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If the MVT what those Does this Use the following applies; fcocdioionst function then are satisfy g(x) find all and how the answer numbers you 0| the know 1 that wh...

Question

If the MVT what those Does this Use the following applies; fcocdioionst function then are satisfy g(x) find all and how the answer numbers you 0| the know 1 that whether the satisfy Mean interval [ ~ below: not the Value the conclusion 1,2] j of the MVT V Excphain

If the MVT what those Does this Use the following applies; fcocdioionst function then are satisfy g(x) find all and how the answer numbers you 0| the know 1 that whether the satisfy Mean interval [ ~ below: not the Value the conclusion 1,2] j of the MVT V Excphain



Answers

Show that $f(x)=1-|x|$ satisfies the conclusion of the MVT
on $| a, b \}$ if both $a$ and $b$ are positive or negative, but not if $a < 0$ and $b > 0$ .

First. Let's verify that we can use the mean values here. Um So we're gonna grab half of X equals E. To the negative two X. And then what we end up getting is Looking from 0 to 3. We see this is going to be continuous and differentiable. So we can consider F of three -F. of zero mm divided by three minus era. So you see the slope that we're looking for is a negative 0.33. And we're gonna solve for where this is the case. So we'll look at F prime of X. Which we know is going to be a negative who E. To the negative, correct. It's important to note that F. F. Three is going to be E. To the negative six -1/3. That's where this is coming from. So we want to determine what value of acts um will allow us to get that same slope and we end up getting, it's gonna be negative one a half times the natural log 1- is the 6th over 6th. And that's gonna be our final answer.

We wish to show that the mean value theorem applies and then we'll see the c value that works. This function is continuous On the interval from 1 to 3. The only place that it would be discontinuous is the X value of zero. That's not in the interval. I could rewrite the function by dividing each of the X and one by the X and the denominator. So that's 11 plus one of our X or one plus X to the negative first. That would mean that the derivative is one minus X to the negative second or one minus one over x square. That would be non differential board X equals zero, but that is not inside of our interval. So it's differentiable where it needs to be now. According to mean value theorem, the derivative must equal the average rate of change on the interval. So let's see where this is true, Derivative is 1 -1 of Rex Square The function at B. If I fill in three, I get four Over 3, 4/3 by filling one I get to over one or two And I'll bottom I get three -1 or two. So that's one minus one over. X squared equals let's see four thirds minus six thirds, negative two thirds over to or one minus one over X squared Equals -1 3rd. We will add the one over. Actually subtract the one over, which will give us negative one over X squared Equals -4/3. And then we will set the cross products equal the negatives. All say our with the top functions or top of the fraction, so negative four X squared equals negative three, Divide by the -4 X squared equals 3/4. And then take the square root. Now if I take the square root I could have positive or negative square root of three fours or positive or negative square 23 divided by two, but the negative is not inside the interval. So your answer is simply The positive Square 123/2.

Yeah. Right. In this problem we're looking for a C that's between A and B. Which in this problem is one and two. So some see between one and two that satisfies this equation up here from the extended mean value there. There's a sea that somewhere between one and 2. Okay, So that's the mean part. Okay. Someone they're sort of like the middle somewhere there, but not it doesn't mean exactly half. Okay. And that C satisfies these properties or this this property here. So, they want us to find what it is. All right. So, first we need the derivatives And the prime of X is two x And is one and be a stupid. Okay, So F prime Fc will be minus one over C squared and G prime of C. Oh, I forgot the extra will be to see. And that is equal to F B, which is one half minus F of a. Which is one can remember. We're using F over on on the right hand side. So this function not that one. This function not that one. We use the derivatives over on this side on the blue side. All right. And then now we're doing G f b Which would be 4 -40 minus job, which would be one minus 3. -4, which is -3. Okay? Um I'll flip that around so I get -1 over to c cubed and then over here I get one half minus one which is minus one half Over 3/1. Which is -1 over to c cubed equals minus 1/6. So to see cubed equals six, C. Cubed Equals three. So c equals the cube root of three. It's the number that satisfies that formula that lies between one and 2.

For the fine crumb. Our goal is to verify. Um Do you function and making sure it satisfies the hypothesis of the mean value theorem on the given interval. And then we want to find all numbers. See that satisfy the conclusion of the main value theory. So here we have F. Of X. Equaling act over exposed to. Okay, We're going to be looking from 1-4 for that. So if we consider the mean value theorem, We see that it is going to be continuous from 1-4. So we can use mean values here and it's differentiable. So considering from one before we have effort for everyone, what we're dealing now is finding the average rate of change or the The slope of the secret line. And we get that it's 1/9. So then let's compare this to the derivative of prime of X. We see that this is going to equal 1/9. Thank you. So that's gonna end up being our final answer.


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