Results 1 to 4 of 4

Math Help - inverse composite function

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    36

    inverse composite function

    hi i need help on how to solve these questions

    verify, algebraically, that f(f{^-1}(x)) = x

    f(x) = x^2 -4

    and

    f(x) = \frac{1}{x-2}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    RRH
    RRH is offline
    Junior Member
    Joined
    Oct 2009
    From
    Florida
    Posts
    43
    Quote Originally Posted by extraordinarymachine View Post
    hi i need help on how to solve these questions

    verify, algebraically, that f(f{^-1}(x)) = x

    f(x) = x^2 -4

    and

    f(x) = \frac{1}{x-2}
    first you need to find the inverse of x^2-4

    y=x^2-4

    x=y^2-4

    x+4=y^2

    \sqrt [2] {x+4}=\sqrt [2] {y^2}

    \sqrt [2] {x}+2=y

    now we plug the inverse back into the original function

    (\sqrt [2] {x}+2)^2-4

    x+4-4=x

    Can you follow the example to solve the second function
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,449
    Thanks
    1864
    Quote Originally Posted by extraordinarymachine View Post
    hi i need help on how to solve these questions

    verify, algebraically, that f(f{^-1}(x)) = x

    f(x) = x^2 -4
    Unless you have some restriction on the possible values of x (on the domain) this function does not have an inverse!
    RRH gets to y^2= x+ 4 and then says that y= \sqrt{x+ 4} but there are two values of y whose square is x+ 4.

    If the original function is " f(x)= x^2- 4 with domain x\ge 0", THEN f^{-1}(x)= \sqrt{x+ 4} with domain x\ge -2.

    If the original function is " f(x)= x^2- 4 with domain x\le 0, THEN f^{-1}(x)= -\sqrt{x+ 4} with domain x\ge -2

    RRH is also wrong when he writes \sqrt{x+ 4}= \sqrt{x}+ 2. For example, if x= 1, \sqrt{1+ 4}= \sqrt{5} which is certainly not equal to \sqrt{1}+ 2= 3.


    and

    f(x) = \frac{1}{x-2}
    Follow the pattern RRH used (which is correct). From y= \frac{1}{x-2} (which has natural domain all real numbers except 2), the inverse function is given by x= \frac{1}{y- 2}, where I have swapped x and y. Now solve for y.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    RRH
    RRH is offline
    Junior Member
    Joined
    Oct 2009
    From
    Florida
    Posts
    43
    Quote Originally Posted by HallsofIvy View Post
    Unless you have some restriction on the possible values of x (on the domain) this function does not have an inverse!
    RRH gets to y^2= x+ 4 and then says that y= \sqrt{x+ 4} but there are two values of y whose square is x+ 4.

    If the original function is " f(x)= x^2- 4 with domain x\ge 0", THEN f^{-1}(x)= \sqrt{x+ 4} with domain x\ge -2.

    If the original function is " f(x)= x^2- 4 with domain x\le 0, THEN f^{-1}(x)= -\sqrt{x+ 4} with domain x\ge -2

    RRH is also wrong when he writes \sqrt{x+ 4}= \sqrt{x}+ 2. For example, if x= 1, \sqrt{1+ 4}= \sqrt{5} which is certainly not equal to \sqrt{1}+ 2= 3.




    Follow the pattern RRH used (which is correct). From y= \frac{1}{x-2} (which has natural domain all real numbers except 2), the inverse function is given by x= \frac{1}{y- 2}, where I have swapped x and y. Now solve for y.
    thank you for the correction I was thinking about that last night ... I have been sloppy with my math lately ... Again I apologize to the original poster to the original poster and to others in the MHF for my mistakes
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding a composite inverse function.
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 2nd 2010, 09:34 AM
  2. Replies: 6
    Last Post: October 20th 2009, 03:26 AM
  3. Composite Function
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 27th 2009, 06:58 AM
  4. [SOLVED] Domain of composite of inverse
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: February 13th 2009, 03:23 AM
  5. Composite Function
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 21st 2006, 08:16 PM

Search Tags


/mathhelpforum @mathhelpforum