Results 1 to 10 of 10
Like Tree5Thanks
  • 2 Post By Plato
  • 1 Post By Plato
  • 1 Post By Plato
  • 1 Post By Plato

Math Help - Inverse functions with restricted domain

  1. #1
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Inverse functions with restricted domain

    I'm wondering how the domain affects the inverse function.

    Taking the example: y=\sqrt{x+3}, x\geq 3

    The inverse is y=x^2-3, x\geq 0

    Why is the inverse restricted to 0 while the other was restricted to -3? I see it on a graph and I understand that it must be because the function stops being a 1 on 1 at x=0 but I'm wondering how we figure that out with functions that are more complex such as a sinus function or something?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,663
    Thanks
    1616
    Awards
    1

    Re: Inverse functions with restricted domain

    Quote Originally Posted by Paze View Post
    Taking the example: y=\sqrt{x+3}, x\geq 3
    The inverse is y=x^2-3, x\geq 0
    Why is the inverse restricted to 0 while the other was restricted to -3? I see it on a graph and I understand that it must be because the function stops being a 1 on 1 at x=0 but I'm wondering how we figure that out with functions that are more complex such as a sinus function or something?
    Use function notation: f(z)=\sqrt{x+3}~\&~g(x)=x^2-3.

    For inverses we require f(g(x))=g(f(x))=x

    But f(g(x))=\sqrt{(x^2-3)+3}=\sqrt{x^2}. What does it take to insure that \sqrt{x^2}=x~?
    Thanks from topsquark and Paze
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Inverse functions with restricted domain

    Hm, it seems to me that the function can take on negative values as well? Seems to me that it equals x by definition? Can you please clarify.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,663
    Thanks
    1616
    Awards
    1

    Re: Inverse functions with restricted domain

    Quote Originally Posted by Paze View Post
    Hm, it seems to me that the function can take on negative values as well? Seems to me that it equals x by definition? Can you please clarify.
    In order for \sqrt{x^2}=x it is necessary that x\ge 0.
    Thanks from Paze
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Inverse functions with restricted domain

    Hm. I must be misunderstanding, as \sqrt{(-1)^2}=1
    sqrt((-1)^2) - Wolfram|Alpha

    What am I misunderstanding?

    EDIT: Woops..Now I see what you mean. -1 is of course not the same as 1. Thank you!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Inverse functions with restricted domain

    What if we do it the other way around and take g(f(x))? Then we get \sqrt{x+3}^2-3 which must mean that x has to be equal to or larger than -3 as well? Or doesn't that matter?

    As far as I remember, f(g(x))=g(f(x))=x for it to be a valid inverse.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,663
    Thanks
    1616
    Awards
    1

    Re: Inverse functions with restricted domain

    Quote Originally Posted by Paze View Post
    What if we do it the other way around and take g(f(x))? Then we get \sqrt{x+3}^2-3 which must mean that x has to be equal to or larger than -3 as well? Or doesn't that matter?
    As far as I remember, f(g(x))=g(f(x))=x for it to be a valid inverse.
    Well if x\ge 0 then you know that x\ge -3. Right?
    Thanks from Paze
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Inverse functions with restricted domain

    Well...According to wolfram I'm wrong (sqrt(-4+3))^2-3 - Wolfram|Alpha

    How is it that this is calculated? I would have thought that we started calculating inside the root to get -1, then attempt to take the root but we would end up with an imaginary number...But perhaps that imaginary number squared is equal to -1? (I'm not very familiar with complex numbers).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,663
    Thanks
    1616
    Awards
    1

    Re: Inverse functions with restricted domain

    Quote Originally Posted by Paze View Post
    Well...According to wolfram I'm wrong (sqrt(-4+3))^2-3 - Wolfram|Alpha
    How is it that this is calculated? I would have thought that we started calculating inside the root to get -1, then attempt to take the root but we would end up with an imaginary number...But perhaps that imaginary number squared is equal to -1? (I'm not very familiar with complex numbers).
    I have no idea what you are talking about.
    The functions f(z)=\sqrt{x+3}~\&~g(x)=x^2-3 are inverses of each other if x\ge 0.
    Therefore, -4 has absolutely to do with any of this because -4<0.
    Thanks from Paze
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member Paze's Avatar
    Joined
    Nov 2012
    From
    Iceland
    Posts
    379
    Thanks
    19

    Re: Inverse functions with restricted domain

    Quote Originally Posted by Plato View Post
    I have no idea what you are talking about.
    The functions f(z)=\sqrt{x+3}~\&~g(x)=x^2-3 are inverses of each other if x\ge 0.
    Therefore, -4 has absolutely to do with any of this because -4<0.
    No wonder. I'm not even sure what I am asking anymore. Thank you for your insight, it has clarified the matter greatly. I will keep studying the subject.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse Functions and Restricted Domains
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 12th 2012, 02:01 PM
  2. Replies: 4
    Last Post: August 10th 2011, 05:24 AM
  3. Replies: 11
    Last Post: January 12th 2011, 06:49 PM
  4. Replies: 1
    Last Post: May 12th 2010, 05:11 AM
  5. Inverse With Restricted Domain
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 19th 2009, 10:28 AM

Search Tags


/mathhelpforum @mathhelpforum