Finding the Domain, foundations for calculus lesson

Type $\sqrt{-4}$ into a calculator set to real numbers and it answers with an error. Type $1/0$ and it does the same. A formula quietly refuses some inputs, and before you graph it, take its limit or differentiate it, you need to know which ones. Finding the domain is the habit of asking, for each operation in the formula, what it needs in order to give a real number.

Key terms

Real-valued function

Definition A real-valued function is a function whose outputs are real numbers. In this course its inputs are real numbers as well, so a formula counts as defined at $x$ only when every step of evaluating it at $x$ produces a real number. Complex numbers are not allowed as intermediate or final values.

In plain words Every number that appears while you compute $f(x)$ , and the final answer, must be an ordinary real number.

Example For $f(x) = \sqrt{x}$ the input $x = 9$ gives the real output $3$ . The input $x = -4$ would need $\sqrt{-4}$ , which is not a real number, so $f$ is not defined at $-4$ as a real-valued function.

Domain

Definition The domain of a function $f$ is the set of all inputs $x$ for which $f(x)$ is defined. A function is not fully described until its domain is known, either because it is stated or because it is understood from the formula.

In plain words The domain is the list of inputs the function accepts.

Example A function given by a table

$x$	$-1$	$0$	$2$
$g(x)$	$5$	$3$	$8$

has domain $\{-1, 0, 2\}$ . Asking for $g(1)$ makes no sense, because $1$ is not an input of $g$ .

Natural domain

Definition When a function is given by a formula and no domain is stated, its natural domain is the set of all real numbers $x$ for which the displayed formula produces a real value. It is found from the formula exactly as written, before any simplification. Canceling a factor does not add the inputs where the factor was zero.

In plain words Use every real number that the formula, as written, can handle. Simplifying first can hide a problem, so read the domain off the original formula.

Example The formula

q(x) = \frac{x^2 - 4}{x - 2}

is undefined at $x = 2$ , where the denominator is zero. For $x \ne 2$ it equals $x + 2$ , but the natural domain of $q$ is still all real numbers except $2$ . The graph is the line $y = x + 2$ with a hole at $(2, 4)$ .

Radicand

Definition The radicand is the expression under a root sign. A real even root, such as $\sqrt{u}$ or $\sqrt[4]{u}$ , is defined exactly when its radicand satisfies $u \ge 0$ . A real odd root, such as $\sqrt[3]{u}$ , is defined for every real $u$ .

In plain words Square roots and other even roots need the inside to be zero or positive. Cube roots accept anything.

Example For $f(x) = \sqrt{6 - 2x}$ the radicand must satisfy $6 - 2x \ge 0$ . Subtract $6$ to get $-2x \ge -6$ , then divide by $-2$ and reverse the inequality, which gives $x \le 3$ . The natural domain is $(-\infty, 3]$ . The endpoint $3$ is included because $\sqrt{0} = 0$ is a real number. By contrast, $\sqrt[3]{x - 5}$ has domain $(-\infty, \infty)$ .

Denominator restriction

Definition Every denominator in a formula must be nonzero at the input. When the whole denominator is an even root, that root must exist and must not be zero, so its radicand must be strictly positive. For example

\frac{1}{\sqrt{u}} \text{ is defined exactly when } u > 0.

In plain words You can never divide by zero. A square root underneath a fraction bar adds a stronger condition than a square root on top, because the root itself must not be zero.

Example For $k(x) = \frac{1}{\sqrt{x - 1}}$ the radicand must be positive, so $x - 1 > 0$ and $x > 1$ . The natural domain is $(1, \infty)$ . At $x = 1$ the root equals $0$ , and dividing by it is impossible, so $1$ is left out, while $\sqrt{x - 1}$ on its own would accept $x = 1$ .

Intersection of conditions

Definition If a formula imposes conditions $C_1, C_2, \ldots, C_n$ on $x$ , one for each root and each denominator, then its natural domain is the set of real numbers that satisfy all of them at the same time. This is the intersection of the solution sets,

\begin{aligned} &\{x \mid C_1\} \cap \{x \mid C_2\} \\[0.4em] &\quad \cap \cdots \cap \{x \mid C_n\}. \end{aligned}

Each endpoint of the result is included or excluded according to every condition that involves it.

In plain words Solve each condition on its own, draw them on one number line, and keep only the part where they all overlap.

Example Compare two formulas built from the same pieces. The first has the root in the numerator,

h(x) = \frac{\sqrt{x + 2}}{x - 3}.

The root needs $x + 2 \ge 0$ , so $x \ge -2$ . The denominator needs $x \ne 3$ . Both hold on

[-2, 3) \cup (3, \infty).

The second has the root in the denominator,

p(x) = \frac{x - 3}{\sqrt{x + 2}}.

Now the numerator is fine for every $x$ , and the denominator needs $x + 2 > 0$ . The domain is $(-2, \infty)$ , which includes $3$ and excludes $-2$ , the opposite choices from $h$ at both points.

Input	$h(x)$	$p(x)$
$-2$	$0$	undefined
$3$	undefined	$0$

Stated domain

Definition A stated domain is a domain given explicitly with the function, such as $f(x) = x^2$ for $0 \le x \le 3$ , or one forced by the context of a problem. A stated domain must be a subset of the natural domain, and when one is given it replaces the natural domain, even if the formula would accept more inputs.

In plain words Sometimes the problem tells you which inputs to use. Then those are the only inputs, whatever the formula could handle.

Example The area of a square with side $s$ is $A(s) = s^2$ . The formula accepts every real $s$ , but a side length must be positive, so in this context the domain is $(0, \infty)$ . The input $s = -3$ gives the number $9$ , yet it does not describe any square.

Check yourself

Find the natural domain of $f(x) = \frac{\sqrt{5 - x}}{x + 1}$ .

Show the answer

The root needs $5 - x \ge 0$ , so $x \le 5$ . The denominator needs $x \ne -1$ . Together the domain is

(-\infty, -1) \cup (-1, 5].

Find the natural domain of $g(x) = \frac{1}{\sqrt{x^2 - 9}}$ .

Show the answer

A root in the denominator needs a strictly positive radicand, so $x^2 - 9 > 0$ , which means $x^2 > 9$ . That holds exactly when $x < -3$ or $x > 3$ , so the domain is

(-\infty, -3) \cup (3, \infty).

A classmate says the domain of $r(x) = \frac{x^2 - 1}{x - 1}$ is all real numbers, because $r(x) = x + 1$ . Are they right?

Show the answer

No. The natural domain comes from the formula as written, and its denominator is zero at $x = 1$ . The domain is all real numbers except $1$ . The simplification $r(x) = x + 1$ holds only for $x \ne 1$ .

Finding the Domain

Key terms

Real-valued function

Domain

Natural domain

Radicand

Denominator restriction

Intersection of conditions

Stated domain

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus