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Use Newton's method to approximate the indicated root of the equation correct to six decimal places.The root of $x^{4}-2 x^{3}+5 x^{2}-6=0$ in the interval $[1,2]$

1.217562

Calculus 1 / AB

Chapter 4

APPLICATIONS OF DIFFERENTIATION

Section 6

Newton's Method

Derivatives

Differentiation

Applications of the Derivative

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Okay, So the root of the equation f of X is equal to X to the fourth minus two X cubed plus five X squared can be estimated, um, using Nunes method. So recall Nunes meth. It says that Well, given the end approximation, the, um, except and plus one is equal to well, except and minus F of x seven divided by F prime of X event. Okay, so we have our function after backs, right? We have to find, um, are derivative. So are derivative. F prime of acts is just equal to while four x cubed, um, minus six x squared. Plus 10 x. Okay. So, um, substituting these values in, um, we get that X sub and plus one is equal to Well, um X sub end minus while x up end to the fourth minus two X up and cubed. Plus five. Um, x up and squared. Um, minus six. Divided by, um, So it was his a minus six year. Right? That's OK, but that I'm divided by, um four x sub and ah, Cube minus six. Xa been squared. Plus 10 x sub. And Okay, so, um, this well becomes, um right. We get this as well. Our numerator here becomes four x amended of fourth minus six X seven Q plus 10 X events where at my exhibition plus two xy been cubed minus five X events where plus six. And then, um, this becomes, well, three X sub and to the fourth minus four x of end cube plus five x of end squared plus sex. Um divided by four X sub and pube minus six X up and squared plus 10. Excellent. Okay, um so if you were to draw the graph of our function um, effort Max, we can observe that 1.2 is a possible close value, a close value to the root. So if 1.2 is, you know, a possible close value, we can let you live. Weaken. Let's let X someone be equal to, um 1.2. Okay, so except one is equal 1.2, then what is except to accept to is equal to the phone we have here says Well, except two is equal to, um right, three except one to the forest minus right. So using this formula here, plugging an ex of one is equal to 1.2, we get that except to is equal to approximately, um, one point to 177 57 Okay. And similarly, we confined except three by then, plugging in our value of except to, um, into our formula here right now. Except to is now, um, exit meant so to find x some three we get, except three is approximately while three times except to in the fourth, minus four, except to queue, but above all right, So you accept three is about 1.2 17 562 Okay. And we observe also that the same method that except four, um is well, 1.2 17 562 Right. So the fact that except through it, except for, um, are both agree to 12345 to 6 decimal places. Well, therefore, the route of our function the root of F of X, um, is equal to X to the fourth minus two X cube plus five X squared, minus six, right. The root adjusted to six decimal places is given right here as 1.217562

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