Domain and Range of a function

Question

PROBLEM: I do understand that for one input of $x$, $f(x)$ can only one output. But I can't understand how can $2$ inputs of $x$ give $1$ output of $y$.

I know for $f(x) = x^2$ , we can have same $f(x)$ for $x$ and $-x$ and for $f(x) = 3 $ is a singleton but take this specific example:

http://www.purplemath.com/modules/fcns.htm

This example says it is a function. What I see from here is this:

$f(-3) = 3$,

$f(-2) = -6$,

$f(-1) = 0$,

$f(0) = 15$,

$f(1) = 0$.

For $f(-1)$ and $f(1)$ to be same I need to have $f(x) = x^2 - 1$ but then $f(-3)$ will not fit into this relation. For $f(0) = 15$ I can take $f(x) = x + 15$ but then none of the other relations fit. So, I can't figure out if there is any relation that exists for "$f(x) = $ something here" and hence it does not seem to be a function. But I know I am wrong but how?

Answer 1

Not all the functions should be mathematical expressions such as $f(x) = ax^2+bx+c$ or $f(x) = \sin(x)$ or etc. We consider functions simply as a mapping from one set to another as in your example; mapping of the elements may or may not result in a mathematical expression.

Note that in your example, even if it gives a mathematical expression, $f(x) = x^2 - 1$ is not the only function where $f(-1) = f(1) = 0$. For example take $f(x) = \cos(x^2-1)-1$. What I mean here is that if we try to consider all the functions as a mathematical expression, we may have hard time doing that because there are many possibilities with given mapping. Moreover, it would be wrong to express a function as in your example with a mathematical expression.

As an example, two sets $A$ and $B$ and a function $f: A \to B$ with $A = \{-2,-1,0,1,2\}$, $B = \{-1,0,3\}$ and $$f(-2) = 3$$ $$f(-1) = 0$$ $$f(0) = -1$$ $$f(1) = 0$$ $$f(2) = 3$$ Notice that $f$ satisfies the conditions of a function $f(x) = x^2-1$ so it seems that we can say $f(x) = x^2-1$. However, stating this without any restrictions is wrong because $f(x) = x^2-1$ is a function $\mathbb{R} \to [-1,+\infty)$ but we only have $5$ elements in domain ($|A| = 5$) and $3$ elements in codomain (also in range, $|B| = 3$). It also takes some irrational values although we don't have any in set $A$. So we can't say $f(x) = x^2-1$ even if the mapping fits $f(x)=x^2-1$.

Answer 2

I think you are looking for a "rule" which matches with

$f(-3) = 3$,

$f(-2) = -6$,

$f(-1) = 0$,

$f(0) = 15$,

$f(1) = 0$.

Is that right?

There doesn't need to be a rule. And this is a pedagogical hiccup. In highschool we often tell students a function is some type of "rule" that connects inputs to outputs. And by college / late highschool we change the definition of a function and don't seem to inform students that we changed the game.

A function simply needs to follow your first claim: For one input there is one output. And if you are unable to come up with some explicit rule that says how you get from inputs to outputs... that doesn't make it not a function. In fact... there are functions that we know to exist but cannot write down the explicit rules for.

But just for fun! We can probably come up with rules that satisfy those points.

How about this one? It's no better than the infinitely many other ones that describe those points... $$f(x)= \begin{cases} 3x+12 \hspace{1.5 cm} x\leq-2 \\ 15(1-x^2) \hspace{1 cm} x>-2 \end{cases}$$