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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) $

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04:25

Daniel Jaimes

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Missouri State University

Baylor University

University of Michigan - Ann Arbor

Lectures

04:40

In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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Use the Intermediate Value…

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Use the intermediate-value…

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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.

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