Is $f(x)=\frac{1}{x}$ invertible?

Question

Let $f(x)=\frac{1}{x}$, then $f^{-1}(x)$ must equal $$y=\frac{1}{x}$$ and after swapping the variables $$x=\frac{1}{y}$$and rearranging to solve for $y$, $$\frac{1}{x}=y$$ that being said, can you conclude that $f^{-1}(x)=\frac{1}{x}$? I know it works since, when you take $f(f^{-1}(x))$ the result is $x$, it's just unheard of to me to have a function whose inverse is itself... Any help is appreciated.

Answer 1

This might no longer look strange when you consider that from a geometric point of view, inverting a function is just flippes its graph on the $45^°$-line (i.e. the graph of the function $f(x)=x$).

Now look at the graph of $1/x$, which is perfectly symmetric to this line. No wonder that its inverse is the same. You will immediately see that the graphs of

$$f(x)=x,\qquad f(x)=-x+k,\qquad f(x)=\sqrt{1-x^2}$$

are all symmetric in the same way (on an appropriate domain), hence they are all their own inverses.

Answer 2

Yes, the function is invertible, and is its own inverse. It's also not the only such function, $f(x)=-x$ is another such example.

Answer 3

Actually, there are more examples of functions that are their own inverse: $$f(x)=-x$$ $$f(x)=-\frac1x$$ $$f(x)=x$$ $$f(x)=\begin{cases}x+0.5\text{ if }x-\lfloor x\rfloor<0.5\\x-0.5\text{ otherwise}\end{cases}$$ $$\ldots$$