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Use Newton's method to apporzimate the indicated root of the equation correct to six decimal places. The negative root of $e^x = 4-x^2$

I do not know what a negative root is nor do I really know what I am supposed to do. I am guessing raise everything by loge.

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    What Listing said was that a "negative root" is a root that happens to be a negative number. A root of an equation is a *solution* to the equation. So they are asking you to find the negative value of $x$ that makes $e^x = 4-x^2$ true.2011-10-27

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The negative root is the one where $x<0$. This equation has two roots, one near $x=-2$ and one near $x=1$. You could look at this plot

You are solving $0=4-x^2-e^x$. For Newton's method, you need to take the derivative, find a starting value like $x_0=-2$, then iterate x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)} until the change in $x$ gets smaller than $10^{-6}$

Newton's method works well when you start close to the root as long as the derivative doesn't get too small. In this case, I find convergence within three cycles.

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    I know all about it but it just doesn't seem to work when I use it.2011-10-28
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e^x=4-x^2
Subtract e^x
4-x^2-e^x=0
Take Derivative
-2x-e^x=0

plug this formula into the graphing calculator:

x-((4-x^2-e^x)/(-2x-e^x))

(NOTE - this formula can be used with any Newton's method problem when needing to find roots: x-(original fx/derivative of fx)

From the graphing screen, hit 2nd + Trace, select Value, put x=-1. The y-value given on the screen is the "new" x-value. Now do 2nd + Trace again this time typing in the "new" x-value, the new given y-value is the next "new" x-value. Keep doing this over and over until the y-values stop growing/shrinking.

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    Welcome to MSE! It really helps to format questions using MathJa$x$ (see FAQ). Reagrds2013-04-07