B Understanding the Relationship Between a Function and Its Inverse

[brotherbobby]

TL;DR Summary

Prove that a function and its inverse are mirror images along the line $y=x$

I could only verify this for a few elementary functions. Does a proof exist? Does it go beyond the realms of high school mathematics?
Many thanks.

 

[fresh_42]

TL;DR Summary: Prove that a function and its inverse are mirror images along the line $y=x$

I could only verify this for a few elementary functions. Does a proof exist? Does it go beyond the realms of high school mathematics?
Many thanks.

It becomes more obvious if you write a function $f\, : \, X\longrightarrow Y\, , \,x\longmapsto y=f(x)$ as subset of $X\times Y,$ namely $f=\{(x,y)\in X\times Y\,|\,y=f(x)\}.$ With this notation, $f^{-1}$ becomes $f^{-1}=\{(y,x)\,|\,f^{-1}(y)=x\}.$

"... mirror the image (graph) at the line $y=x$ ..." means exchanging $x$ and $y.$

 

[FactChecker]

You can do it with high school math. As @fresh_42 indicates, you should not use examples. You should use the definition of an inverse and apply it to a general, abstract, function y=f(x). What does it mean to say that the function and its inverse are mirror images along the line y=x?

 

[brotherbobby]

What does it mean to say that the function and its inverse are mirror images along the line y=x?

I don't get you. Any hints?

 

[PeroK]

I don't get you. Any hints?

Draw the x and y axes, the line $y = x$, and a one-to-one function $y = f(x)$. Project every point on the graph of $y = f(x)$ at a right angle through the line $y =x$, the same distance on the other side.

If the projected graph a graph of the inverse function $y = f^{-1}(x)$?

 

[FactChecker]

I don't get you. Any hints?

I would say that sets A and B are mirror images along the line y=x means that $(r_1,r_2) \in A$ if and only if $(r_2,r_1)\in B$. For a function, $f$ defining set $A=\{(x,f(x))| x\in \mathbb R\}$, and its inverse, $f^{-1}$ defining set $B=\{(y,f^{-1}(y))| y \in \mathbb R\}$, what does that mean?

 

[brotherbobby]

Draw the x and y axes, the line $y = x$, and a one-to-one function $y = f(x)$. Project every point on the graph of $y = f(x)$ at a right angle through the line $y =x$, the same distance on the other side.

If the projected graph a graph of the inverse function $y = f^{-1}(x)$?

I copy and paste below the graph of the line $y = 3x-5$ and its inverse $y = \dfrac{x+5}{3}$

Yes, one line does look like the mirror image of the other across the line $y=x$. Is there a reason why should this always be the case?

 

[fresh_42]

I copy and paste below the graph of the line $y = 3x-5$ and its inverse $y = \dfrac{x+5}{3}$

View attachment 340061

Yes, one line does look like the mirror image of the other across the line $y=x$. Is there a reason why should this always be the case?

Yes, as mentioned earlier. Assuming the existence of an inverse, we map $y\longmapsto x$ instead of $x\longmapsto y.$ The roles of $x$ and $y$ are simply exchanged.

 

[PeroK]

I copy and paste below the graph of the line $y = 3x-5$ and its inverse $y = \dfrac{x+5}{3}$

View attachment 340061

Yes, one line does look like the mirror image of the other across the line $y=x$. Is there a reason why should this always be the case?

Yes, because the point $(x, f(x))$ is mapped to the point $(f(x), x)$, which can be rewritten as $(x, f^{-1}(x))$. If you'll excuse the loose notation.

 

[FactChecker]

Yes, one line does look like the mirror image of the other across the line $y=x$. Is there a reason why should this always be the case?

That is what the problem asks you to prove. You need to take the answers already given seriously and show that you are making an effort in that direction.

 

[brotherbobby]

That is what the problem asks you to prove. You need to take the answers already given seriously and show that you are making an effort in that direction.

Yes, I got it now. Let me state the problem and again and solve it below.

The key to the question is - what do we mean by the inverse of a function? If $y=f(x)$ is a function, then its inverse given by $y=f^{-1}(x)$ is a different function, but related to the original by the fact that the new function now maps the original $y$ to the $x$. So if $(x,f(x))$ is a point in the original function, $(f(x),x)$ is a point in the inverse.
Now what does reflection about the line $y=x$ do? It interchanges the values of a point from $(x,y)\rightarrow(y,x)^{\text{how?}\Large{\color{red}*}}$. But that is just the inverse of a function where $y=f(x)$.
Hence we can say that an inverse of a function is the mirror image of the original function about the line $y=x.$
Please let me know if my reasoning is correct.
Thanks to @PeroK and @FactChecker.

$\Large{\color{red}{*}}$ This can be shown using geometry. If $(a,b)$ be one point and $(b,a)$ the other, the slope of the line segment joining them is $-1$, which makes it perpendicular to the line $y=x$. The midpoint of the line segment is $\left(\dfrac{a+b}{2}, \dfrac{a+b}{2}\right)$, which falls on the line $y=x$. Hence the line $y=x$ is the perpendicular bisector of the line segment joining $(a,b)\rightarrow (b,a)$, which means one point is a mirror image of the other about $y=x.$

 

[Mark44]

@brotherbobby, after you have completed this exercise, there's something you should consider. The idea that the graphs of $y = f(x)$ and $y = f^{-1}(x)$ (for an invertible function f) are mirror images of each other across the line y = x is one that is presented in most textbooks on precalculus. However, it's of very little consequence in subsequent studies of calculus, IMO.
What is more important, in my view, is that for a one-to-one function f, the equations $y = f(x)$ and $x = f^{-1}(y)$ are identical. For example, consider $y = f(x) = \sqrt {x - 1}$. An equivalent equation is $x = f^{-1}(y) = y^2 + 1$. Note that the domain of f is the set $\{x | x \ge 1\}$ and the range is $\{y | y \ge 0\}$. The domain of the inverse is exactly the same as the range of the original function, and the range of the inverse is exactly the domain of the original function.
Looking at a function and its inverse in this way lets you see each point on the graph of a function in two different ways. The first (with y = f(x)) lets you calculate the y value in terms of a given x value. The second (using the inverse) lets you calculate an x value in terms of a given y value.

Where this point of view comes into play is in calculus, and specifically in using integration to find the area under a curve. If you're asked to find the area under the graph of $y = \sqrt{x - 1}$ between x = 1 and x = 10, one integral that expresses this area is $\int_{x=1}^{10} \sqrt{x - 1} ~dx$. Here, the typical area element is vertical, running from the x-axis up to the graph of $y = \sqrt{x - 1}$.
Another integral that represents the same area is $\int_{y = 0}^3 (10 - (y^2 + 1))~dy$. In this example, the typical area element is horizontal, and runs from the graph of $x = y^2 + 1$ to the line x = 10##. By the way, both integrals evaluate to the same number, namely 18.

This example might be beyond your present knowledge, but my point is that all this business of reflecting a graph across the line y = x is of little value in courses that follow precalculus.