Rational Function Range-Finding, with and without Calculus
Date: 06/01/2012 at 16:55:47 From: Ibrahim Subject: Finding the range of a rational function (special case) Finding the range of a rational function is not always easy or intuitive -- particularly when the degree of the polynomial in the numerator is greater than that in the denominator. An example would be f(x) = (x^2)/(x + 1) How can I find the range of this function without using a graphing calculator? When we draw a graph of the function, a gap appears between two symmetrical curves. How can I find where that curve starts or ends algebraically, i.e., without a graphing calculator? If I knew that, then I could find the range relative to the y-axis. In the example above, the vertical asymptote is x = -1. Because the degree of the polynomial in the numerator is greater than that in the denominator by one, the "horizontal" asymptote is a line. By long division, we find that the line is y = x - 1. Now the domain is all R except -1. But the range? No clue. I've worked every part of the function except the range. I just don't know how to find it. I've searched everywhere for an algebraic method to find the range of this special case of rational functions, but I haven't found any. Even my teacher in school, who has a PhD in math, couldn't give me an answer.
Date: 06/01/2012 at 17:33:40 From: Doctor Peterson Subject: Re: Finding the range of a rational function (special case) Hi, Ibrahim. I suspect your teacher has no answer in one of two senses: either you are not yet fully prepared for an analytic approach to this particular problem; or, more likely, no general method will work for any rational function of this sort. Having said all that, this particular example is actually rather easy. I can think of several ways to go about it, and they all work for this one, but wouldn't work for a harder case (higher degree). The first thing I think of is calculus. You can find out, without too much trouble, where the graph is horizontal, which gives the location and values of the local minimum (x = 0, y = 0) and the local maximum (x = -2, y = -4). Then you can see the range that your graphing calculator showed you -- namely, (-inf, -4] U [0, inf) Assuming you don't know the little bit of calculus that is needed to do that, my next thought is to try inverting the function, as the range of a function is the domain of its inverse. Of course, the inverse here will not be a function. But if it is possible to solve for x (giving multiple solutions), you can then find for what values of y there is a solution. Those values of y are the range of our function. That can be done in this case. Solve for x by multiplying both sides by (x + 1), which will give -- treating y as a constant -- a quadratic equation in x. Apply the quadratic formula, and you have two values of x for any y for which the discriminant is not negative. So your range will be the set of y for which that discriminant is greater than or equal to zero. One more thing you could do is to sketch a graph by hand and guess that (0, 0) is the local minimum on the right branch of the graph. Then you can show that this is true by considering the sign of y for small positive x, and for small negative x: When x > 0, x^2/(x + 1) is positive When -1 < x < 0, x^2/(x + 1) is positive Since when x = 0, x^2/(x + 1) = 0, that's the minimum! Then you can use symmetry (or another guess) to find the local maximum. I've deliberately left a lot for you to work out, so you can get a chance to do some serious thinking for yourself. I think you'll enjoy it. But, just to repeat: you (and if I may presume intent, your teacher) are right that in general there is no way to algebraically find the range in all cases. But these ideas are all good tools to have in your pocket when you need to try, or just want a challenge. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 06/01/2012 at 17:56:17 From: Ibrahim Subject: Thank you (Finding the range of a rational function (special case)) Thank you very much, Doctor Peterson. Your answer was very clear and helpful. About what my teacher had told me, now I see that I wasn't quite ready for it (since I haven't studied calculus). However, I still get your point about looking for where the graph is horizontal. Thanks again!
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