Question
Verify Rolle's theorem for the following function on the indicated intervals

$\text{f}(\text{x})=\sin2\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$

Answer

The given function is $\text{f}(\text{x})=\sin2\text{x}$

Since, $\sin2\text{x}$ is everywhere continuous and differentiable.

Therefore
 $\sin2\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big],$ and differentiable on $\Big(0,\frac{\pi}{2}\Big)$

Also,

$\text{f}\Big(\frac{\pi}{2}\Big)=\text{f}(0)=0$

Thus, f(x) satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.

We have

$\text{f}(\text{x})=\sin2\text{x}$

$\Rightarrow\text{f}'(\text{x})=2\cos2\text{x}$

$\Rightarrow\text{f}'(\text{x})=0$

$\Rightarrow2\cos2\text{x}=0$

$\Rightarrow\cos2\text{x}=0$

$\Rightarrow\text{x}=\frac{\pi}{4}$

Thus, $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that $\text{f}'(\text{c})=0.$

Hence, Rolle's theorem is verified .

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