## Finding the range of rational functions algebraically

how to find the range of rational functions like f(x) = $\frac{1}{{x}^{2}-4}$ algebraically , i graphed it and seen that (-1/4,0] can not be in range . generally i am interested in how to find the range of functions and rational functions in particular
 PhysOrg.com mathematics news on PhysOrg.com >> Pendulum swings back on 350-year-old mathematical mystery>> Bayesian statistics theorem holds its own - but use with caution>> Math technique de-clutters cancer-cell data, revealing tumor evolution, treatment leads
 Recognitions: Homework Help Science Advisor How do you think it might relate to the maxima and minima of the function? Where are those for your example?

 Quote by haruspex How do you think it might relate to the maxima and minima of the function? Where are those for your example?

yes i got those points by differentiating but can't we find it without using derivatives ? i mean any algebraic method ?

Recognitions:
Homework Help

## Finding the range of rational functions algebraically

 Quote by vrmuth yes i got those points by differentiating but can't we find it without using derivatives ? i mean any algebraic method ?
I can't think of any algebraic methods, and if I were to use logical reasoning, it would still involve some kind of crude link to limits and derivatives.
 Recognitions: Science Advisor By inspection |x| = 2 is critical and x = 0 is a local minimum.

 Quote by mathman By inspection |x| = 2 is critical and x = 0 is a local minimum.
minimum ? it's max. there is no minmimum for this function .but how can i find the range though ? i mean how it's useful to find out the range exactly ?

Recognitions:
 Recognitions: Homework Help I took a little while to think about it, and yes, there is an algebraic solution to the problem. For the function $$y=\frac{1}{x^2-4}$$ let the range be denoted R, which is also the y-value, so we have $$R=\frac{1}{x^2-4}$$ Now, we want to solve for x: $$Rx^2-4R=1$$ $$x^2=\frac{1+4R}{R}$$ $$x=\pm\sqrt{\frac{1+4R}{R}}$$ Now, x exists (and thus a correspondent range exist) whenever $$\frac{1+4R}{R}\geq 0$$ and clearly the opposite of that is, if $$\frac{1+4R}{R}< 0$$ then the domain does not exist, thus the range does not exist. Now, if we solve this inequality, for R>0 (we can see that $R\neq 0$) $$1+4R<0$$ $$R<\frac{-1}{4}$$ And we obviously can't have that both R>0 and $R<\frac{-1}{4}$ so we scrap that. Now if R<0 $$1+4R>0$$ $$R>\frac{-1}{4}$$ Which gives us the intersection $$\frac{-1}{4} Recognitions: Science Advisor  Quote by Mentallic I took a little while to think about it, and yes, there is an algebraic solution to the problem. For the function [tex]y=\frac{1}{x^2-4}$$ let the range be denoted R, which is also the y-value, so we have $$R=\frac{1}{x^2-4}$$ Now, we want to solve for x: $$Rx^2-4R=1$$ $$x^2=\frac{1+4R}{R}$$ $$x=\pm\sqrt{\frac{1+4R}{R}}$$ Now, x exists (and thus a correspondent range exist) whenever $$\frac{1+4R}{R}\geq 0$$ and clearly the opposite of that is, if $$\frac{1+4R}{R}< 0$$ then the domain does not exist, thus the range does not exist. Now, if we solve this inequality, for R>0 (we can see that $R\neq 0$) $$1+4R<0$$ $$R<\frac{-1}{4}$$ And we obviously can't have that both R>0 and $R<\frac{-1}{4}$ so we scrap that. Now if R<0 $$1+4R>0$$ $$R>\frac{-1}{4}$$ Which gives us the intersection $$\frac{-1}{4} There are no values of x where -1/4 < y < 0. As I have indicated, the range is everything but this interval. Recognitions: Homework Help  Quote by mathman There are no values of x where -1/4 < y < 0. As I have indicated, the range is everything but this interval. In my solution I wrote  Quote by Mentallic [tex]x=\pm\sqrt{\frac{1+4R}{R}}$$ Now, x exists (and thus a correspondent range exist) whenever $$\frac{1+4R}{R}\geq 0$$ and clearly the opposite of that is, if $$\frac{1+4R}{R}< 0$$.