Domain of $\frac{1}{\frac{1}{x}}$

Question

Let $f(x)=\frac{1}{x}$, then we have $f^{-1}(x)=\frac{1}{x}$. So $f(f^{-1})=\frac{1}{\frac{1}{x}}=x$.

My question is, what is the domain of $f(f^{-1})(x)$? is it everything? or everything but zero?

Then if we plot the graph of $\frac{1}{\frac{1}{x}}$ how will it look like? Will it look just like $y=x$ or $y=x$ with the origin excluded?

I tried to plot using google, the result looks exactly like $y=x$ without the origin removed. But why? I think according to the rule when we compose two functions in this case, the origin must be removed.

Thanks for clarifying my confusion.

Answer 1

https://www.wolframalpha.com/input/?i=f%28x%29%3D1%2F%281%2Fx%29

You are right that the graph is $y=x$ with the origin removed.

Answer 2

We consider the function $f : \mathbb{R}^* \to \mathbb{R}^*$, $x \mapsto 1/x$ and basically compose it with itsself. This results in a function $f \circ f : \mathbb{R}^* \to \mathbb{R}^* \to \mathbb{R}^*$. (In general, if $f : A \to B$ is a function and $g : B \to C$ is a function, then $g \circ f$ is a function $A \to C$.) Here, we have $f(f(x))=x$ for every $x \in \mathbb{R}^*$. But we cannot say $f(f(0))=0$ since $f(0)$ is not defined. So in fact, the graph of $f \circ f$ is just the usual line, but without the origin.

Answer 3

Notice that the domains of, for example, $f(x)=\frac{x^2}{x}$ and $g(x)=x$ is different.

Answer 4

It's a problem of definition of the function. Every function is not given by the relation $f$ but by three things:

1) a set $A$, the domain;

2) a set $B$, the codomain;

3) a relation $R$ which is a subset of $A\times B$ and has the property that it is a function, i.e. for each $x\in A$ there is exactly one couple $(x,y)\in R$.

So a function is really a triple $(A,B,R)$.

In your case the functions $f(x)=1/(1/x)$ and $g(x)=x$ are different because the domain is different. You can see them as given by the (different) triplets

\begin{equation} f:(\mathbb{R}\setminus\{0\}, \mathbb{R},R_1)\qquad\text{and}\qquad g:(\mathbb{R}, \mathbb{R},R_2) \end{equation} where \begin{equation} R_1=\{(x,y)\in(\mathbb{R}\setminus\{0\})\times\mathbb{R}:x=y\}\\ R_2=\{(x,y)\in\mathbb{R}\times\mathbb{R}:x=y\} \end{equation}

Answer 5

A quotient is defined when both its numerator and denominator are defined and the denominator is nonzero.

The quotient $\frac1{\frac1x}$ is defined when $1$ is defined (always true) and $\frac1x$ is defined, and $\frac1x$ is nonzero (always true).

The quotient $\frac1x$ is defined when $1$ is defined (always true) and $x$ is defined (always true), and $x$ is nonzero (false when... $x=0$).

Hence, $\frac1x$ is undefined at $x=0$, hence $\frac1{\frac1x}$ is undefined at $x=0$.