$f(x) = 1 / (x-1)$ is not a function because for $x = 1$ there is a vertical asymptote which means infinte number of values of $y$ for $x = 1$.
It is a function for $\mathbf{R}- \{1\}$.
I want someone to just tell me if I am getting it right.
$f(x) = 1 / (x-1)$ is not a function because for $x = 1$ there is a vertical asymptote which means infinte number of values of $y$ for $x = 1$.
It is a function for $\mathbf{R}- \{1\}$.
I want someone to just tell me if I am getting it right.
When you define a function you need to specify what domain and codomain it has. The function cannot have this functional equation if the expression is not defined for a member of its domain (i.e. division by $0$). So you did not specify a function, you specified an equation that the function f is supposed to satisfy. There are multiple solutions to this equation so this does not define a single function. If you want to be more specific you could define $f$ to be the function with the maximal domain in $\mathbb R$ with codomain $\mathbb R$ that satisfies this equation. In this case, yes f is the function with domain $\mathbb R \setminus \{1\}$, codomain $\mathbb R$ and functional equation $f(x)=\frac{1}{x-1}$ for all $x$ of the domain.
It is a function, with an expected domain of $(-\infty, 1) \cup (1, +\infty)$ instead of all $\Bbb{R}$. When you define a function, you need to specify what your domain is.