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Answer:

$c=\frac{\pi}{2} \in(0, \pi)$

Hint:

$f(0)=f(\pi)$ , so there exists at$c \in(0, \pi)$

Given:

$f(x)=4^{\sin x}$ on $[0, \pi]$

Explanation:

Given function is $f(x)=4^{\sin x}$ on $[0, \pi]$

We know that sine function is continuous and differentiable on R.

Let’s find the values of the function at an extreme

$\\ \Rightarrow \quad f(0)=4^{\sin (0)} \\\\ f(0)=4^{0}\\\\ \therefore f(0)=1\\\\ \Rightarrow f(\pi)=4^{\sin (x)}\\\\ \Rightarrow \quad f(\pi)=4^{0} \\\\ \therefore f(\pi)=1\\\\$

We have$f(0)=f(\pi)$, so there exists at$c \in(0, \pi)$ , such that$f^{\prime}(c)=0$

Let’s find the derivative of f(x)

$\Rightarrow \quad f^{\prime}(x)=\frac{d\left(4^{\operatorname{lin} x}\right)}{d x}\\ \Rightarrow \quad f^{\prime}(x)=4^{\sin x} \log 4 \frac{d(\sin x)}{d x} \ \ \ \ \left[\because \frac{d(\sin x)}{d x}=\cos x\right]\\ \Rightarrow \quad f^{\prime}(x)=4^{\sin x} \log 4 \cos x$

We have $f ' (c) = 0$

$\\ \Rightarrow \quad 4^{\text {sinc }} \log 4 \cos c=0\\\\ \Rightarrow \quad \cos c=0\\\\ \Rightarrow \quad c=\frac{\pi}{2} \in(0, \pi)$

Hence Rolle’s Theorem is verified

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