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Use Newton's method to find all roots of the equation correct to six decimal places.$(x-2)^{2}=\ln x$

$x=1.412391$ and $x=3.057103$

Calculus 1 / AB

Chapter 3

Derivatives

Section 8

Linear Approximations and Taylor Polynomials

Harvey Mudd College

Baylor University

University of Nottingham

Idaho State University

Lectures

04:40

In mathematics, a derivati…

30:01

In mathematics, the deriva…

01:48

Use Newton's method t…

01:36

01:38

00:57

02:35

01:32

05:31

06:44

04:10

05:52

03:54

06:48

06:33

Finding all roots Use Newt…

01:27

Solve. Where appropriate, …

01:59

Approximate the root speci…

00:20

00:53

Find all the real-number r…

01:08

01:16

Hello. Very. But you are going to solve the problem number 15 for years and method, which is using to solve nonlinear equations. So we have to follow some steps to get the results. Okay. Still number one, we have to get the derivative F dash X for the new function. Step # two We have to use initial solution or initial value. That will be yes sir. And the question initial value for X as you know, and or we can assume it's okay. So, so on the third step is substituting and newton formula X M plus one equal X n minus F of X and over a dash of X. And Okay, so the question we have to get the roots Oh, the functions ffx equal sorry? Uh X managed to all squared minus Alex X. I used to all squared minus ling X. Okay, so it's all be uh very simple to get the result and you can first simplifications that the main function. Okay, so the drift of that function, it will be equal. F dash X will be equal. X two, X minus two minus one over X. Okay, so when we get the graph of that main function, it will be like that. Okay, so we can that there are two intersecting points between the main function and access. Okay, so one of them here and the second is here. Okay, so one of them needs to or located between one and 2 and the second one? Uh is there's 23. Okay, so we can we have to form our table and we can start with one As into initial solution for the first solution. So we can get at at the 4th and 5th iteration, it's almost the same value. So the first one is equal 1.41 2391. That's first solution. And for the second one we can use three initial solution and it gives us at the 3rd and 4th, nutrition almost the same value and the fortune after substituting with that value, it will be equal zero. So the second one will be Almost equal 3.0 sorry three point oh five 71 zero. Well, okay, so thanks for watching and this you later in the Naked Newton Method application.

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