Finding the range of rational functions algebraically

vrmuth

how to find the range of rational functions like f(x) = [itex]\frac{1}{{x}^{2}-4}[/itex] 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

haruspex

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

vrmuth

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

Mentallic

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.

mathman

By inspection |x| = 2 is critical and x = 0 is a local minimum.

vrmuth

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 ?

mathman

You're right. It is local max (-1/4). As |x| -> 2 from below, f(x) -> -∞. However |x| -> 2 from above, f -> +∞. Finally as |x| -> ∞, f -> 0.

Net result: range has two parts (-∞,-1/4) for |x| < 2, and (0,+∞) for |x| > 2.

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}[/tex]

let the range be denoted R, which is also the y-value, so we have

[tex]R=\frac{1}{x^2-4}[/tex]

Now, we want to solve for x:

[tex]Rx^2-4R=1[/tex]

[tex]x^2=\frac{1+4R}{R}[/tex]

[tex]x=\pm\sqrt{\frac{1+4R}{R}}[/tex]

Now, x exists (and thus a correspondent range exist) whenever

[tex]\frac{1+4R}{R}\geq 0[/tex]

and clearly the opposite of that is, if

[tex]\frac{1+4R}{R}< 0[/tex]

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 [itex]R\neq 0[/itex])

[tex]1+4R<0[/tex]

[tex]R<\frac{-1}{4}[/tex]

And we obviously can't have that both R>0 and [itex]R<\frac{-1}{4}[/itex] so we scrap that. Now if R<0

[tex]1+4R>0[/tex]

[tex]R>\frac{-1}{4}[/tex]

Which gives us the intersection [tex]\frac{-1}{4}<R<0[/tex] as required.

mathman

There are no values of x where -1/4 < y < 0. As I have indicated, the range is everything but this interval.

Mentallic

In my solution I wrote