What a function is: domain, range and notation

One input, one output

A function is a rule that takes an input and gives back exactly one output. Lesson 0.2.1 wrote this as $f(x)$: the output of the rule $f$ at the input $x$.

A table of values is the simplest way to see one. Here is $f(x) = x^2 - 2x$:

Two inputs may share an output: $f(-1) = f(3) = 3$. What a function may not do is give two outputs for one input. "The number whose square is $x$" is not a function, because for $x = 4$ it could be 2 or $-2$.

The output carries its own units, set by the rule. A car at a steady 60 km per hour has gone $d(t) = 60t$ km after $t$ hours. The input is a time in hours and the output a distance in km, so $d(2.5) = 150$ km.

A function can be given by a formula, a table, a graph or a piece of code. All four answer the same question: which output goes with each input?

The domain: which inputs are allowed

The domain is the set of inputs the function accepts. Unless told otherwise, it is every real number for which the formula makes sense.

Two things break a formula in this module: dividing by zero, and taking the square root of a negative number. Logarithms add a third in module 1.4.

Take $g(x) = \sqrt{x - 3}$. A table shows where it starts working:

At $x = 2$ the inside is $-1$, which has no square root. Algebra finds the edge exactly: the inside must be at least 0, so $x - 3 \ge 0$, which gives $x \ge 3$ (lesson 0.3.3).

The domain is the interval $[3, \infty)$. The square bracket keeps 3 itself, because $g(3) = \sqrt{0} = 0$ is fine.

Now take $h(x) = \frac{1}{x - 2}$. The bottom is zero at $x = 2$, so the domain is every real number except 2. In set-builder notation (lesson 0.2.1) that is $\{x \in \mathbb{R} : x \neq 2\}$.

Close to 2 the outputs grow very large, positive on one side and negative on the other: $h(2.1) = 10$, $h(2.01) = 100$ and $h(1.9) = -10$.

Three more cases complete the picture.

• Both rules at once. In $\frac{1}{\sqrt{x - 3}}$ the root needs $x - 3 \ge 0$ and the bottom needs $\sqrt{x - 3} \neq 0$. Together, $x - 3 > 0$: the domain is $(3, \infty)$, with 3 left out.
• Nothing to break. $x^2 - 2x$ has no fraction and no root, so its domain is all of $\mathbb{R}$.
• Limits from the situation. A square of side $s$ cm has area $A(s) = s^2$ cm². The formula accepts $s = -2$, but a side cannot be negative, so the domain is $s \ge 0$.

The range: which outputs come out

The range is the set of outputs the function produces as the input runs over the whole domain. On a graph, the domain is the graph's shadow on the $x$-axis and the range is its shadow on the $y$-axis.

A graph suggests the range; algebra confirms it. A number $y$ is in the range when the equation $f(x) = y$ has a solution in the domain. Rearrange it for $x$, as in lesson 0.3.2.

• $\sqrt{x - 3} = y$. Squaring both sides gives $x = y^2 + 3$: the output 5, for example, comes from $x = 28$. A square root is never negative, so this works for every $y \ge 0$ and no other. The range is $[0, \infty)$.
• $\frac{1}{x - 2} = y$. Multiply by $x - 2$ and divide by $y$: $x = 2 + \frac{1}{y}$. That works for every $y$ except 0, so the range is every real number except 0. The graph never touches the $x$-axis.
• $x^2 - 2x$. Multiplying out $(x - 1)^2$ gives $x^2 - 2x + 1$, so $x^2 - 2x = (x - 1)^2 - 1$. A square is never below 0, so the output is never below $-1$, and it is $-1$ at $x = 1$. The range is $[-1, \infty)$; lesson 1.2.2 turns this trick into a method.

A constant function such as $c(x) = 5$ has the smallest possible range: the single number 5.

Lesson 0.2.1 wrote $f: A \to B$, read "$f$ takes inputs from $A$ to outputs in $B$". The set $A$ is the domain.

The set $B$ only promises where outputs land; the range is the part of $B$ actually reached. For $g: [3, \infty) \to \mathbb{R}$ with $g(x) = \sqrt{x - 3}$, outputs land in $\mathbb{R}$, but the range is only $[0, \infty)$.

Graphs

The graph of $f$ is every point $(x, f(x))$ drawn on a grid. Two places on it have names.

• Where the graph meets the $x$-axis, $f(x) = 0$. Those inputs are the zeros (or roots) of $f$: for $x^2 - 2x$, 0 and 2. A function can have no zeros: $x^2 + 4$ has range $[4, \infty)$, so it never reaches 0.
• Where the graph meets the $y$-axis is the point $(0, f(0))$, and $f(0)$ is the $y$-intercept. It exists only when 0 is in the domain, so $\sqrt{x - 3}$ has none.

A graph also shows at a glance whether a curve is a function at all. Two points straight above each other would mean two outputs for one input.

These ideas carry the rest of the module. Lesson 1.1.2 moves and reshapes graphs, lesson 1.1.3 runs functions backwards, and case study 1.1.5 draws option payoffs as graphs.

Common mistakes

1. Mixing up domain and range. The domain is about inputs; the range is about outputs.
2. Getting the edge wrong. $\sqrt{x - 3}$ accepts 3, but $\frac{1}{\sqrt{x - 3}}$ does not: a root in a bottom must be strictly positive.
3. Forgetting the situation. A length or a time cannot be negative, even if the formula accepts it.
4. Trusting a sample as the whole range. Evaluating at many points shows the outputs on that interval only. Solve $f(x) = y$ to be sure.
5. Reading off a $y$-intercept that is not there. $f(0)$ exists only when 0 is in the domain.
6. Using math on arrays. math.sqrt takes one number. For a NumPy array use np.sqrt.
7. Accepting two outputs for one input. If a vertical line meets a curve twice, the curve is not a function of $x$.