Root-FindingDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/