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Given the polynomlal 2r" 3022 547satsfles the Mean Value Theorem on the Interval [0, 5]:255306-405Find the average or mean slope ofthe function on this interva...

Question

Given the polynomlal 2r" 3022 547satsfles the Mean Value Theorem on the Interval [0, 5]:255306-405Find the average or mean slope ofthe function on this interval: Round the nearest thousandth:PreviewBy the Mean Value Theorem;, we Know there exists at least one c in the open interval (0, 5) such that f '(c) equal to this mean slope: If there more than one answer enter the smaller value: Round the nearest thousandth: Preview

Given the polynomlal 2r" 3022 547 satsfles the Mean Value Theorem on the Interval [0, 5]: 255 306- 405 Find the average or mean slope ofthe function on this interval: Round the nearest thousandth: Preview By the Mean Value Theorem;, we Know there exists at least one c in the open interval (0, 5) such that f '(c) equal to this mean slope: If there more than one answer enter the smaller value: Round the nearest thousandth: Preview



Answers

Use the graph of f given in Exercise 7 to estimate the values of $c$ that satisfy the conclusion of the Mean Value Theorem for the interval $[1,7]$ .

Applying the mean value theorem from 1 to 7. We want to find out at what points we have tangent lines that match that dotted slope there. And it looks like the first one that we have it somewhere around here. And that might be about a little over three. Maybe we could say 3.13 point two, and then the next one is right around here. They had a peril. So at about exes six or you might say it's 6.1. So those were the two C values that satisfy the, uh, I mean value them.

We are going to see if the mean value theorem applies and then if it does what the value of C is. Since we do have a polynomial function, the function itself is continuous on the closed interval and it was differentiable on the open interval. Yeah. Yeah. Mhm. Now what we need to find is the X. Value where the derivative equals the average rate of change. The derivative is three x squared minus 10 X plus four. Yeah. Okay. The function at three would be 27 minus 45 Plus 12 -2. 27 -45 plus 12 minus two. That would be negative 18 plus 12 -6 -2 were negative. Act so track what I get when I feel in one which is one -5-plus 4 -2, negative two. And that's all over 3 -1. Now the minus of negative two. I can put plus the positive. So I have three X squared -10 plus four equals six over two or three. Okay at this point I could subtract the three over that will give me three X squared minus 10 X plus one equals zero. Mhm. Yeah That will factor to three X. Actually it's not going to factor cleanly. So we're going to actually use the quadratic here. That would be negatively 10 minus square root B squared 100 minus four A. C. That would be minus 12 Over 286. So that's 10 minus the square root plus or minus the square root of 88 Over six. Square root of 88. Well perfect factor perfect Square four goes in. So that's too radical. 22 and we could reduce that to five plus or minus The square root of 22 All over three. Now which of those are inside the interval. If I add the square root of 22 I'll end up with a number that is Oh well that's less than three. If i subtract the 22, I will end up with a value that is smaller than one. So the only one that works is the positive one, Five plus square to 22/3.

I wish to determine where this function satisfy the mean value theorem. We know that it does it does satisfy the mean value theorem because although it would be discontinuous and non differentiable at one that's outside the interval, so it is equal to for me it is differentiable on the open interval and it is continuous on the closed by definition, mean value theorem says the derivative at some value inside the interval will equal the average rate of change from the beginning to end. F A B minus F. Of a over b minus a. Now I'm going to take the derivative at X and just realize that that X is the C value. The derivative is one, Well the relative to zero, but the derivative of three Over X -1 could be, We could rewrite the function is three times the quantity X -1 to the negative first. The derivative would use the chain rule. We bring the power out front and subtract one from the old power. So make that x minus one to the negative second. And then we multiply by the derivative inside. But that's one. The function at seven would be nine plus 3/4. So that's 9, 9 and 3/4 or 39/4. If we fill in two, that's gonna be two plus two plus three, that's seven And that's all over 7 -2. So we have 1 -3 quantity. There's three over the quantity X -1 squared. And that's going to equal 11. Force over five multiply the top and bottom by four. And that would be 11 20th. So I have one minus three over X -1 Squared equals 11 20th. Let's subtract the one from both sides to give ourselves negative three over x minus one squared equals -9/20. Now, if we use cross products We could find that we have negative 60 equals negative nine quantity x minus one quantity squared Divide by the -9. That's 60/9. And at that point we would take the square root. So if we take the square root, that's the square root of 60/9. At that point, all we have to do is add the one, which would give us the square root of 60/9 Plus one as our solution and that is inside the interval. So that is our solution.

Let's see how the main value theorem applies to this function. We have the function f of X equals X cubed minus two X squared. We're looking at it on the interval from 0 to 1. Well, let's first see if this function meets the conditions needed to apply the mean value theorem. First of the function is continuous on the interval, that's a polynomial is continuous on the entire set of real numbers. So it is definitely continuous on the close interval from 0 to 1. The next is that the function is defensible on that same interval which changed. The derivative has to exist on the interval from 0 to 1. Well, this is a polynomial function. We know the derivative is a polynomial function. Let's go ahead and find it anyway. So the derivative of X cubed minus two x squared. It's gonna be three x squared minus four X again, that is a continuous function exists for all real numbers until the derivative exists from the interval from 0 to 1. And so if the original function is differential on the interval from 0 to 1. So the main value theorem tells us that somewhere between zero and one. There is a point where the derivative has the same value as the slope of the end points of the function, but the ends of the interval. So see what f of A and f of b r and to find out what that slope is. So if I evaluate the function at zero, which is our value of a I get zero cubed minus three times zero squared is zero. If I evaluate this function at one, I get one cubed minus two times one square, which is one minus two, which is negative one. And so there is a value see between zero and one that I can plug into the derivative three C squared, minus foresee, and that's gonna be equal to the slope between those two points, so be is one and a zero. So I have a negative one minus off of a zero over one minus zero. So I want to find where the derivative has a value off negative one. So I need to solve the quadratic equation. Three C Square minus four c equals negative one. That same is the equation three C squared, minus four C plus one equals zero, and we're going to go ahead and factor this. This can be factor three C minus one times C minus one equals zero. We can check that real quick outside terms. Negative three seat and site terms like one. See that's too negative for C. And so I get to solutions I get one solution from three C minus one equals zero, and the other one comes from the other factor. C minus one people in zero. It's like a tomb. Also values for C See is 1/3 or C is positive one. Now, at least one of these is guaranteed to be inside the interval, the open interval from 0 to 1, and that's going to be the one at C equals 1/3. This is the end point. We're not given any information about the possibility of a value of C ad and then point. It's not that it's not one we're interested in, but the value at 1/3 is the one that mean Value Theorem guarantees us to exist


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