Range and Codomain, precalculus lesson

A fuel pump display can show any amount from 0.00 to 999.99, yet a pump filling a 50-liter tank at 2 dollars per liter never shows more than 100.00. The display is a container for possible outputs, and the amounts that actually appear form a smaller set. Every function comes with these two sets, and the smaller one answers a practical question. For which numbers $y$ does the equation $f(x) = y$ have a solution?

Key terms

Codomain

Definition A function $f$ from a set $A$ to a set $B$ , written $f\colon A \to B$ , assigns to each element $x$ of $A$ exactly one element $f(x)$ of $B$ . The set $A$ is the domain of $f$ and the set $B$ is its codomain. The codomain is chosen as part of the definition of the function, and the only requirement is that every output lies in it. When a real function is given by a formula alone, its domain is the natural domain of the formula and its codomain is taken to be $\mathbb{R}$ , the set of all real numbers.

In plain words The codomain is the set where the outputs are allowed to land. Think of it as the label on a container, which says what may go inside, not what is actually inside.

Example The rule $f(x) = x^2$ as a function $f\colon \mathbb{R} \to \mathbb{R}$ has codomain $\mathbb{R}$ , even though no output is negative. The same rule from $\mathbb{R}$ to $[0, \infty)$ is also a valid function, because every square lies in $[0, \infty)$ . The same rule from $\mathbb{R}$ to $[1, \infty)$ is not a function at all, since the input $0$ has the output $0$ , which is outside $[1, \infty)$ .

Image

Definition Let $f\colon A \to B$ be a function. For an element $x$ of $A$ , the element $f(x)$ is called the image of $x$ under $f$ . For a subset $S$ of $A$ , the image of $S$ is the set of all outputs of inputs from $S$ ,

f(S) = \{\, f(x) \mid x \in S \,\}.

So a number $y$ belongs to $f(S)$ exactly when $y = f(x)$ for at least one $x$ in $S$ .

In plain words The image of a number is where the function sends it. The image of a set is everything its inputs reach, with each output listed once, however many inputs share it.

Example Take $f(x) = x^2$ and $S = \{-2, -1, 0, 1, 2\}$ . The five inputs have the images $4, 1, 0, 1, 4$ , so

f(S) = \{0, 1, 4\}.

This set has only three elements, because $-2$ and $2$ share the image $4$ , and $-1$ and $1$ share the image $1$ .

Range

Definition The range of a function $f\colon A \to B$ is the image of its whole domain,

\begin{aligned} f(A) = \{\, &y \in B \mid y = f(x) \\[0.4em] &\text{for some } x \in A \,\}. \end{aligned}

A number $y$ is in the range exactly when the equation $f(x) = y$ has at least one solution $x$ in the domain $A$ . Some older books use the word range for the codomain. Here the range always means the set of outputs that actually occur.

In plain words The range is the set of outputs the function really produces. To test a number $y$ , try to solve $f(x) = y$ for an input $x$ that is allowed.

Example Let $f(x) = x^2 + 3$ with domain $\mathbb{R}$ . Solving $x^2 + 3 = y$ gives $x^2 = y - 3$ , which has a real solution exactly when $y - 3 \ge 0$ . For $y = 7$ the solutions are $x = 2$ and $x = -2$ , so $7$ is in the range. For $y = 1$ the equation becomes $x^2 = -2$ , which has no real solution, so $1$ is not. The range is $[3, \infty)$ .

Range as a subset of the codomain

Definition For every function $f\colon A \to B$ the range is a subset of the codomain,

f(A) \subseteq B,

because each output is an element of $B$ by definition. The two sets may be equal, or the range may be strictly smaller. A function whose range equals its codomain is called onto, or surjective.

In plain words The codomain is the container and the range is what is actually in it. The container can have room to spare.

Example The function $f\colon \mathbb{R} \to \mathbb{R}$ with $f(x) = x^2$ has range $[0, \infty)$ , a proper part of its codomain. A negative number such as $-4$ is in the codomain but is never an output, since $x^2 = -4$ has no real solution. By contrast, the function $g\colon \mathbb{R} \to \mathbb{R}$ with $g(x) = x^3$ is onto, because every real number $y$ is the output of the input $x = \sqrt[3]{y}$ .

Projection of a graph

Definition The graph of a real function $f$ with domain $D$ is the set of points $(x, f(x))$ with $x$ in $D$ . Projecting a point $(x, y)$ onto the $y$ -axis keeps only its second coordinate $y$ . The range of $f$ is the projection of its graph onto the $y$ -axis, that is, the set of second coordinates of all points on the graph. Equivalently, a number $c$ is in the range exactly when the horizontal line $y = c$ meets the graph at least once. In the same way, the domain is the projection of the graph onto the $x$ -axis.

In plain words Slide every point of the graph sideways onto the $y$ -axis. The part of the axis that gets covered is the range.

Example The graph of $f(x) = \sqrt{4 - x^2}$ , with domain $[-2, 2]$ , is the upper half of the circle of radius $2$ centered at the origin. Its heights run from $0$ at the ends up to $2$ at the top, so its projection onto the $y$ -axis is $[0, 2]$ . Algebra confirms the picture. Every value is at least $0$ because it is a square root, and at most $2$ because $4 - x^2 \le 4$ . Each $c$ in $[0, 2]$ is reached, since $\sqrt{4 - c^2}$ is in the domain and

\begin{aligned} f\left(\sqrt{4 - c^2}\right) &= \sqrt{4 - (4 - c^2)} \\[0.4em] &= \sqrt{c^2} \\[0.4em] &= c. \end{aligned}

The last step uses $c \ge 0$ .

Attained and unattained bounds

Definition Let $f$ have domain $D$ . A number $M$ is an upper bound of $f$ if $f(x) \le M$ for every $x$ in $D$ , and a number $m$ is a lower bound of $f$ if $f(x) \ge m$ for every $x$ in $D$ . A bound is attained if it is itself a value of $f$ , that is, it equals $f(c)$ for some $c$ in $D$ , and unattained otherwise. An attained upper bound is the largest element of the range, the maximum value of $f$ , and an attained lower bound is the minimum value. The unattained bounds that matter are the ones the values approach. An upper bound $M$ is unattained but approached if $f(x) \ne M$ for every $x$ in $D$ while for every $\varepsilon > 0$ some $x$ in $D$ has $f(x) > M - \varepsilon$ . Such an $M$ is not in the range, and the range has no largest element. Lower bounds that are unattained but approached are defined the same way.

In plain words A range can creep up on a number forever without ever touching it. When the range is an interval, that number is an open end and is written with a parenthesis in interval notation.

Example Consider

f(x) = \frac{x^2}{x^2 + 1}, \qquad x \in \mathbb{R}.

Since $0 \le x^2 < x^2 + 1$ , every value satisfies $0 \le f(x) < 1$ . The lower bound $0$ is attained at $x = 0$ . The upper bound $1$ is never reached, yet the values come as close to it as you like.

$x$	$f(x)$
$0$	$0$
$1$	$0.5$
$3$	$0.9$
$10$	$\approx 0.990$
$100$	$\approx 0.9999$

Solving $f(x) = c$ confirms this. Multiplying by $x^2 + 1$ gives $(1 - c)x^2 = c$ . For $c = 1$ this says $0 = 1$ , so $1$ is never a value. For $c \ne 1$ it gives

x^2 = \frac{c}{1 - c},

which has a real solution exactly when $0 \le c < 1$ . The range is $[0, 1)$ . Questions about whether a bound is attained come back in calculus as extreme values.

Effect of a restricted domain

Definition If $S$ is a subset of the domain $D$ of $f$ , the restriction of $f$ to $S$ is the function with domain $S$ and the same rule. Its range is the image $f(S)$ , and

f(S) \subseteq f(D).

Restricting the domain keeps the range the same or makes it smaller, never larger. The new range has to be found from all of $S$ , not only from the values at the endpoints of $S$ , because the graph can turn around in between, and an endpoint left out of $S$ can leave a bound unattained.

In plain words Fewer inputs give at most the same outputs. Cut the graph down to the allowed inputs, then project what is left onto the $y$ -axis.

Example The function $f(x) = x^2$ has a different range on each of these domains.

Domain	Range
$\mathbb{R}$	$[0, \infty)$
$[-1, 3]$	$[0, 9]$
$(1, 3]$	$(1, 9]$

On $[-1, 3]$ the endpoint values are $f(-1) = 1$ and $f(3) = 9$ , yet the range is $[0, 9]$ and not $[1, 9]$ , because the input $0$ lies in $[-1, 3]$ and gives the output $0$ . On $(1, 3]$ the function is increasing, so its values lie in $(1, 9]$ , and each $c$ in $(1, 9]$ is reached at $x = \sqrt{c}$ . The bound $1$ is unattained because the input $1$ was left out.

Check yourself

Find the range of the function below on its natural domain.

g(x) = \frac{2x + 1}{x - 3}

Show the answer

The natural domain is all $x \ne 3$ . Solving $g(x) = y$ gives

\begin{aligned} y(x - 3) &= 2x + 1, \\[0.4em] \text{so} \quad x(y - 2) &= 3y + 1. \end{aligned}

For $y \ne 2$ this has the solution

x = \frac{3y + 1}{y - 2},

which is never $3$ , since setting it equal to $3$ leads to $3y + 1 = 3y - 6$ , that is $1 = -6$ . For $y = 2$ the equation becomes $0 = 7$ , with no solution. The range is $(-\infty, 2) \cup (2, \infty)$ .

The function $h\colon [0, 4] \to \mathbb{R}$ is given by $h(x) = (x - 1)^2$ . Name its codomain, find its range, and say whether $h$ is onto.

Show the answer

The codomain is $\mathbb{R}$ . Every value is at least $0$ , and $h(1) = 0$ . For $x$ in $[0, 4]$ the distance from $x$ to $1$ is at most $3$ , so every value is at most $9$ , and $h(4) = 9$ . The endpoint value $h(0) = 1$ is neither the smallest nor the largest value. Each $c$ in $[0, 9]$ is reached at $x = 1 + \sqrt{c}$ , which lies in $[1, 4]$ . So the range is $[0, 9]$ , and $h$ is not onto, since a number such as $-1$ is never an output.

Find the range of $k(x) = 3 - \frac{1}{x}$ on $[1, \infty)$ and say which bounds are attained.

Show the answer

For $x \ge 1$ we have $0 < \frac{1}{x} \le 1$ , so $2 \le k(x) < 3$ . The lower bound $2$ is attained at $x = 1$ . Each $c$ in $[2, 3)$ is reached at $x = \frac{1}{3 - c}$ , which is at least $1$ . The upper bound $3$ is never attained, because $\frac{1}{x} > 0$ . The range is $[2, 3)$ .

Range and Codomain

Theory, part 9

Practice, part 10

Key terms

Codomain

Image

Range

Range as a subset of the codomain

Projection of a graph

Attained and unattained bounds

Effect of a restricted domain

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus