#### Question

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

#### Solution

The given function is \[f\left( x \right) = 2\sin x + \sin2x\] .

Since

\[\sin x \text { and }\sin2x\] are everywhere continuous and differentiable,

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

Also,

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

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

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

We have

\[f\left( x \right) = 2\sin x + \sin2x\]

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

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

\[ \Rightarrow 2\cos x + 2\cos2x = 0\]

\[ \Rightarrow \cos x + \cos2x = 0\]

\[ \Rightarrow \cos x + 2 \cos^2 x - 1 = 0\]

\[ \Rightarrow 2 \cos^2 x + \cos x - 1 = 0\]

\[ \Rightarrow \left( \cos x + 1 \right) \left( 2\cos x - 1 \right) = 0\]

\[ \Rightarrow \cos x = - 1, \cos x = \frac{1}{2}\]

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

\[ \Rightarrow x = \pi, \frac{\pi}{3}\]

Thus,

\[c = \frac{\pi}{3} \in \left( 0, \pi \right)\] such that \[f'\left( c \right) = 0\] .

Hence, Rolle's theorem is verified.

Is there an error in this question or solution?

Solution Verify Rolle'S Theorem for the Following Function on the Indicated Interval F(X) = 2 Sin X + Sin 2x on [0, π] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.