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

Answer

The given function is $\text{f}(\text{x})=\text{e}^{\text{x}}\sin{\text{x}}$
Since $\sin\text{x}\ \&\ \text{e}^{\text{x}}$ are everywhere continuous and differentiable.
Therefore, being a product of these two, f(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
$\text{f}(\text{x})=\text{e}^{\text{x}}\sin{\text{x}}$
$\Rightarrow\text{f}'(\text{x})=\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})=0$
$\Rightarrow\sin\text{x}+\cos\text{x}=0$
$\Rightarrow\tan\text{x}=-1$
$\Rightarrow\text{x}=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
Since $\text{c}=\frac{3\pi}{4}\in(0,\pi)$ such that f'(c) = 0
Thus, Rolle's theorem verified.

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