Q.Check if the relation shown in Fig.\ 6.34 is a function, where $A=\{3,-2,1,0,2,4\}$, $B=\{9,7,-6,3,2\}$, and the arrows join $3\to7,\ 1\to9,\ 0\to-6,\ 2\to3,\ 4\to2$ (element $-2$ of $A$ has no arrow).

A function $f$ from a set $A$ to a set $B$ (written $f:A\to B$) is a rule that associates to every element $x\in A$ a unique element $y\in B$, written $y=f(x)$. Two things can break this: an element of $A$ left with no image at all, or an element of $A$ given two different images — either failure means the relation is not a function.

$A$ is called the domain (the full set of allowed inputs), $B$ is the co-domain (the set outputs are guaranteed to land inside), and the range is the smaller set $f(A)=\{y\in B\mid y=f(x)\text{ for some }x\in A\}$ of values actually produced — always a subset of the co-domain, though it may equal it. A function can be shown in six equivalent ways: a verbal description, an arrow diagram between two sets, a set of ordered pairs, an algebraic rule/formula, a table of values, or a graph — with domain and range read off each form appropriately (e.g. on a graph, domain is the horizontal spread/projection onto the $x$-axis, and range the vertical spread/projection onto the $y$-axis).

When the domain sits in $R$, the vertical line test decides whether a curve represents a function at all: if any vertical line meets the curve more than once, that $x$-value has two outputs, so it fails. To find a domain algebraically, list every structural restriction the formula carries (a denominator can't be zero, a square root's argument can't be negative, a logarithm's argument must be positive) and intersect them; to find a range, either track how the formula's value changes as $x$ sweeps the domain, or set $y=f(x)$ and solve for $x$ in terms of $y$ to see which $y$-values are actually achievable.