Problem

Use the Intermediate Value Theorem to show that t…

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Problem 54 Hard Difficulty

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

$ \ln x = x - \sqrt{x} $, $ (2, 3) $

Answer

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Video Transcript

Okay, we want to find a route of this equation. Um where we want to show that there is one in the interval from 2-3 using the intermediate value there. So root usually means the value of some function is zero. We actually don't have that. So we have to make that. Let's make the function natural log of X minus x plus route X. Because if that were zero then this equation would be true. Right? We can get from there to that equal zero by adding or subtracting stuff from both sides. Okay. Um So we have that. Uh let's look at uh this function is defined on the endpoints. Natural log is defined for two and three. So let's look at the values of FF two and Ffs and because we have a natural log we need a calculator, I'm not going to be able to show that on the screen because I don't have a good computer calculator. But whenever you're typing into a calculator, just make sure all the parentheses work out correctly. So FF two, Let's see plus the root of two um Is approximately and we just need an approximate value 0.107 And then f of three is approximately negative 0.169. And this negative is very important. So what's going to happen, let's sketch this. I don't know what this function looks like, but between two and three The value at two is some positive number. The value at three is some negative number and the function is continuous continuous In the interval 2- three. so there's no way to get from 2-3 without lifting your Pencil or pen without passing through zero. So that means the intermediate value theorem says there exists Some value of x in the interval 2-3 such that FFX is zero where zero is just, I can pick any value, so I could have picked .05. That's also a value between .107, a negative .169. And there would have to be a value because I can't get from one to the other continuously without going through all the values in the middle.

Stark L.

University of Washington

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 5

Continuity

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