(a) Prove that the equation has at least one real root.
(b) Use your calculator to find an interval of length 0.01 that contains a root.
$ \ln x = 3 - 2x $
It's a problem of fifty eight of this tour. Calculus eighth edition Section two point five party proved that equation has at least one real route. Part B. Use your calculator to find and a roll of length point on one that contains a route. The function A dick question, given his Ellen X minus, equals three minus two x And what we do is we rearrange the function to have a form of effort X equals to zero. So moving the terms and the right to the left and we call all the terms on the left effects and finding the route means meaning meaning, meaning that we find an ex value that thanks this function equal to zero. Um, we will not exactly calculate this route, but we will first prove that it exists in party and then never down where it might be in part B of a party. Ah, we take a look at the function and we see that it's a log rhythmic function. Plus a linear function is a constant palli. The sum of all these functions since they are independently continuous provides us a function that is ah, totally continuous on its own domain. Allen of X has a certain domain. I've only zero not included to infinity s o. The domain that we're looking at is strictly X is greater than zero. Ah So with that in mind, we know that the functions continuous on this X is greater than zero domain and will only concern ourselves with that domain. So we will take a look at the extreme values for the function. First looking at exes, the function is an expert zero from the right next and do toothy ln function, this is approaching negative. Infinity is your princess from the right on this term just goes to zero and this is negative three. But this goes too negative affinity which are which dominates the function. And then as we approach infinity for axe the function ah also purchase Infinity is the natural argue function and the linear function does go towards infinity and so were we have shown is in the dark domain that we have of interest from zero not included to infinity. The function takes on all the values between native into the infinity because we know it's a continuous function on this domain and through the Internet about this here. It must have taken the value zero, which is between these two numbers. And so in that case, we could confirm that this function does have the route at least one from control zero to infinity. Now, in part B, we're trying to narrow down this interval. Since it is a very large interval, we want to narrow it down to an interval. That is length point one, you know, first, start off by some trial and error. Look, ATT. Some smaller numbers taking on values that say of us of one Ah, which in our case would be Alan of one zero plus two times one just to ministry. So this is negative one. And then after of two to this inn into a wall of from one to two. This will be Ellen of two plus two temperatures for minus three. Okay. And if we do Eleanor to your calculator, I add form. In this three, we get about one point sixteen and we see here that between one and two, the function changes values from a negative value to bother to value. So the function must have taken the value zero between this interval from one to was a lot more trial and error. We can narrow down the interval two very small length of point one. And if we take a look at the value of the function at one point three four and one point three five after much Charl, they're We have narrowed down to this small Inderal, where at one one point three four on the function takes on the value of about native zero point out to seven. And for one point two five, it's actually going to be a very small number because the value is very close to one point three five O. R. Zero is very close to one point three five on this day that it takes on this point Oh, Oh oh One oh five. So we know that zero is somewhere between these two because the function changes from a negative value to positive value. So with a lot of Charles, Harry can use a calculator to determine this exact interval, and you could even expand further, adding more decimals. Teo narrow down exactly what the value of the route is. But for the the purpose of this problem, we have just found the interval of link point on one. That disk contained this route