verify that the function satisfies the three hypotheses of rolles theorem on the given interval. then find…

Explanation:

Step1: Check continuity

The function ( y = \sin\left(\frac{x}{2}\right) ) is a trigonometric function. Trigonometric functions ( y=\sin(u) ) where ( u=\frac{x}{2} ) is a linear - function are continuous everywhere. So ( f(x)=\sin\left(\frac{x}{2}\right) ) is continuous on the closed interval (\left[\frac{\pi}{2},\frac{3\pi}{2}\right]).

Step2: Check differentiability

The derivative of ( y = f(x)=\sin\left(\frac{x}{2}\right) ) using the chain rule ( (u = \frac{x}{2}, y=\sin(u)) ), ( y^\prime=f^\prime(x)=\frac{1}{2}\cos\left(\frac{x}{2}\right) ). The derivative ( f^\prime(x)=\frac{1}{2}\cos\left(\frac{x}{2}\right) ) exists for all ( x\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right) ). So ( f(x) ) is differentiable on the open interval (\left(\frac{\pi}{2},\frac{3\pi}{2}\right)).

Step3: Check ( f(a)=f(b) )

Calculate ( f\left(\frac{\pi}{2}\right)=\sin\left(\frac{\pi/2}{2}\right)=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} ) and ( f\left(\frac{3\pi}{2}\right)=\sin\left(\frac{3\pi/2}{2}\right)=\sin\left(\frac{3\pi}{4}\right)=\frac{\sqrt{2}}{2} ). So ( f\left(\frac{\pi}{2}\right)=f\left(\frac{3\pi}{2}\right) ).

Step4: Apply Rolle's Theorem

Since ( f(x) ) satisfies the three hypotheses of Rolle's Theorem (( f(x) ) is continuous on (\left[\frac{\pi}{2},\frac{3\pi}{2}\right]), differentiable on (\left(\frac{\pi}{2},\frac{3\pi}{2}\right)) and ( f\left(\frac{\pi}{2}\right)=f\left(\frac{3\pi}{2}\right))), then there exists ( c\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right) ) such that ( f^\prime(c) = 0 ). Set ( f^\prime(x)=\frac{1}{2}\cos\left(\frac{x}{2}\right)=0 ). Then (\cos\left(\frac{x}{2}\right)=0). Let ( t=\frac{x}{2} ), so ( \cos(t) = 0 ) implies ( t=\frac{\pi}{2}+k\pi,k\in\mathbb{Z} ). Substituting back ( t = \frac{x}{2} ), we have ( \frac{x}{2}=\frac{\pi}{2}+k\pi), or ( x=\pi + 2k\pi). For ( x\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right) ), when ( k = 0 ), ( x=\pi ).

Answer:

(\pi)