Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin3\text{x}\text{ on }[0,\pi]$

Answer

The given function is $\text{f}(\text{x})=\sin3\text{x}$
Since $\sin3\text{x}$ everywhere continuous and differentiable,
$\sin3\text{x}$ is continuous on $[0,\pi]$ and differentiable on $(0,\pi).$
Also,
$\text{f}(\pi)=\text{f}(0)=0$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in(0,\pi)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\sin3\text{x}$
$\Rightarrow \text{f}'(\text{x})=3\cos3\text{x}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow3\cos3\text{x}=0$
$\Rightarrow\cos3\text{x}=0$
$\Rightarrow3\text{x}=\frac{\pi}{2},\frac{3\pi}{2},....$
$\Rightarrow\text{x}=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}$
Since, $\text{c}=\frac{\pi}{4}\in(0,\pi)$ such that f'(c) = 0
Hence, Rolle's theorem is verified.

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