Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=2\sin\text{x}+\sin2\text{x}\text{ on }[0,\pi]$

Answer

Here,$\text{f}(\text{x})=2\sin\text{x}+\sin2\text{x}\text{ on }[0,\pi]$
We know that sine function is continuous and differentiable every where, so f(x) is continuous is $[0,\pi]$ and differentiable is $(0,\pi).$
Now,
$\text{f}(0)=2\sin0+\sin0=0$
$\text{f}(\pi)=2\sin\pi+\sin2\pi=0$
$\Rightarrow\text{f}(0)=\text{f}(\pi)$
So, Rolle's theorem is applicable, so there must exist a point $\text{c}\in(0,\pi)$ such that f'(c) = 0.
Now,
$\text{f}(\text{x})=2\sin\text{x}+\sin2\text{x}$
$\text{f}'(\text{x})=2\cos\text{x}+2\cos2\text{x}$
Now,
$\text{f}'(\text{c})=0$
$2\cos\text{c}+2\cos2\text{c}=0$
$\Rightarrow2(\cos\text{c}+2\cos^2\text{c}-1)=0$
$\Rightarrow(2\cos^2+2\cos\text{c}-\cos\text{c}-1)=0$
$\Rightarrow(2\cos\text{c}-1)(\cos\text{c}+1)=0$
$\Rightarrow\cos\text{c}=\frac{1}{2},\cos\text{c}=-1$
$\Rightarrow\tan\text{c}=1$
$\text{c}=\frac{\pi}{3}\in(0,\pi),\text{c}=\pi$
Hence, Rolle's theorem is verified.

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