# Newton's method with negative roots and variable in the exponent

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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A negative root $x$ of a function $f$ is a value $x \in \mathbb{R}$ such that $x < 0$ and $f(x)=0$ –  Listing Oct 27 '11 at 21:11
Also the question is not really different from the one you just asked ( math.stackexchange.com/questions/76421/newtons-method-problem ), didn't you learn anything from the answer? :-/ –  Listing Oct 27 '11 at 21:15
I have no idea what you just said. –  user138246 Oct 27 '11 at 21:27
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. –  Arturo Magidin Oct 27 '11 at 22:26

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 don't understand what is happening, I keep using it and I am getting numbers smaller than 1 eventually. –  user138246 Oct 27 '11 at 22:28
You should read about Newton's method. The basic idea is that you have a point on the graph that is close to the root. The derivative gives the tangent line to the graph, and you project that tangent down to the x axis for your next guess at the root. Then continue until you are as close as you want to be. –  Ross Millikan Oct 28 '11 at 0:48
I know all about it but it just doesn't seem to work when I use it. –  user138246 Oct 28 '11 at 12:38

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 MathJax (see FAQ). Reagrds –  Amzoti Apr 7 '13 at 20:52