Finding the range of rational functions algebraically

[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

 

[haruspex]

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

 

[vrmuth]

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 ?

 

[Mentallic]

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.

 

[mathman]

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

 

[vrmuth]

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 ?

 

[mathman]

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 ?

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

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}&lt; 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&lt;0

R&lt;\frac{-1}{4}

And we obviously can't have that both R>0 and R&lt;\frac{-1}{4} so we scrap that. Now if R<0

1+4R&gt;0

R&gt;\frac{-1}{4}

Which gives us the intersection \frac{-1}{4}&lt;R&lt;0 as required.

 

[mathman]

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}&lt; 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&lt;0

R&lt;\frac{-1}{4}

And we obviously can't have that both R>0 and R&lt;\frac{-1}{4} so we scrap that. Now if R<0

1+4R&gt;0

R&gt;\frac{-1}{4}

Which gives us the intersection \frac{-1}{4}&lt;R&lt;0 as required.

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

 

[Mentallic]

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

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}&lt; 0.

 

[UnD3R0aTh]

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}&lt; 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&lt;0

R&lt;\frac{-1}{4}

And we obviously can't have that both R>0 and R&lt;\frac{-1}{4} so we scrap that. Now if R<0

1+4R&gt;0

R&gt;\frac{-1}{4}

Which gives us the intersection \frac{-1}{4}&lt;R&lt;0 as required.

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

 

[Mentallic]

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

I had written that precisely a year ago

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

So the range is
y&gt; 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.

 

[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?

 

[Mentallic]

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?

Yes and yes.

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?

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.