Restrictions on Variables in DivisionDate: 8/4/96 at 18:25:6 From: Shannon Subject: Restrictions on Division In simplifying rational expressions I have to identify the "restrictions" on several expressions, but I'm having trouble. I know that: 1. The restrictions are in the denominator, not the numerator 2. It's not possible to have a term in the denominator containing a variable equal to zero. If it does, it becomes a restriction. For some reason I'm not doing it right. I have provided two sample questions. I hope you can help. Thanks. a) y^3 + 8 b) x^3 + 64 ____________ ________________ y^2 + 3y + 2 ax + 4b + 4a +bx Date: 8/4/96 at 20:39:12 From: Doctor Paul Subject: Re: Restrictions on Division The two 'rules' you wrote down are correct. Here's the idea: The 'restrictions' to which you refer occur because the denominator of the function equals zero. At this point the function is 'sick' and many of the rules we use to analyze functions do not apply. This is because it is 'illegal' to divide something by zero. There are explanations in the Dr. Math archive: http://mathforum.org/dr.math/problems/divide.zero.html http://mathforum.org/dr.math/problems/harmon2.12.96.html http://mathforum.org/dr.math/problems/sky.html In your first equation: y^3 + 8 ------------- y^2 + 3y + 2 you want to be able to identify the points where this function is 'sick'. The function will only be sick when you try to divide by zero. In other words, it will only be sick when the denominator is zero. So to find the points where your function is sick we set the denominator equal to zero and solve: y^2 + 3y + 2 = 0 This factors nicely: (y + 1)(y + 2) = 0 So when y = -1 or when y = -2, your function is sick. These would be the two points that have restrictions for this equation. Question two: From the above we know that all we have to do is set the denominator equal to zero: ax + 4b + 4a +bx = 0 x(a + b) + 4(a + b) = 0 (x + 4) (a + b) = 0 so when x = -4 or when a = -b, your function is sick. Note that if x = -4, then the values of a and b don't matter... whatever they are, your function will have a zero in the denominator. I hope this clears things up. -Doctor Paul, 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/