\[ \Rightarrow f'\left( x \right) = 3\cos3x\]

\[\therefore f'\left( x \right) = 0\]
\[ \Rightarrow 3\cos3x = 0\]
\[ \Rightarrow \cos3x = 0\]
\[ \Rightarrow 3x = \frac{\pi}{2}, \frac{3\pi}{2}, . . . . \]
\[ \Rightarrow x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\]

Thus,\[c = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \in \left( 0, \pi \right)\] such that \[f'\left( c \right) = 0\] .

​Hence, Rolle's theorem is verified.