The value of '$c$ ' in Rolle's Theorem for the function $f(x)=\cos \frac{x}{2}$ on the interval $[\pi ,3\pi ]$ is:

Rolle's Theorem states that if $f(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, there exists $c\in (a,b)$ such that $f^{\prime }(c)=0$.Given $f(x)=\cos \frac{x}{2}$ on $[\pi ,3\pi ]$:$f(\pi )=\cos \frac{\pi }{2}=0$$f(3\pi )=\cos \frac{3\pi }{2}=0$ Since $f(\pi )=f(3\pi )=0$, Rolle's theorem applies.Find $f^{\prime }(x)$:
$$f^{\prime }(x)=-\frac{1}{2}\sin \frac{x}{2}$$
Set $f^{\prime }(c)=0$:
$$-\frac{1}{2}\sin \frac{c}{2}=0⟹\sin \frac{c}{2}=0$$
$$\frac{c}{2}=n\pi ⟹c=2n\pi$$
For $c\in (\pi ,3\pi )$, we take $n=1$:
$$c=2\pi$$