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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