# Finding the range of rational functions algebraically

1. Oct 11, 2012

### vrmuth

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

2. Oct 12, 2012

### haruspex

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

3. Oct 12, 2012

### vrmuth

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

4. Oct 12, 2012

### 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.

5. Oct 12, 2012

### mathman

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

6. Oct 12, 2012

### 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 ?

Last edited: Oct 13, 2012
7. Oct 13, 2012

### 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.

8. Oct 13, 2012

### Mentallic

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 $$\frac{-1}{4}<R<0$$ as required.

9. Oct 14, 2012

### mathman

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

10. Oct 14, 2012

### Mentallic

In my solution I wrote

11. Oct 15, 2013

### UnD3R0aTh

can u plz explain to me how did u reach the conclusion that R will be greater than zero? solving the inequality only yields R>-1/4

12. Oct 15, 2013

### Mentallic

I had written that precisely a year ago

Since $-1/4 < R < 0$ is where the range of the function does not exist, everything else is where the range does exist.

So the range is
$y> 0 \cup y\leq -1/4$

(edit: $y\neq 0$ because we can't divide by 0)

If there's something in my earlier solution that you don't understand, just point it out.

Last edited: Oct 15, 2013
13. Oct 15, 2013

### UnD3R0aTh

lol well i'm glad u're still alive :P, yes there is something i do not get, the range of that function would be the domain of the inverse function correct? 1+4R/R > or equal to 0 will give us all the possible range correct?

solving this only gives us range bigger than -1/4, and range does not equal to 0, but as u already know this function has range above than zero, the question is how did u deduce that from the problem/inequality?

14. Oct 15, 2013

### Mentallic

Yes and yes.

Oh you're right about excluding R=0 from the solution, I missed that. You're not solving the inequality correctly though.

Remember that when you multiply through by R, if R<0 then you need to change the reverse the sign of the inequality

So for R>0
$1+4R\geq 0$

but for R<0
$1+4R\leq 0$

Now when you solve these two inequalities, you only keep the intersections that makes sense. For example, if we assume R<0 but then solve the inequality and find that R>1, there are no solutions for R such that both of these can hold true at the same time (we say that the intersection of the sets is the empty set). If however we assume R<0 and solve the inequality to find R>-1, then our solution set is -1<R<0.