Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.
$ \sin x = x^2 - x $, $ (1, 2) $
this problem Number fifty six of the Stewart character. This eighth edition section two point five used the intermediate value theorem to show that there is a root of the given equation. In the specified interval Sign of X is equal to X squared of my specs. The interval is from one to two and we're going to re arrange us to have it look like a function Ah, on one side, equal to zero. We're just going to check sign on both sides and Rick Lucy, this is equal to zero and we consider all the terms of the will Send it to be a function F and And our goal is to confirm that there is a root a root meaning that there's a value for exit makes this function left side equal to zero. We want to prove that this is true, that there exists at least one route and we'll use intermediate value theorem. Unscrew this and the Internet media value theorem states that for a continuous function Ah, between an interval from a to B that the function will take on every value between the function evaluated, eh on the function evaluated at B so For that reason, we're going to calculating and figure out what the function value is at the end. Points first as one gives this one squared minus one minus sign of one and using a calculator, we consider that this is approximately the zero point eight four at the other end point effort, too. I mean, we have two squared, which is four minus two, my sign of two. And again, in our calculated, we get an approximate value of one point one. So we've confirmed the F one eyes, a negative number, meaning that is zero. And that, too, is a positive number. We need that it's greater than zero, and the reason that we single out zero's because we want our function tio have take on the value of zero. And as long as this function is continuous, the intermediate Valium states that the function will take on every value from negative point eight four up until one point one, including the value of zero. The function is X squared minus X minus annex individually X squared is continuous as a polynomial X is continuous as a linear function sign of exes continues is a trick in magic function. The sum of continuous functions is also continuous somewhere the difference. So this function is confirmed to be continuous. It will take on all the values between these two values, including zero. And in that way we have confirmed that there is a least one route for this one should have.