Results 1 to 4 of 4

Math Help - Proof of a Function

  1. #1
    Junior Member
    Joined
    Mar 2014
    From
    Canada
    Posts
    35

    Proof of a Function

    If (a, b) = f, then (b, a)= f^-1. The claim has been made that the graph of the inverse of a function is the mirror image of the function, reflected in the line y = x. To prove this you must

    a) Show that the line that passes through the points (a, b) and (b, a) intersect the line y = x at a right angle.

    b) Show that the distances between each point and the line y = x are the same. (To do this you might have to determine the point of intersection of the line passing though (a, b) and (b, a) and the line y = x).

    Please help! I know that you can't really graph it on here, but it'd be so much help if you explained what the question means and how to solve it. I really don't understand it, so any sort of help would be really appreciated. Thanks!
    Last edited by eleventhhour; March 29th 2014 at 09:44 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1

    Re: Proof of a Function

    Quote Originally Posted by eleventhhour View Post
    If (a, b) = f, then (b, a)= f^-1. The claim has been made that the graph of the inverse of a function is the mirror image of the function, reflected in the line y = x. To prove this you must

    a) Show that the line that passes through the points (a, b) and (b, a) intersect the line y = x at a right angle.

    b) Show that the distances between each point and the line y = x are the same. (To do this you might have to determine the point of intersection of the line passing though (a, b) and (b, a) and the line y = x).

    Please help! I know that you can't really graph it on here, but it'd be so much help if you explained what the question means and how to solve it. I really don't understand it, so any sort of help would be really appreciated. Thanks!
    I think that you are complicating the question.
    Let $\ell$ be the line $x-y=0$.
    By definition $(a,b)\in f$ if and only if $(b,a)\in f^{-1}$, assuming the inverse exists.

    The slope of $\ell$ is $1$ and $\dfrac{b-a}{a-b}=-1$ that shows the perpendicular aspect.

    If $P: (p,q)$ is any point in the plane then the distance from $P$ to $\ell$ is $\dfrac{|p-q|}{\sqrt{2}}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Feb 2014
    From
    United States
    Posts
    657
    Thanks
    335

    Re: Proof of a Function

    Quote Originally Posted by eleventhhour View Post
    If (a, b) = f, then (b, a)= f^-1. The claim has been made that the graph of the inverse of a function is the mirror image of the function, reflected in the line y = x. To prove this you must

    a) Show that the line that passes through the points (a, b) and (b, a) intersect the line y = x at a right angle.

    b) Show that the distances between each point and the line y = x are the same. (To do this you might have to determine the point of intersection of the line passing though (a, b) and (b, a) and the line y = x).

    Please help! I know that you can't really graph it on here, but it'd be so much help if you explained what the question means and how to solve it. I really don't understand it, so any sort of help would be really appreciated. Thanks!
    Let's take an example

    $f(x) = x^3 \implies f(a) = b = a^3.$

    You must understand that a function may not have an inverse, but the function shown above does.

    So $f^{-1}(b) = a \implies f^{-1}(a^3) = a \implies f^{-1}(x) = \sqrt[3]{x}.$

    $g(x)\ is\ the\ inverse\ of\ f(x) \implies f(x)\ is\ the\ inverse\ of\ g(x)\ and\ g(f(x)) = x = f(g(x)).$

    Let's see how this works with my example.

    $f(x) = x^3 \implies f^{-1}(x) = g(x) = \sqrt[3]{x}.$

    $f(g(x)) = \left(\sqrt[3]{x}\right)^3 = x = \sqrt[3]{x^3} = g(f(x)).$

    The line joining $(a,\ a^3)\ and\ (a^3,\ a)\ has\ slope\ of \dfrac{a^3 - a}{a - a^3} = -\ \dfrac{a^3 - a}{a^3 - a} = -\ 1.$ So it is perpendicular to $y = x.$

    The equation of the line joining $(a,\ a^3)\ and\ (a^3,\ a)\ is:$

    $\dfrac{y - a^3}{x - a} = - 1 \implies y - a^3 = - x + a \implies y = a^3 + a - x.$ That intersects the line y = x at:

    $x = y = a^3 + a - x \implies 2x = a^3 + a \implies x = \dfrac{a^3 + a}{2} = y.$

    The distance to $(a,\ a^3)$ from $\left(\dfrac{a^3 + a}{2},\ \dfrac{a^3 + a}{2}\right)$ is $\sqrt{\left(a - \dfrac{a^3 + a}{2}\right)^2 + \left(a^3 - \dfrac{a^3 + a}{2}\right)^2}.$

    The distance to $(a^3,\ a)$ from $\left(\dfrac{a^3 + a}{2},\ \dfrac{a^3 + a}{2}\right)$ is $\sqrt{\left(a^3 - \dfrac{a^3 + a}{2}\right)^2 + \left(a - \dfrac{a^3 + a}{2}\right)^2} = \sqrt{\left(a - \dfrac{a^3 + a}{2}\right)^2 + \left(a^3 - \dfrac{a^3 + a}{2}\right)^2}.$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2014
    From
    Canada
    Posts
    35

    Re: Proof of a Function

    Quote Originally Posted by Plato View Post
    I think that you are complicating the question.
    Let $\ell$ be the line $x-y=0$.
    By definition $(a,b)\in f$ if and only if $(b,a)\in f^{-1}$, assuming the inverse exists.

    The slope of $\ell$ is $1$ and $\dfrac{b-a}{a-b}=-1$ that shows the perpendicular aspect.

    If $P: (p,q)$ is any point in the plane then the distance from $P$ to $\ell$ is $\dfrac{|p-q|}{\sqrt{2}}$.
    Thanks, I think I understand a bit better now. But, how do you know that the slope of ℓ is 1?
    Also, I'm not really sure about the last part (b). How does that prove that the distances between each point are the same?
    Oh, and I think the question wants it to be graphed as well. How would you do that? Would you just choose two random points?

    Thanks again. sorry for the probably basic questions. I'm only in grade 11, so this is a bit complicated for me haha
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof that binomial function is a probability function?
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: December 28th 2011, 02:26 PM
  2. Replies: 1
    Last Post: December 30th 2010, 04:23 PM
  3. Function proof
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 20th 2009, 10:08 AM
  4. Replies: 0
    Last Post: September 14th 2009, 07:13 AM
  5. Another proof using the phi-function
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: April 29th 2009, 12:28 PM

Search Tags


/mathhelpforum @mathhelpforum