If a function is not defined at a point then will its inverse be defined at that point?

Question

I have a function $y=\frac{x-2}{x-3}$ clearly the function is not defined at $x= 3$

But for its inverse $ \frac{3x-2}{x-1},$ can we say that it is defined at $x= 3$?

Answer 1

Sure. The domain of the inverse of $f$ (if it exists) is the range of $f$. And the fact that $3$ belongs or not to the domain of $f$ is independent of the fact that it belongs or not to the range of $f$.

Answer 2

Yes your map in indeed defined at $x=3$.

In general, for $A \subseteq B$ and a map $f : A \to B$, there is no relationship between the fact that $f$ is defined at $b \in B \setminus A$ and the fact that $f^{-1}$ is defined at $b$.

$f^{-1}$ will be defined at $b \in B$ providing that:

1. $b$ belongs to $f(A)$.
2. The inverse image of $b$ under $f$ is unique.

Answer 3

$$y=\frac{x-2}{x-3}$$

Let's study it's range.

$$y=\frac{x-2}{x-3}=\frac{x-3+1}{x-3}=1+\frac{1}{x-3}$$

Hence $1$ is not the domain of $f^{-1}$ but $3$ is.

Answer 4

Just to fill in the contrapositive which the other questions don't mention. A function whose entire partial function is a bijection from a set onto itself does have the property that its inverse is defined wherever it is defined.