Sketching a Polynomial
Date: 04/04/2002 at 23:53:03 From: Shikhar Subject: Wavy curve method for solving inequalities Dear Dr. Math, When solving inequalities of the form P(X)/Q(X) > 0 or P(X)/Q(X) < 0, we have to find the roots of P(X) and Q(X) and plot the roots on the real line in increasing order (excluding common roots). After that a curve is drawn starting beyond the extreme right root along the real line and going toward the left. This curve is said to be changing its position alternately between consecutive roots. 1. It is not clear to me why the curve should change its position or sign at roots. Can't the curve have a positive value for all roots? 2. Is the curve symmetrical about the real line? 3. Why exclude common roots?
Date: 04/05/2002 at 16:04:25 From: Doctor Peterson Subject: Re: Wavy curve method for solving inequalities Hi, Shikhar. I am not familiar with this method, but it makes sense to me, with some clarifications. Essentially, you are sketching the polynomial PQ, whose sign is the same as P/Q for any given x, paying attention only to where it is positive and negative, not to how far it goes from the axis. Imagine the function factored: P(x) (x - p1)(x - p2)...(x - pm)(product of quadratic factors) f(x) = ---- = -------------------------------------------------------- Q(x) (x - q1)(x - q2)...(x - qn)(product of quadratic factors) The quadratic factors, which can't be factored further, will each be always positive (or always negative - your description leaves out the fact that the highest degree terms of P and Q must be positive in order for your curve to start above the axis at the right). So we can ignore them as far as the sign is concerned. The p1 ... pm and q1 ... qn are the roots of P and Q. Now, as you go along the x axis, at some point you will pass a root, say pk. As you do so - if there is only one factor (x - pk) - that one factor will change sign, and no others. Therefore, f(x) will change sign there. But if there are two identical factors (either in P or Q), they will both change sign, and f(x) will not. However, you really shouldn't completely ignore them, because they will cause f(x) to be zero (for a root of P) or undefined (for a root of Q), which will affect your solution. Instead, you should show your curve just touching the axis without crossing it. Therefore, you have to not only cancel common factors (ignoring common roots), but also "cancel" pairs of identical roots within either polynomial. So if a root appears an even number of times in P and Q together, it will not show up as a crossing on your curve (though it will be a zero or undefined point); if it appears an odd number of times, it will act as if it appeared once. You should try actually graphing several functions like this to see how they compare to the curve you are sketching. You will find that roots shared by P and Q have no effect at all (other than forming a "hole discontinuity" in f(x), since the function is not defined there); roots appearing twice in the factorization of P will make the curve tangent to the axis; roots of Q will produce a vertical asymptote (undefined value), and for double roots of Q the curve will have the same sign on both sides of the asymptote. All in all, I think I would rather see you learning to sketch rational functions (as I just described in the last paragraph) and apply the result to the inequality, rather than learn a related, but less accurate, trick for handling the inequality alone. Curve sketching gives you a fuller understanding of how the function behaves, and is not much more work than what you have been taught. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum