Is there an explanation why the reflection of $f(x)$ through $y = x$ is its inverse?

Question

e.g. The function $e^x$ reflected through $y=x$ is $\ln x$. Is this always true OR just in some cases?

Answer 1

Yes, but it is the graph of the function that is reflected, not the function itself.

The graph of a function $f$ is the set of all pairs $(x,y)$ with $y=f(x)$. If $f$ has an inverse function $g$, then $y=f(x)$ is equivalent to $x=g(y)$, so when $(x,y)$ belongs to the graph of $f$, then $(y,x)$ belongs to the graph of $g$ and vice versa.

Interchanging $x$ and $y$ in the pair $(x,y)$, that is replacing it by $(y,x)$, can be described as reflection throught the diagonal. And that explains the phenomenon in general.

Answer 2

Reflection with respect to the diagonal $y=x$ means substitution of $x$ with $y$. So, start from $x$, apply $f$ and get $y$, then substitute $y$ with $x$, obtaining $x$. Conversely, start with $y$, apply substitution and get $x$, then apply $f$ to obtain $y$. As you can see, substitution is right and left inverse of $f$, hence is its inverse.

Answer 3

Consider the reflection matrix:

$$\begin {bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$$

This matrix reflects a point over the y = x axis. Now consider a point on a function, (x,f(x)). With respect to the origin, it can be expressed as a vector:

$$\begin {bmatrix} x \\ f(x) \\ \end{bmatrix}$$

Now apply the transformation matrix to the vector:

$$\begin {bmatrix} f(x) \\ x \\ \end{bmatrix}=\begin {bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\begin {bmatrix} x \\ f(x) \\ \end{bmatrix}$$

Now, in order to normalize this point (i.e. make x-coordinate x and the y coordinate a function of x), we need to apply the inverse operation to both coordinates:

$$\begin {bmatrix} f(x) \\ x \\ \end{bmatrix}=\begin {bmatrix} x \\ f^{-1}(x) \\ \end{bmatrix}$$$$

Answer 4

Here is a summary of my comments on Harald Hanche-Olsen's ans.

To proof g is the inverse of f iff they reflect each other along $y=x$.

First, we need $y′\equiv g(x′)$, g is the inverse of f. W/ the original y values of f we could get back the x's w/ g. So, $x′=y$ & $y′=x$.

Second, we need to let $(x_2,y_2)$ be the image of $(x_1,y_1)$ (above y=x) w/ dist. d from $(x_3,x_3)$. We have $y2=x3−d/\sqrt{2}=x1$. Similarly for $x_2=y_1$.

So $(x_0,y_0)$ of f, corresponds $(y_0,x_0)$ of g, which is also where it's image locate.

Same for other points of f.

So proofed.