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Math Help - Finding the Domain of an Inverse Function

  1. #1
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    Finding the Domain of an Inverse Function

     f(x) = 1/4x - 4 ,<br />
 -3 < x < 1 (theyre both equal as well as less than)<br />

    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.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Lolcats View Post
     f(x) = 1/4x - 4 ,<br />
 -3 < x < 1 (theyre both equal as well as less than)<br />

    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 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 [f(-3),f(1)]
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  3. #3
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    Quote Originally Posted by Lolcats View Post
     f(x) = 1/4x - 4 ,<br />
-3 < x < 1 (theyre both equal as well as less than)<br />

    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
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  4. #4
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    Oh jeez, thanks! that should have been apparent.
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  5. #5
    MHF Contributor Calculus26's Avatar
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    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!!
    Attached Thumbnails Attached Thumbnails Finding the Domain of an Inverse Function-graphs.jpg  
    Last edited by Jhevon; January 20th 2010 at 08:45 PM. Reason: language!
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  6. #6
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    you make a good point but i must say that i interpret 1/4x to be x/4 based on the notation
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by emathinstruction View Post
    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
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  8. #8
    MHF Contributor Calculus26's Avatar
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    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
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Calculus26 View Post
    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!
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  10. #10
    MHF Contributor Calculus26's Avatar
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    Now I have to wait till 1:30 to catch the replay--someone better ask a good question on this forum
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Calculus26 View Post
    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
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