Relations and Functions: Practice

A relation is any set of ordered pairs, which you can picture as a set of points in the plane. It might or might not be a function.

$\text{relation} = \text{a set of pairs } (x, y)$

The extra requirement of one output per input is what upgrades a relation to a function; it is not part of being a relation.

A function pins the output down: once the input is fixed, there is a single output.

$\text{input} \mapsto \text{exactly one output}$

The property that each output comes from only one input is a different idea (one-to-one), and it is not required for a relation to be a function.

With $y$ as the input, the rule is $x = y^2$, so square the input.

$x = (-3)^2 = 9$

Read this way the relation is a function of $y$: each value of $y$ produces one value of $x$.

The only danger is a zero denominator, so find where it vanishes.

$x - 6 = 0 \;\Rightarrow\; x = 6$

Every other input gives a real output, so the natural domain is all real numbers except $6$.

Domain is about inputs, range is about the outputs produced, and codomain is the declared set the outputs live in.

$\text{domain} = \{\, \text{allowed inputs} \,\}$

So the domain is the set of inputs the function is allowed to take.

The range collects what actually comes out as the input runs over the domain.

$\text{range} = \{\, f(x) : x \text{ in the domain} \,\}$

That is the set of outputs actually produced, which can be smaller than the declared codomain.

A function needs each input to appear with only one output. Check for a repeated input.

$1 \mapsto 2, \quad 3 \mapsto 2, \quad 5 \mapsto 2$

Here the inputs $1, 3, 5$ are all different, so it is a function; the repeated output $2$ is fine. The other sets each repeat an input.

There is no denominator and no root, so nothing can go wrong for any input.

$5x + 2 \text{ is defined for every real } x$

The natural domain is all real numbers.

The relation $x = 5$ is the vertical line through $5$: every point on it has first coordinate $5$ and any second coordinate.

$x = 5 : \; y \text{ can be any real number}$

So the one input $x = 5$ has infinitely many outputs, which fails the definition of a function of $x$.

For $y = x^3$, each $x$ gives one $y$, and each $y$ comes from a single cube root.

$y = x^3 \;\Longleftrightarrow\; x = \sqrt[3]{y}$

So it passes both the vertical and horizontal line tests. The others fail at least one reading.

By convention, an unstated domain means the natural domain.

$\text{no domain stated} \;\Rightarrow\; \text{use the natural domain}$

That is the largest set of inputs for which the formula returns a real number, found by excluding zero denominators and negatives under even roots.

Two different inputs sharing one output is allowed. Look for distinct inputs with equal outputs.

$f(2) = 4 = f(-2)$

The inputs $2$ and $-2$ both give $4$. The pair $f(2) = 4$ and $f(2) = 5$ would instead be one input with two outputs, which a function forbids.