Results 1 to 5 of 5

Math Help - Simple question about functions

  1. #1
    Member
    Joined
    Mar 2010
    Posts
    75

    Simple question about functions

    Suppose that:
    1) g={(0,0),(1,1/2),(2,1),(3,3/2),(4,2)}
    2) f={(0,0),(1/2,0),(1,1),(3/2,1),(2,2)}

    Can someone explain the logic behind figuring out whether a function is an inverse function?

    I know that #1 is an inverse function and #2 is NOT an inverse function.

    IS the correct way to figure this out to simply flip the x and y values and then see if it is still one-to-one and onto?

    Also,
    Can someone explain how I would find f-of-g?

    THANK YOU.
    Last edited by matthayzon89; April 13th 2010 at 07:48 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2010
    Posts
    466
    Thanks
    4
    Suppose that:
    1) g={(0,0),(1,1/2),(2,1),(3,3/2),(4,2)}
    2) f={(0,0),(1/2,0),(1,1),(3/2,1),(2,2)}

    Can someone explain the logic behind figuring out whether a function is an inverse function?
    Do you mean whether the inverse of a function is a function?

    I know that #1 is an inverse function and #2 is NOT an inverse function.
    Maybe you mean the inverse of #1 is a function and the inverse of #2 is not a function?

    IS the correct way to figure this out to simply flip the x and y values and then see if it is still one-to-one and onto?
    If all you're checking is to see whether the inverse of a function is a function, then just flip the x and y values to see whether you have a function (you don't have to check 1-1 nor onto).

    To review:

    inverse(f) = {<y x> | <x y> in f}

    So, inverse(f) is a function if and only if {<y x> | <x y> in f} is a function.

    {<y x> | <x y> in f} is a function
    if and only if
    for all x. y, z, if <y x> and <y z> in {<y x> | <x y> in f} then x = z.

    Can someone explain how I would find f-of-g?
    Do you mean f-comp-g?

    f-comp-g = {<x y> | exists z such that <x z> in g & <z y> in f}.

    So [f-comp-g](x) = f(g(x)). (I.e., determine g(x) = y, then determine f(y).)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2010
    Posts
    75
    Quote Originally Posted by MoeBlee View Post
    Do you mean whether the inverse of a function is a function? YES

    Maybe you mean the inverse of #1 is a function and the inverse of #2 is not a function? YES

    If all you're checking is to see whether the inverse of a function is a function, then just flip the x and y values to see whether you have a function (you don't have to check for 1-1 adn onto).

    To review:

    inverse(f) = {<y x> | <x y> in f}

    So, inverse(f) is a function if and only if {<y x> | <x y> in f} is a function.

    {<y x> | <x y> in f} is a function
    if and only if
    for all x. y, z, if <y x> and <y z> in {<y x> | <x y> in f} then x = z.

    I DONT UNDERSTAND THIS
    Can you use the conditions and numbers that I have provided to create an example?

    Do you mean f-comp-g?

    f-comp-g = {<x y> | exists z such that <x z> in g & <z y> in f}.

    So [f-comp-g](x) = f(g(x)). (I.e., compute g(x) = y, then compute f(y).)

    I ALSO DON'T UNDERSTAND THIS please try to help explain given the information I know b/c the notation and the way you are explaining this is above my level of understanding, this is all really new to me, thank you for trying though
    Read the bolded
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Feb 2010
    Posts
    466
    Thanks
    4
    By the way, as to 1-1 and onto, probably what you have in mind are these facts:

    If f is a 1-1 function then inverse(f) is a function.

    If f is a function and inverse(f) is a function, then f is a 1-1 function.

    If f is a 1-1 function onto S, then inverse(f) is a 1-1 function from S.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Feb 2010
    Posts
    466
    Thanks
    4
    Let's review the definitions:

    first coordinate in <x y> = x.

    second coordinate in <x y> = y.

    f is a relation iff every member of f is an ordered pair.

    I.e., every member of f is of the form <x y> for some x and y.

    f is a function
    iff
    f is a relation and for all x, y, z, if <x y> in f and <x z> in f, then y=z.

    I.e., every member of f is of the form <x y> for some x and y, and there is no pair <x y> and <x z> in f unless y=z.

    So, in your examples, both g and f are functions, and inverse(g) is a function. But inverse(f) is not a function because both <0 0> and <0 1/2> are in inverse(f), also because both <1 1> and <1 3/2> are in inverse(f).

    /

    I wrote "f-comp-g = {<x y> | exists z such that <x z> in g & <z y> in f}."

    That means f-comp-g is the set of ordered pairs <x y> such that there exists a z such that <x z> in g and <z y> in f.

    I.e., f-comp-g is formed by taking all the pairs <x y> where for some z there is a pair <x z> in g and <z y> in f.

    In other words, look for any z that is the second coordinate in a pair in g and is the first coordinate in a pair in f.

    In other words, find all the z's that link from the "end" in a pair in g to the "start" of a pair in f. Then the "comp" is the set of ordered pairs that have z as a "link" as just described.

    For example,
    suppose g = {<1 3> <4 2> <5 1> <6 2>}
    and
    f = {1 0> <3 2> <9 5>}.

    Then f-comp-g = {<1 2> <5 0>}
    because
    3 is the second coordinate in <1 3> in g and the first coordinate in <3 2> in g, and 1 is the second coordinate in <5 1> in g and the first coordinate in <1 0> in f.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Really simple question too hard for a simple mind
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 5th 2010, 08:03 AM
  2. Easy Question - Simple Interest, Simple Discount
    Posted in the Business Math Forum
    Replies: 0
    Last Post: September 21st 2010, 08:22 PM
  3. Simple Functions Question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 11th 2010, 12:40 PM
  4. Replies: 4
    Last Post: May 4th 2010, 09:08 AM
  5. Simple functions - zero of f(x)
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: September 28th 2009, 04:42 AM

Search Tags


/mathhelpforum @mathhelpforum