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# Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers $c$ that satisfy the conclusion of Rolle's Theorem.$f(x) = \sin(x/2)$, $[\pi /2, 3\pi /2]$

## $f(x)=\sin (x / 2),[\pi / 2,3 \pi / 2] . \quad f,$ being the composite of the sine function and the polynomial $x / 2,$ is continuous anddifferentiable on $\mathbb{R}$, so it is continuous on $[\pi / 2,3 \pi / 2]$ and differentiable on $(\pi / 2,3 \pi / 2) .$ Also, $f\left(\frac{\pi}{2}\right)=\frac{1}{2} \sqrt{2}=f\left(\frac{3 \pi}{2}\right)$$f^{\prime}(c)=0 \Leftrightarrow \quad \frac{1}{2} \cos (c / 2)=0 \Leftrightarrow \cos (c / 2)=0 \Leftrightarrow c / 2=\frac{\pi}{2}+n \pi \Leftrightarrow c=\pi+2 n \pi, n$ an integer.Only $c=\pi$ is in $(\pi / 2,3 \pi / 2),$ so $\pi$ satisfies the conclusion of Rolle's Theorem.

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right here. We're being asked to verify that the function have satisfied the three hypothesis of role serums on the given interval. And then we are being asked to find the numbers seized as satisfied the conclusion for roll student. So here the function is sine of X over two on the closed interval from pi over 2 to 3.2. And so now to recall that the three conditions of role term is continuity on the close interval from A to B, the French ability on the open interval from A to B and the endpoint must equal each other on the functions alphabetical, complete. So, um, to ignore metric functions are all continuous. So we know that this is a given. Um, they're also all defensible on a closed interval. And since we are on an interval, we know that this is also I mean, it's not it doesn't have to be on interval. It is differentiable everywhere because it is a smooth function. So those are just a property is a trick geometric function and, um, the last point after bank with F B, we have to actually figure those out. So if we plug in the values F part of it too into our functions sine of x over to We have to get route to over to And if you do, if you do it again for three pi over two, you also get good to over two. So we know that this last condition is also satisfied. So once used to be conditions are satisfied. We can no. We know that there is a c that satisfies the conclusion to roll student. And the conclusion is there is a C such that the derivative if prime of C is equal to zero. So all we have to do is we can actually find these numbers sees by setting the derivative this function equals zero. So if we take the derivative f prime of X, we actually have to apply the chain rule. You have to do the inside do every time, the outside to able to keep in the same inside. So, um, the derivative of X over two is one half and the derivative of Sinus cosign would keep co sign and then we keep the inside the same. And all we do now is set this function equal to zero and This is a trick geometric function. Um, yeah, we know that. CO sign the cosine function equals zero. Just a coastline function equals zero when it is at PI over two. So we're we want to find the values where x X X over two is equal to pi over two. And since close sign is a periodic function, we know that is a high periodic function. So we're gonna at Cape I. So, K is any integer number 12 from any, uh, because any integer number, so it could be negative. 12112345 etcetera. So if you solve for X, you get X is equal to hi. When you multiply two on both sides, you get X equals pi plus to Cape I Oh, and since our interval, it is only from pi over 2 to 3 pi over two. Um, the only, uh um the only number that actually satisfies, uh, just the that lays in this interval is actually just pie the access to go to pie because since our function is telling us that it has to be two kpi period, um, according to the way I function of setup. Um, so the next the next point in which we will get a value of zero is when we are at, um okay. Equals one. So there'll be two pi three pi and three pi is not in our interval. So the only interval that lies is when x is equal to pi. And so this is the sea that gives us a primary tactical zero mhm.

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