Root-Finding

Date: 02/09/98 at 17:24:40
From: Anonymous
Subject: Root-Finding

Dr. Math,

Hello. I'm currently enrolled in a Numerical Analysis class. There is 
this one problem which I've made several attempts at, and being that 
it seems so simple, it is driving me more insane.

We are given a polynomial of 3rd degree:

  X^3 + 5X - 2 = 0

and we are asked to show that this function has EXACTLY one root over 
the interval [0,1].

Now this is what I have done so far, unfortunately to no avail:

  I first found the function values at each endpoint: 0 and 1;
  Thus:

   f(0) = -2; and f(1) = 4;

This would indicate to me that since the values are of differing 
signs, that the function does cross the x-axis AT LEAST once. I have 
to show that it crosses EXACTLY once and where.

Well, I then used simple algebra by factoring the left side:

   X(X^2 + 5) - 2 = 0

then I got X = 2; and X^2 = -3 {Cmplx: so disregard}

Obviously X = 2 is not a root, since when I plug it back into the 
original function, I don't get 0.

So I'm wondering, where am I going wrong? This is NOT a homework 
problem. I am in the process of reviewing for my first exam in the 
class coming next Thursday.

I would appreciate any assistance whatsoever,

Ibrahim Siddiqui

Date: 02/10/98 at 14:50:48
From: Doctor Wolf
Subject: Re: Root-Finding

Hi Ibrahim,

  Some quick observations about your problem:

> X^3 + 5X - 2 = 0; and we are asked to show that this function
> has EXACTLY one root over the interval [0,1].

First, you're correct in your observation that f(0) = -2 and
f(1) = 4 implies the existence of at least one root in (0,1)

Since it appears to be a relatively simple problem, why not appeal
to calculus: since the derivative (3X^2 + 5) is positive in (0,1),
the function is strictly increasing on this interval, and so can
have just the one root. (Visualize the graph ...)

I hope this has helped. Don't hesitate to drop in again with
this or another problem ...
  
-Doctor Wolf,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/