Is there a general geometrical connection between $f(x)$ and $f(1/x)$?

Question

What is the relationship between graphs of a function $f(x)$ and the function $f\left(\frac{1}{x}\right)$?

If I consider $f(x)=x$, $x^2$, and $x^2-x$ and then $g(x)=f(1/x)$, especially in the last example ($x^2-x$) there is no answer clear to me.

Answer 1

In order to make the "point at infinity" idea work you have to do the following (nothing original here, this is just the simplest instance of the so called projective spaces).

Instead of just one real number $x$ consider pairs of real number $(x,y)$ except you

• exclude the pair $(0,0)$ and
• consider equal two points when one can be obtained from the other by multiplication for a real nonzero number, i.e. $(x,y)=(x^\prime,y^\prime)$ when $x^\prime=\lambda x$ and $y^\prime=\lambda y$ (same $\lambda$) for some $\lambda\neq0$.

Call $\mathbb{P}$ the set constructed in this way. Every point in the real line has a corresponding point in $\mathbb{P}$ simply as $x\mapsto (x,1)$. But $\mathbb{P}$ has an extra-point which doesn't "come from" the real line, namely the point $\infty=(1,0)$. This is the "point at infinity".

Given a function $f(x)$ for real $x$ you can define a function on $\mathbb{P}$ setting $$ F(a,b)=f(a/b)\qquad \text{if $b\neq0$.} $$ In general $F$ will not be defined at $\infty$ but in some cases you can extend $F$ to a function continuous at $\infty$, namely when $\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=L<\infty$.

(Mind that in this post I'm using the symbol $\infty$ with two different meanings)

Now the transformation $x\mapsto1/x$ on the real line can be read as $(x,1)\mapsto(1,x)$ on $\mathbb{P}$ which now makes sense also for $x=0$, thus $0=(0,1)$ and $\infty$ are interchanged.

Therefore, if you think the graph of $f(x)$ as "centered at $0$", the graph of $f(1/x)=F(1,x)$ is nothing but the graph of $F$ "centered at $\infty$".