# Thread: Proof of a Function

1. ## 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!

2. ## Re: Proof of a Function

Originally Posted by eleventhhour
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 $\dfrac{|p-q|}{\sqrt{2}}$.

3. ## Re: Proof of a Function

Originally Posted by eleventhhour
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 $\sqrt{\left(a^3 - \dfrac{a^3 + a}{2}\right)^2 + \left(a - \dfrac{a^3 + a}{2}\right)^2} = \sqrt{\left(a - \dfrac{a^3 + a}{2}\right)^2 + \left(a^3 - \dfrac{a^3 + a}{2}\right)^2}.$

4. ## Re: Proof of a Function

Originally Posted by Plato
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