Natural and Specified Domains, precalculus lesson

A calculator shows an error when you ask for $1/0$ or for $\sqrt{-4}$ . A formula has the same blind spots, and the inputs it can handle make up its domain. Finding the domain first keeps you from computing answers that do not exist, and calculus keeps asking for it, for example when it decides where a function is continuous.

Key terms

Real-valued formula

Definition A formula in $x$ is real-valued at a real number $a$ when substituting $a$ for $x$ and carrying out every operation in the formula produces a real number. In a formula built from addition, subtraction, multiplication, division and roots, only two steps can fail. Division by $0$ is undefined, and an even root of a negative number is not a real number. Logarithms and some trigonometric functions add further restrictions later in the course.

In plain words Plug the number in and work it out. If every step gives an honest real number, the formula is fine at that input.

Example The formula $\sqrt{x - 2}$ is real-valued at $x = 6$ , where it gives $\sqrt{4} = 2$ , and at $x = 2$ , where it gives $\sqrt{0} = 0$ . At $x = 1$ it would need $\sqrt{-1}$ , which is not a real number, so the formula is not real-valued there.

Natural domain

Definition When a function is given by a formula and no domain is stated, its domain is taken to be its natural domain, the set of all real numbers at which the formula is real-valued. In set-builder notation the natural domain of $f(x)$ is

\{x \in \mathbb{R} \mid f(x) \text{ is a real number}\}.

In plain words If nobody tells you the domain, use every real number the formula can handle.

Example The polynomial $f(x) = x^3 - 4x$ is real-valued for every real $x$ , so its natural domain is $\mathbb{R}$ . The function

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

fails only at $x = 3$ , so its natural domain is $(-\infty, 3) \cup (3, \infty)$ .

Specified domain

Definition A domain can be stated together with the formula, as in $f(x) = x^2$ for $0 \le x \le 2$ . The domain is then exactly the stated set, which must be a subset of the natural domain of the formula. Two functions with the same formula but different domains are different functions.

In plain words A stated domain replaces the set of inputs the formula could handle. It can shrink that set but never enlarge it.

Example Let $f(x) = x^2$ with domain $[0, 2]$ . Then $f(1.5) = 2.25$ , but $f(3)$ is undefined, even though $3^2 = 9$ is a perfectly good number, because $3$ is not in the stated domain. On the other hand, the formula $\frac{1}{x}$ cannot be given the domain $[-1, 1]$ , since $0$ lies in that interval but not in the natural domain.

Denominator restriction

Definition A quotient

\frac{P(x)}{Q(x)}

is real-valued at $a$ exactly when $P(a)$ and $Q(a)$ are real numbers and $Q(a) \ne 0$ . So every input that makes a denominator equal to $0$ must be excluded from the domain. A numerator equal to $0$ causes no problem.

In plain words You can never divide by zero, so throw out every input that turns a denominator into zero.

Example For

h(x) = \frac{x + 1}{x^2 - 4},

the denominator is $0$ when $x^2 = 4$ , that is at $x = 2$ and $x = -2$ . The natural domain is $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$ . The input $x = -1$ is allowed, and $h(-1) = \frac{0}{-3} = 0$ .

Even-root restriction

Definition Let $n \ge 2$ be an even integer. For a real number $u$ , the real $n$ th root $\sqrt[n]{u}$ is defined exactly when $u \ge 0$ , and then it is the unique nonnegative real number whose $n$ th power is $u$ . So a formula containing $\sqrt[n]{E}$ with $n$ even is real-valued only where $E \ge 0$ . When the even root is the whole denominator, as in $\frac{1}{\sqrt[n]{E}}$ , the condition becomes $E > 0$ , because the root is $0$ exactly when $E = 0$ . If the root is only part of a denominator, keep the condition $E \ge 0$ and separately require the whole denominator to be nonzero. For example, the formula $\frac{1}{\sqrt{x} + 1}$ is real-valued for every $x \ge 0$ , including $x = 0$ , since $\sqrt{x} + 1 \ge 1$ .

In plain words Square roots, fourth roots and other even roots need something nonnegative underneath. When the root is the whole denominator, the number underneath must be strictly positive.

Example For $f(x) = \sqrt{6 - 2x}$ we need $6 - 2x \ge 0$ . Subtracting $6$ gives $-2x \ge -6$ , and dividing by $-2$ reverses the sign, so $x \le 3$ . The natural domain is $(-\infty, 3]$ . As a check, $f(3) = 0$ and $f(1) = 2$ , while $f(4)$ would need $\sqrt{-2}$ . For

k(x) = \frac{1}{\sqrt{x - 1}}

the radicand must be positive, so the natural domain is $(1, \infty)$ .

Odd roots

Definition Let $n \ge 3$ be an odd integer. For every real number $u$ there is exactly one real number whose $n$ th power is $u$ , and it is the real $n$ th root $\sqrt[n]{u}$ . So an odd root adds no restriction of its own, and negative numbers under an odd root are allowed.

In plain words Cube roots and other odd roots work for every real number, because a negative number cubed is negative.

Example Since $(-2)^3 = -8$ , we have $\sqrt[3]{-8} = -2$ . The function $f(x) = \sqrt[3]{x - 5}$ has natural domain $\mathbb{R}$ , and $f(-3) = \sqrt[3]{-8} = -2$ . The denominator rule still applies to odd roots, so

g(x) = \frac{1}{\sqrt[3]{x}}

has natural domain $(-\infty, 0) \cup (0, \infty)$ .

Excluded values

Definition The excluded values of a formula are the real numbers at which it is not real-valued, that is, the real numbers outside its natural domain. When a formula has several parts, each part gives its own condition, and the natural domain is the set of numbers that meet all of the conditions, the intersection of the separate sets.

In plain words List every danger in the formula, write down the condition each one needs, and keep only the inputs that pass every condition.

Example Take

f(x) = \frac{\sqrt{x + 2}}{x - 1}.

The square root needs $x + 2 \ge 0$ , which gives $[-2, \infty)$ . The denominator needs $x \ne 1$ . The intersection is

[-2, 1) \cup (1, \infty).

The excluded values are all $x < -2$ together with $x = 1$ .

Contextual domain

Definition When a function models a real situation, its contextual domain is the set of inputs that make sense in that situation. It is a subset of the natural domain and is often much smaller.

In plain words The formula might accept an input that the real world does not, like a negative length.

Example A rectangle has perimeter $20$ meters and width $x$ meters. Its length is $10 - x$ meters, so its area is

A(x) = x(10 - x) \text{ square meters.}

The formula works for every real $x$ , but a real rectangle needs $x > 0$ and $10 - x > 0$ , so the contextual domain is $(0, 10)$ . The value $A(12) = -24$ is a real number, yet no rectangle has it as its area.

Preserving restrictions after cancellation

Definition The natural domain is read from the formula as it is given, before any simplification. If a common factor $x - a$ is cancelled from a numerator and a denominator, the simplified formula gives the same values as the original one at every input where the original is defined. The original is still undefined at $x = a$ , so the restriction $x \ne a$ belongs to the function and must be kept and written down next to the simplified formula.

In plain words Cancelling makes the formula look nicer, but it does not change which inputs the original function accepts.

Example For $x \ne 3$ ,

\begin{aligned} f(x) &= \frac{x^2 - 9}{x - 3} \\[0.8em] &= \frac{(x - 3)(x + 3)}{x - 3} \\[0.8em] &= x + 3. \end{aligned}

The function $f$ is still undefined at $3$ , because its denominator is $0$ there, so $f$ is not the same function as $x + 3$ . Its graph is the line $y = x + 3$ with a hole at $(3, 6)$ . Inputs close to $3$ give outputs close to $6$ , while $f(3)$ itself does not exist.

$x$	$2.9$	$2.99$	$3$	$3.01$	$3.1$
$f(x)$	$5.9$	$5.99$	undefined	$6.01$	$6.1$

In calculus such a hole can be filled on purpose, which gives a continuous extension, a new function that is also defined at $3$ .

Check yourself

Find the natural domain of the following function.

f(x) = \frac{\sqrt{4 - x}}{x + 2}

Show the answer

The square root needs $4 - x \ge 0$ , so $x \le 4$ , and the denominator needs $x \ne -2$ . The natural domain is $(-\infty, -2) \cup (-2, 4]$ .

Is the function below the same as $g(x) = x + 2$ ?

f(x) = \frac{x^2 + 2x}{x}

Show the answer

No. For $x \ne 0$ the formula simplifies to $f(x) = x + 2$ , but $0$ is not in the domain of $f$ , since its denominator is $0$ there. So $f(0)$ is undefined while $g(0) = 2$ . The graph of $f$ is the line $y = x + 2$ with a hole at $(0, 2)$ .

Find the natural domain of the following function.

h(x) = \sqrt[3]{x - 7} + \frac{1}{\sqrt{x + 1}}

Show the answer

The cube root accepts every real number. The square root in the denominator needs $x + 1 > 0$ , so $x > -1$ . The natural domain is $(-1, \infty)$ .

Natural and Specified Domains

Theory, part 7

Practice, part 8

Key terms

Real-valued formula

Natural domain

Specified domain

Denominator restriction

Even-root restriction

Odd roots

Excluded values

Contextual domain

Preserving restrictions after cancellation

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus