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Find all numbers € that satisfy the conclusion of the Mean Value Theorem for the following function and interval. Enter the values in increasing order and ent...

Question

Find all numbers € that satisfy the conclusion of the Mean Value Theorem for the following function and interval. Enter the values in increasing order and enter N in any blanks you don't need to use: f(e) = 24x? + l62 + 10, [-1,1]NNNSubmit answer

Find all numbers € that satisfy the conclusion of the Mean Value Theorem for the following function and interval. Enter the values in increasing order and enter N in any blanks you don't need to use: f(e) = 24x? + l62 + 10, [-1,1] N N N Submit answer



Answers

Average values Find the average value of the following functions on the given interval. Draw a graph of the function and indicate the average value. $f(x)=x^{n} ;[0,1],$ for any positive integer $n$.

Before we begin, we could confirm that a graph of L N of X is continuous and differential from 1 to 4. So we're good on the conditions and then we'll go ahead, set up. But the slope of the, uh, seeking line is that's four minus up with one all over four minus one. And that, in other words, would be. But the slope is connecting these two points that suit get line and let's see, we'll end up getting, um, f of 10 in the bottom. That's the same thing. It's three, so really on the left side, there will end up with Ellen of four over three. And that's that seeking slope on the right hand side. We want to find out witnesses equal to what? See value. Does the tangent line have the same slope? In other words, what is the sea for F prime toe? Have that, um, and so taking the derivative that's gonna get us one over X, the derivative of Atlanta Becks, and then just cross multiplying here or taking the reciprocal of both sides. Alternatively, we'll end up getting three over l and four, and that would be our see value. You just double check at that fits in between. We need to also make sure it's a value between one and four. So we divided by L on the fourth. It is. It's about 2.16. So that would be somewhere around here. Well, that tells us that will have a slow tangent to the graph, which is parallel to that red line. And so this is our si valley.

The following problem. We want to verify the intermediate value theorem. Um So we're going to find the numbers, see um that's given to us for the indicated value of and so we see that if we take F of X. Okay, Which is equal to X squared. That's ex That's one. Um We know that we're going to be looking between The interval of negative 2-3, Half of negative two gives us three, and that's the three gives us 13. So that means that six must be in here somewhere. And sure enough, we see that if we drag it over here That it's about 1.79 um is how we would end up getting six is our answer, but through the intermediate value for and we see that six has to be somewhere in this interval.

We need to find the average value off the function. Ffx is equal to X rays to the part of underrated by end in the interval, 0 to 1 for any positive and teacher, and we know that the average value of a function is given us one divided by B minus. Ain't able running from a to B for ffx DX, which will be equal to one divided by one minus Rigaudeau integral running from 0 to 1 extremes to the part of one divided by N bx, which will be equal to one multiplied with X rays to the part of Underrated by n plus one divided by one divided by n plus one, with limits running from 0 to 1, which will become equal to X rays to the part off one plus n divided by end, divided by one plus in divided by n limits earning from 0 to 1, which will be equal to X rays to the part off. One plus indicated by N multiplied with n divided by one plus n With limits starting from 0 to 1. Now, this will be equal to one. The race to the part of one plus and divided by N multiplied, with end divided by one plus F

Yeah, we're gonna start by verifying the mean value theorem. A kicker. Let's give me fx. It's equal to three X. Where That's 5-plus 5. We're looking between negative one and 1 and we see that it is going to be continuous and differentiable over this. So we consider F of one -F of negative one divided by one minus negative one. So we end up getting excellent to be too. Then let's consider what we have is have prime of X. That's kind of giving us uh six x. That's jim. And ultimately we wanted to equal to for a slope we can subtract the two over. We get six X equals zero and that ended just giving us X equals zero. So with that we consider our function and see the X equals zero. The slope is going to be too If you just took us five is the tangent line. So their final answer.


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