Results 1 to 15 of 15

Math Help - Invertible functions

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    12

    Invertible functions

    i try do it but only first one #_# 2 and 3 i dont know how can do #_#


    Thanks.
    Attached Thumbnails Attached Thumbnails Invertible functions-5.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by liptonpc View Post
    i try do it but only first one #_# 2 and 3 i dont know how can do #_#


    Thanks.
    remember, f^{-1} undoes f. This means, the output becomes the input. Let y = f^{-1}(x), then

    x = 3y^2 - 1

    solve for y

    (Now you should notice a problem here...this function does not have an inverse. not unless the domain was restricted some way)


    for (iii) \frac fg (x) = \frac {3x^2 - 1}{x + 5}

    now just plug in x = 1 and evaluate
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    12
    f^{-1}(x)=\sqrt\frac {x + 1}{3}<br />
    is it correct ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by liptonpc View Post
    f^{-1}(x)=\sqrt\frac {x + 1}{3}<br />
    is it correct ?
    technically you need a +/- in front of that. which would make f^{-1} not a function.

    only bijective functions are invetible, and f isn't bijective
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2010
    Posts
    12
    sorry i dont understand #_#
    so the result is ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by liptonpc View Post
    sorry i dont understand #_#
    so the result is ?
    f does not pass the horizontal line test and is hence not invertible. it has no inverse. not unless the domain was restricted somehow.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2010
    Posts
    12
    f^{-1}(x)=\sqrt\frac {x + 1}{3}
    when i check
    f(f^{-1}(x))=x

    so why it not inverse?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by liptonpc View Post
    f^{-1}(x)=\sqrt\frac {x + 1}{3}
    when i check
    f(f^{-1}(x))=x

    so why it not inverse?
    because f^{-1}(x) \ne \sqrt{\frac {x + 1}3}, rather, f^{-1} (x) = ~{\color{red} \pm}~ \sqrt {\frac {x + 1}3}

    which means f^{-1} is not a function at all, since that does not pass the vertical line test.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Jan 2010
    Posts
    12
    if f(x)=x^2-9
    f^{-1}(x)=\pm(x+9)^{\frac{1}2}
    and same thing it not inverse?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by liptonpc View Post
    if f(x)=x^2-9
    f^{-1}(x)=\pm(x+9)^{\frac{1}2}
    and same thing it not inverse?
    correct.

    now on the other hand, if they had restricted the domain and said, f(x) = x^2 - 9 for x \ge 0, then you could say f^{-1}(x) = \sqrt {x + 9} and that's the inverse function.

    we don't need the negative square root, because the domain of our original function only has non-negative numbers.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jan 2010
    Posts
    29
    Quote Originally Posted by liptonpc View Post
    if f(x)=x^2-9
    f^{-1}(x)=\pm(x+9)^{\frac{1}2}
    and same thing it not inverse?

    You use the vertical line test to check if something is a function. Then, if it is a function, you use the horizontal line test to see if it has an inverse. If it does not pass the horizontal line test, then it isn't an invertible function

    The reason if because the (fake) inverse function that you would get would spit out 2 values for (almost) every x value when talking about parabolas, as shown by the \pm in the problem above, and when that happens it cant be a function
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Jan 2010
    Posts
    12
    yeah #_# because i usually think x\ge0 i forgot they didt say this ^^
    thanks ^^
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Jan 2010
    Posts
    12
    but if dont ask about inverse or not. Just ask Find f^{-1}(x)
    so i have to answer
    f^{-1}(x)=\pm(x+9)^{\frac{1}2}
    or f^{-1}(x)=(x+9)^{\frac{1}2}
    ???
    Follow Math Help Forum on Facebook and Google+

  14. #14
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by liptonpc View Post
    but if dont ask about inverse or not. Just ask Find f^{-1}(x)
    so i have to answer
    f^{-1}(x)=\pm(x+9)^{\frac{1}2}
    or f^{-1}(x)=(x+9)^{\frac{1}2}
    ???
    haha, asking to find f^{-1} is asking to find the inverse. if your function is not one-to-one (or onto), either by itself or by restricting its domain, then it does not have an inverse. the question makes no sense.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Jan 2010
    Posts
    12
    ^^ thanks u ^^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. bit theory, functions (one-one,onto, invertible)
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 30th 2011, 08:03 AM
  2. Why is T not invertible?
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 22nd 2011, 02:38 PM
  3. Show that if M is invertible, M^t is also invertible
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: August 15th 2010, 02:40 PM
  4. [SOLVED] Self-Invertible
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: July 16th 2010, 03:38 PM
  5. invertible?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: November 16th 2008, 01:42 PM

Search Tags


/mathhelpforum @mathhelpforum