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.