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Derermine whcthct thc Mcan Value Thevccr spplic {0 Ihc fLncticn [4 cprn intcrval (-6,4} such that f' (c) = F(-6) MVT does not apply; No value ol € ExIsts...

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Derermine whcthct thc Mcan Value Thevccr spplic {0 Ihc fLncticn [4 cprn intcrval (-6,4} such that f' (c) = F(-6) MVT does not apply; No value ol € ExIsts MVTapplies €App€MYT Joolles c >

Derermine whcthct thc Mcan Value Thevccr spplic {0 Ihc fLncticn [4 cprn intcrval (-6,4} such that f' (c) = F(-6) MVT does not apply; No value ol € ExIsts MVTapplies € App€ MYT Joolles c >



Answers

Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of $c$ guaranteed by the theorem. $$f(x)=\frac{x^{2}+x}{x-1},\left[\frac{5}{2}, 4\right], \quad f(c)=6$$

Able to one minus the absolute value of fulfills the mean value zero only SP and a our booth positive or both. Negative. So here we have our mean value here. Um, and what we can do is take the derivative of acid X. But since it is an absolute value function, then we can split this up into when X is greater than or equal to zero. R A derivative is negative one and one hour X is less than zero hour. Derivative is equal toe one. So now we can apply, um, our derivatives and put them into F time at sea, in the immune value through him. So we have negative one and then assuming, um, both A and B are positive. We have one month minus minus. Ah, one minus a over B minus A. And we can simplify that to one minus B minus one plus a over be minus a um And then we can cross out one minus one, which gives us a minus B over B minus A. And we know that this is equal to negative one. So that is true for X is greater than are equal to zero and then for X is less than zero. We have our derivatives, you little one. So we can do one plus B or one minus negative B AA minus one plus a or one minus nine of a on. But then we want to distribute that. So we have one plus B minus one minus a again. We can cross out the ones, Um, and that leaves us with be minus a over B minus a. Which is you goto one. So, um and we can see that if we change the sign of either be over A that would just switch on the order of thes expressions and it would not be equal to the dirtiness.

For this problem first, let's verify the intermediate value theorem applies here. So, for the intermediate value theorem to work and must meet the following three criteria. First, the curve is the function Y equals F of X. Now, if we look here, we do have an fx that's given by x squared minus six X plus eight second, which is continuous on the interval A. B. The function must be continuous. Well, our function is a second degree polynomial which is continuous for all real numbers in its domain, so 0 to 3. That'll work just fine. And W is a number between F of A and F A B. This is what we must verify for this problem. Now, W is going to be given by this number right here. FFC is equal to W. So are W number is zero, and it must be between F zero and F three. If it doesn't fall between these two numbers, then the intermediate value will not. The intermediate value theorem will not apply here. So first it solved for F zero, F zero is going to equal zero squared minus six times zero Plus eight, which is just eight. Now, F F three is given by three squared -6 times three plus eight. Which is just don't worry about that. Nine minus 18 plus eight. Or It's gonna be a -1. And so let's check that zero fall between negative one and eight. Yes, it does. Doesn't matter if F zero is the bigger number of the smaller number, as long as the value W which is zero, of course falls between the two. So it will work. And we can prove that FFC does coach W. So there is some value. See that when plugged into this function will give us an output of zero. So let's find it. Well, let's first try Our other integers so we can try f of 1 1st. And that's going to give us Spirit. Mind, did the function up here, it's going to give us one squared -6 times one plus eight. And when we saw for that we'll get one minus six negative five plus eight. That's going to give us a positive three. So that's not quite it. Let's try to Now, since that's the only other integer in our given domain, two squared minus six times two plus eight. Which is going to give us four minus 12 Plus eight. And that's simply going to be four four plus eight minus 12. That's gonna be zero. So we can prove that C is equal to two, which we solved four by plugging in. And we proved with the intermediate value theorem that it is possible that it's guaranteed that there is a value of C. Such that zero would have been an output.

For this problem, we are asked to verify the intermediate value theorem and then find the value of seat when it's guaranteed by the theorem. So the theorem has three constraints that must be met. First, the curve is the function Y equals F of X, which is true. Were given F of X is equal to x cubed minus x squared plus x minus two. Next the function must be continuous on the interval A to B. Now ffx is a third degree polynomial, so that means that it's continuous for all real numbers for on its domain, so 0 to 3, that will work fine. And w is a number between F of A and F F B. So we must find that fc equals four. We must find that four lies between f f zero and f F three. So, first, let's find f zero f zero is going to equal zero cubes minus zero squared plus zero minus two, which is just going to give us negative too. Next F of three is equal to three square three cubed minus three squared plus three minus two. Which is gonna give us 27 minus nine plus three minus 2. 27 minus nine is 18 plus three is 21 minus two. That's gonna give us 19. And so four lies between negative two and 19. So all three criteria have been met. So we can verify that there is some number that some numbers see such that F F C is equal to four in the situation now because we've already looked at zero and three, it's probably a good idea to test out one and two to see if either will fit uh the great we'll give us four. So F of one is going to give us one cubed minus one squared plus one minus two. Which is going to equal to it's gonna be one minus +10 plus one. And then So it's going to give us a negative one so that's not quite it. And then we'll try FF two and that's going to give us two cubed minus two squared plus two minus two. And let me scroll down. Sorry about that. We scroll down a little bit. We can see that's gonna be 8 -4 Plus 2 -2. So 8 -4 is going to give us four Plus 2, 6 -2 back to four. So it's confirmed that F of two is equal to four, so C is equal to four. Sorry sorry? No no no C is equal to 22 And we proved it that that that's true through the intermediate value theorem that it's possible.

So here we use the intermediate value fear. Um, so we're gonna do first is we're gonna go ahead and evaluated at the end points so f of five halfs belt equal. Five halves squared, plus five halves. All that over five halves minus one. So that gives us 25/4 plus five halves. All that over three halves. And that's gonna equal 35/6. And this value is less than six. So western FFC. Next we plug in at the four, you get four square plus four. All that over four minus one. Thank you. US 20/3. This value is greater than six. So then what we're doing here is we've verified that with immediate value. The're, um there is some value some see between the interval of five halves to four. Such that f s illegal. Six e. Verify this by showing that at the left end interval, you have some value. That is less than six. So let's put six up here somewhere. So you have six. Maybe around here you have five over to somewhere around here, and you had the value for this, which is less than six. So you got something over here. And then at four, you have some value that's above six. So that you connect these dots together. You see, you will pass through the 0.6. So there's some values see there that makes this function equal to six.


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