Expert: Paul Klarreich - 1/16/2008

Question
Hello!
For Newton's Method, I can work easily if it was just one equation but this problem here, gave me too and I really do no t know what to do.

1. Apply Newton's Method to approximate the x-value of the indicated point(s) of intersection of two graphs. Continue the process until two successive approximation differe by less than 0.001.
             A) f(x) = 2x + 1
                g(x) = sqrt (x + 4)

Here's a graph that I tried drawing: http://i6.tinypic.com/7xe6ybb.jpg

Note: The red line (g) has a slight curve to it so it's not a straight linear line.  

THANK YOU VERY MUCH!

Answer
Questioner:   Elaine
Category:  Calculus
Private:  No
 
Subject:  Newton's Method
Question:  Hello!
For Newton's Method, I can work easily if it was just one equation but this problem here, gave me too  (TWO?) and I really do not know what to do.

>> Yes, you do.  YOU just explained it.  Just rewrite the problem so there is one FUNCTION, not two.

1. Apply Newton's Method to approximate the x-value of the indicated point(s) of intersection of two graphs. Continue the process until two successive approximation differe by less than 0.001.
            A) f(x) = 2x + 1
               g(x) = sqrt (x + 4)

Here's a graph that I tried drawing: http://i6.tinypic.com/7xe6ybb.jpg

Note: The red line (g) has a slight curve to it so it's not a straight linear line.  

THANK YOU VERY MUCH!
..................................
HI, Elaine,

At your pt of int, the difference between f and g will be zero. So write:

h(x) = g(x) - f(x) = 2x +1 - sqrt(x + 4)

and solve the equation :  h(x) = 0 by N.M.

which says:

h(x1) - h(x0) = h'(x0)(x1 - x0)

We hope h(x1) = 0, so that is:

 - h(x0) = h'(x0)(x1 - x0)

x1 - x0 =  - h(x0)/h'(x0)

x1 = x0 - h(x0)/h'(x0)  << That's N.M.

Now get h'(x) = 2 - 1/2sqrt(x + 4)

and choose x0 = 1 (from your graph.)

I think you can do the rest. It's:

x1 = x0 - [2x0 +1 - sqrt(x0 + 4)]/[2 - 1/(2sqrt(x0 + 4))]

Excel is good for this.