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# Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers $c$ that satisfy the conclusion of the Mean value Theorem.$f(x) = x^3 - 3x + 2$, $[-2, 2]$

## $$c=\pm \frac{2 \sqrt{3}}{3}$$

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Okay, So now the question is asking us to verify that the function f satisfies the high passes of the main bout on the given interval. And then we're being has to find the number suits of satisfies, the conclusion of the means that Kira and so we can go ahead and structures. So since FX physical, the execute by industry experts to we know right off the back, that's is a form of probably normal function in this case, specifically cubic function and all pine on your functions are continuous and differential on all the old numbers. It's out of the property off, probably normals. So those two are a given. And now the next question is to actually find the number sees that satisfies the conclusion of me like that and recall that the mean value serum is just simply that the derivative of the function, so that there's some value of C in the derivative of effort backs such that the average slope from negritude too. It's the same. So in this case will be after us too. Minus of negative too. Nana will be all over two minus a negative, too. And if you calculate the vary the effort too. It gives us a gives us four. You can check this yourself and then half of negative too. There's just a value of Syria now when we flood that when we plug these values and we get or minus, you know, and in two minus and negative to mention ad two plus two, this gives us four before is equal to or now we take the derivative of effort Beckwitt respect. We take the derivative, but we're going toe put it in terms of C So we three c squared minus three is equal to one and then we get ah. Then we get a value of three. C squared is equal to four. I'm gonna go to the next page, so we have three c square eagle four. Divide both sides by three. You got to see square to go for third and then you could take the square root on both sides to get rid of the square of great value. Get C equals poor third and this can be simply applied to to over three because the squared of for it's just too. However, on this is also plaster minus. This is very important. When you take a square, This is the part of minus. So this is also posted myself here. Talk about that. And then since we don't want to usually right discovered at the bottom, we just multiply. It was three times or three, which will give us two. Three all over three. And if you recall, our values were in the interval from negative too. Two two and positive too. Three over three is lesson one is less than two, obviously, and that end it on a negative too. Is this is greater than negative too? Said this. Both of these numbers actually fall within the interval. So both positive to three of a three and negative to it. To be over three is the correct answer. Yeah.

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