Finding Ranges of inverse functions

1. Jun 5, 2008

thomas49th

Hi, I know that the range of an inverse function is the domain of the function, but how do you work it out if you dont know it? You could sketch a graph but doesn't that get tricky for hard complicated functions. For example:

$$\frac{3x}{x+1} - \frac{x+7}{x^{2}-1}, x > 1$$

If i find the inverse function, who would i then find the range of it?

Thanks

Last edited: Jun 5, 2008
2. Jun 5, 2008

Defennder

I think you meant the range of an inverse function is the domain of the function. Now looking at the expression of the function given, what values of x are not allowed ie. would cause the function to be undefined? The domain would be then R excluding those disallowed values.

3. Jun 5, 2008

thomas49th

well first we need to simplify it dont we. apparently it simplifies to:
$$3 - \frac{4}{x-1}$$

so the inverse of $$3 - \frac{4}{x-1}$$
is $$x^{2} - 4x + 5$$

is that right? So it can take any value cant it?

Thanks :)

4. Jun 5, 2008

Defennder

How did you get that quadratic expression? More importantly what is your question? You want to find the range of the inverse function or the domain of the inverse function?

5. Jun 5, 2008

thomas49th

f(x) = $$3 - \frac{4}{x-1}$$

find $$f^{-1}$$

state the domain of $$f^{-1}$$

so finding the inverse would be:

x = $$3 - \frac{4}{x-1}$$

x² - x = 3x - 3 - 4

thus that's how i got the quadratic. Is it wrong?

EDIT: AHH sorry i want to find the domain of the inverse which is easy as it's the range of the original, whch is x >1. Still, is my inverse function correct?

Last edited: Jun 5, 2008
6. Jun 5, 2008

Defennder

In general you don't choose x = f(x) where f(x) is the function which you want to find the inverse of. That would be confusing it with the original x variable in the expression. Choose y = f(x). Then express x in terms of y. When you're done, change the y in the final expression back to x, and that is your inverse function.

EDIT: By change back to x, I mean replace y with x.

Last edited: Jun 5, 2008
7. Jun 5, 2008

thomas49th

so it's $$y = -\frac{6}{x-3}$$

is that right :)

Thanks

8. Jun 5, 2008

Defennder

Not quite, you made a mistake somewhere. But you don't have to find the inverse function to begin with. You already know that the domain of the inverse, which you want to find, is the range of the original function. What is the range of $$3-\frac{4}{x-1} \ ,x>1$$?

Do you know how to graph that function? You can see the range easily if you know how to sketch it.

9. Jun 5, 2008

thomas49th

Is the inverse function:
$$y = -\frac{6}{x-3}$$

Erm i guess to graph it i could just put some sample values in (say from -5 to 5 for x) and see what comes out.

I just a used computer software to graph it. is it's range y < 6 ??

Thanks :)

10. Jun 5, 2008

konthelion

There's essential two ways to find the domain of the inverse.

1. Solve for the inverse $$f^{-1}$$. Then find the vertical asymptote(there's one in the inverse function). Find the domain of $$f^{-1}$$
2. Find the range of the original function f. Find the horizontal asymptote in f. Find the range of the f= domain of $$f^{-1}$$

I do not get the same inverse function as you did. Show how you got that answer.

Even if you get the inverse function to be this, how did you get the range to be y<6? There's a horizontal asymptote at y=0

Last edited: Jun 5, 2008
11. Jun 5, 2008

thomas49th

y = 3 - 4/x-1

y(x-1) = 3x - 3 - 4
yx - y = 3x - 7
yx - 3x = y + 7
x(y-3) = y + 7
x = y+7/y-3
= x+7/x+3

is that right now?
Cheers :)

12. Jun 5, 2008

konthelion

Error 1: from (2) to (3), that -7 change to +7 for no reason.
Error 2: (5) to (6), put it in parentheses, and divide by the whole thing.

Last edited: Jun 5, 2008
13. Jun 5, 2008

thomas49th

y(x-1) = 3x - 3 - 4
yx - y = 3x - 7
yx - 3x = y - 7
x(y-3) = y - 7
x = y-7/y-3
= x+7/x-3

and if i divide it i get 1 + 10/x-3

is that right now?

Thanks :)

14. Jun 5, 2008

konthelion

No, but you're close. Check again.

$$y=\frac{7-x}{3-x}=1+?$$ ? is the remainder from the division over 3-x

15. Jun 5, 2008

thomas49th

1 + 4/x-3

that right?

16. Jun 5, 2008

konthelion

Yes, that's correct. Now, you want to find the horizontal asymptote. Since the inverse function $$y=\frac{x-7}{x-3}=1+\frac{4}{x-3}$$
Vertical asymptote: is when the denominator is 0 and solve for x.
But we want to find the horizontal asymptote of the inverse function. To do that, since the highest coefficient of the number is equal to the coefficient of the denominator, you can just divide coefficients.

What is the horizontal asymptote of the inverse function? Once, you've find this, the range of $$f^{-1}$$ will be
(Likewise, you could've saved all this trouble, if you found the vertical asymptote of the original function to find the domain of f = range of $$f^{-1}$$

i.e. $$y=3-\frac{-4}{x-1}$$. Set the denominator to zero, you will get the domain of f = range of $$f^{-1}$$

Last edited: Jun 5, 2008
17. Jun 5, 2008

thomas49th

so the verticle asymptote is x - 3 = 0 so x = 3
as for the y asymptote is it the co-efficients of x which is both 1 so the asymptote is y = 1???

and i've got a sneeky trick for you. the inverse function is a reflection in the axis y = x of the original function.

Thanks :)

18. Jun 5, 2008

konthelion

Alright, good job. Now, what's the range of the inverse function?

19. Jun 5, 2008

thomas49th

range: y > 1

because as x tends to infinity y tends to 1. And as the domain says x>1 we can ignore the curve in the other quadrant

Right?

20. Jun 5, 2008

konthelion

Yes, you are correct. I forgot that the original condition was x>1 :P