Question

Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem  f(x) = sin x − sin 2x − x on [0, π] ?

Answer 1

We have ,

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

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

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

Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some 

\[c \in \left( 0, \pi \right)\]  such that

\[f'\left( c \right) = \frac{f\left( \pi \right) - f\left( 0 \right)}{\pi - 0} = \frac{f\left( \pi \right) - f\left( 0 \right)}{\pi}\]

Now,\[f\left( x \right) = \sin x - \sin2x - x\]

\[f'\left( x \right) = \cos x - 2\cos2x - 1\],\[f\left( \pi \right) = - \pi\],\[f\left( 0 \right) = 0\]

∴ \[f'\left( x \right) = \frac{f\left( \pi \right) - f\left( 0 \right)}{\pi - 0}\]

\[\Rightarrow \cos x - 2\cos2x - 1 = - 1\]

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

\[ \Rightarrow \cos x - 4 \cos^2 x = - 2 \]

\[ \Rightarrow 4 \cos^2 x - \cos x - 2 = 0\]

\[ \Rightarrow \cos x = \frac{1}{8}\left( 1 \pm \sqrt{33} \right)\]

\[ \Rightarrow x = \cos^{- 1} \left[ \frac{1}{8}\left( 1 \pm \sqrt{33} \right) \right]\]

Thus, \[c = \cos^{- 1} \left( \frac{1 \pm \sqrt{33}}{8} \right) \in \left( 0, \pi \right)\] such that \[f'\left( c \right) = \frac{f\left( \pi \right) - f\left( 0 \right)}{\pi - 0}\].

Hence, Lagrange's theorem is verified.