# Thread: Finding the Domain of an Inverse Function

1. ## Finding the Domain of an Inverse Function

$\displaystyle f(x) = 1/4x - 4 , -3 < x < 1 (theyre both equal as well as less than)$

And it's asking to find the interval of [A,B].

I know how to find the inverse of the function, and I also know that the range of the function f(x) would be the domain of the inverse. However I don't know how to find that out.

2. Originally Posted by Lolcats
$\displaystyle f(x) = 1/4x - 4 , -3 < x < 1 (theyre both equal as well as less than)$

And it's asking to find the interval of [A,B].

I know how to find the inverse of the function, and I also know that the range of the function f(x) would be the domain of the inverse. However I don't know how to find that out.
you mean $\displaystyle f(x) = \frac 14x - 4$? this is a straight line. just plug in the end points of your range of x-values, that will give you the range of y-values that you hit. that is, the range of f is $\displaystyle [f(-3),f(1)]$

3. Originally Posted by Lolcats
$\displaystyle f(x) = 1/4x - 4 , -3 < x < 1 (theyre both equal as well as less than)$

And it's asking to find the interval of [A,B].

I know how to find the inverse of the function, and I also know that the range of the function f(x) would be the domain of the inverse. However I don't know how to find that out.

well cant you find the range of that function by plugging in the endpoints of your domain interval? and then, as you said, the range of the function is the domain of the inverse function

4. Oh jeez, thanks! that should have been apparent.

5. I took the difficult route assuming f(x) = 1/(4x) -4

The problem is you have a singularity at x = 0

For [-3,0) the range of f is (-inf, f(-3)) = (-inf,-49/12) = domain of f^(-1)

For (0,1] the range is (f(1),infinity)=(f(-15/4),infinity) = domain of f ^(-1)

Domain of f^(-1) is the union of the 2 domains

See graphs in attachment

it all if you have the simple linear function!!

6. you make a good point but i must say that i interpret 1/4x to be x/4 based on the notation

7. Originally Posted by emathinstruction
you make a good point but i must say that i interpret 1/4x to be x/4 based on the notation
i must say i myself find the notation vague, because i have seen posters interpret it both ways. i would much prefer if parentheses are used for clarification: (1/4)x as opposed to 1/(4x) and forget typing things like 1/4x. Of course, the better option is for posters to learn LaTeX

8. Well thats stupid notation 1/4x if you ask me --just waisted time I could have been watching the Daily Show -- gotta go I can still catch the Colbert Report

9. Originally Posted by Calculus26
Well thats stupid notation if you ask me --just waisted time I could have been watching the Daily Show -- gotta go I can still catch the Colbert Report
haha, don't dispair. i am sure someone will find your post edifying. enjoy the show!

10. Now I have to wait till 1:30 to catch the replay--someone better ask a good question on this forum

11. Originally Posted by Calculus26
Now I have to wait till 1:30 to catch the replay--someone better ask a good question on this forum
Haha, I'm sure there are some out there. check the "unanswered threads" subforum. and watch the language will ya, there are kids here. i got in trouble for using the d-word once, so i think it's a bad word around here