Question

Verify Lagrange's Mean Value Theorem for the following function:
$f\left(x\right)=2\sin x+\sin 2x$ on $[0,\pi ]$

Answer 1

Given, $f\left(x\right)=2\sin x+\sin 2x,x\in [0,\pi ]$
$f\left(x\right)$ is continuous is $[0,\pi ]$
$f\left(x\right)$ is differentiable in $(0,\pi )$

Thus, both the conditions of Lagrange's man value theorem are satisfied by the function $f\left(x\right)$ in $[0,\pi ]$, therefore, there exists at least one real number $c$ in $[0,\pi ]$ such that
$f^{\prime }\left(c\right)=\frac{f\left(\pi \right)-f\left(0\right)}{\pi -0}$
$f\pi =2\sin \pi +\sin 2\pi =0$
$f\left(0\right)=2\sin 0+\sin 0=0$

Differentiating $f\left(x\right)$ w.r.t. $x$, we get
$f^{\prime }\left(x\right)=2\cos x+2\cos 2x$

Now, $2\cos x+2\cos 2x=0$
$\Rightarrow 2{\cos }^{2}x+\cos x-1=0$ $(\because \cos 2x=2{\cos }^{2}x-1)$
$\Rightarrow 2{\cos }^{2}x+2\cos x-\cos x-1=0$
$\Rightarrow 2\cos x(\cos x+1)-1(\cos x+1)=0$
$\Rightarrow (2\cos x-1)(\cos x+1)=0$
$2\cos -1=0$
or $\cos x+1=0$
$2\cos x=1$ or $\cos x=-1$

$\cos x=\frac{1}{2}$ or $\cos x=-1$

$\Rightarrow x=\frac{\pi }{3},\pi$

$\therefore x=\frac{\pi }{3}$ $\because \frac{\pi }{3}ϵ(0,\pi )$

Thus Lagrange's mean value theorem is verified.