What does it mean when $f(a)$ is defined?

Question

I am in Calculus I and my calculus jargon is lacking so please if you can explain, for lack of better terms, explain it simply.

Answer 1

To broaden the scope a bit;

Usually functions are defined from a domain to a codomain.

If the function, $f$, takes values from a set $X$ and ends up in a set $Y$ we write that $f: X \to Y$.

We say that $f$ is well defined on $X$ when $f(a)$ is uniquely given in $Y$ (that is, it takes on exactly one value for $a$).

Now, given a well defined function $f: X \to Y$, to say that $f(a)$ is defined is to say that $a \in X$ and nothing more.

Hope this clarifies some things.

Answer 2

"$f(a)$ is defined" means $f$ is a function, and $a$ is in the domain of $f$

For example suppose $f(x) = \frac 1x$

$f(1)$ is defined. $f(0)$ is not defined.

Answer 3

When talking about a function, we really should always remember to include in our description of the function the Domain (the set of all inputs we wish to consider), the Codomain (the set to which we are mapping, which may or may not include all outputs we actually can get) and the actual rule we follow to map an input to an output.

The function $f$ with domain $\Bbb R\setminus \{0\}$ and codomain $\Bbb R$ given by $f(x)=\frac{1}{x}$ is properly defined for all nonzero real numbers $x$. $f(0)$ however is not defined (division by zero is not defined for real numbers).

To be pedantic, I would argue that the function $g$ with domain $\{1,2,3\}$ and codomain $\{1\}$ given by $g(x)=1$ is defined for each of $\{1,2,3\}$. Despite the rule $g(x)=1$ being easy to interpret for any and every input $x$, I would argue that $g$ is not defined for any value different from $1,2,3$. $g(4)$ is not defined in this case. Our definition of the function only told us what happens for inputs from $\{1,2,3\}$ and did not tell us anything about how $g$ "should" act on inputs outside of that.

Such functions can be extended to have a larger domain, but this would no longer be the same function strictly speaking.

Answer 4

According to the definition of function, given $f:A\subseteq\mathbb{R}\to B\subseteq\mathbb{R}$ we say that $\forall a\in A \quad f(a)$ is defined when

$$\exists!b\in B : \quad b=f(a)$$