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Write the formula for Newton's method and use the given initial approximation to compute the approximations $x_{1}$ and $x_{2}$.$$f(x)=x^{2}-2 x-3 ; x_{0}=2$$

$x_{1}=3.5, x_{2}=3.05$

Calculus 1 / AB

Chapter 4

Applications of the Derivative

Section 8

Newton's Method

Derivatives

Differentiation

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All right, so here are function. F of X given to us is X squared minus two x minus three, and our initial value X not is equal to So what does Newton's method say? Well, Nunes Method says that Xsa M plus one is equal to x of n minus f of x A been divided by f prime of x event. So to use that formula, we're gonna have to find while the derivative off affects so f of X is X squared minus two x minus tree, then F prime vax. A crime of X is just equal to well X squared is just two X and the derivative of minus two. X is minus two. So and fruit of a constant zero salt are derivative is two x minus two and then, well, gonna find exit one. And except to so, x of one is just equal to except one X. Someone is equal to well except end in this case, except zero, which is to accept one is equal to two minus well f off to take to plug it in f of X, we get what two squared is four minus four us that zero minus streets that's minus three. So we have two minus a minus three. So two minus a minus three divided by F prime of to so are derivative. Function is to X minus two. So we take our to implement plug it in there and we get two times two is four minus two is two. So we've got except one. You go two to minus a negative three halves or two plus three halves. Which would be, um, two plus 1.5 sets. 3.5, right? Yeah. So 3.5 there is our ex sub one And now rest defying exit too, except to So it's x two or X. It, too, is just again, except M plus one is good, except and so except two is equal to X. Someone just found was 3.5. So, except to is equal to three point five minus what? We're minus f of X event. So now our exit bend is, except one, which is 3.5. So we've got, um, except to is even a 3.5 minus f off 3.5. They 3.5. We plug it back into our again our original function after VAX, which is X squared minus two X minus three. So plug in 3.5. There he is, 3.5 square minus two times 3.5 minus three. And you work that out and you should get, I believe, 2.25 So 3.5 minus 2.25 divided by F prime of X event. So again are derivative. Function is to X minus two and you plug in 3.5. There you get 3.5, um, times two minus two, 3.3 times two is ah, seven writes that stuff minus two, which is five to get 3.5, divided by 2.25 Divided by five in this 2.2 5/5. Very weight. Teoh 0.45 3.5 minus 0.45 is equal to three points. 05 There you go. So are with our initial approximation of to, um we can see that, um, we get to it our better approximation with 3.5, and then we use that and plug that right back into our formula for a murder iterated former here. And we get 3.5 as our ex ah to

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