The implied domain of a formula
A formula arrives with no list of allowed inputs. Which numbers can you actually put into it?
Where you are 4 in Domain and range
Why this exists
Most functions in algebra arrive as a single line, such as \displaystyle f(x) = \sqrt{10 - 2x}, with nothing said about which inputs are allowed. Yet some are not. Put in x = 7 and the formula asks for \displaystyle \sqrt{-4}, which is not a real number, so the function has no output there.
Listing the allowed inputs every time would be tedious, so mathematics uses a convention instead: when no domain is stated, the domain is every real number the formula can accept. That set is the implied domain. Finding it means hunting down every input that breaks the formula, and in this course a formula can break in only two ways.
Definition
The implied domain of a function given by a formula (also called its natural domain) is the set of all real numbers x for which the formula gives a real number as output.
In this course, two things stop a formula from giving a real output:
- a denominator equal to 0, because division by zero has no answer;
- an even root of a negative number, such as \displaystyle \sqrt{-4} or \displaystyle \sqrt[4]{-1}, because no real number has a negative square or fourth power.
Everything else is safe for every real input: adding, subtracting, multiplying, whole-number powers, and odd roots such as \displaystyle \sqrt[3]{x}. So a polynomial has the implied domain of all real numbers, (-\infty, \infty).
A domain can also be set in two other ways, and both narrow the implied one.
- A stated domain is chosen by whoever defines the function and written beside it. The function f(x) = x^2 for 0 \le x \le 3 is a different function from f(x) = x^2, because it accepts fewer inputs.
- A context domain comes from what the variable means. If x counts people, x = 2.5 makes the formula work but makes no sense; that is a reasonable domain for a real situation.
When a domain is stated, it wins. It must sit inside the implied domain, because nobody can choose an input at which the formula gives no output.
Intuition
Read the formula as a row of checkpoints. Every denominator turns away the inputs that make it 0. Every even root turns away the inputs that make its inside negative. An input belongs to the domain only if it passes every checkpoint, so the domain is where all the conditions hold at once.
Figure 1 shows this for \displaystyle h(x) = \dfrac{\sqrt{x + 3}}{x - 5}, the function of Example 3. The root lets through x \ge -3, the denominator lets through everything except 5, and the domain is the overlap.
The marks on a number line match Interval and set-builder notation: a filled dot, like a square bracket, means the endpoint is in; an open circle, like a parenthesis, means it is out. You have read domains off pictures before, in Domain and range from a graph; here the picture is drawn from the formula.
Results
The implied domain, piece by piece. For a formula in x:
- a polynomial accepts every real number;
- a denominator q(x) removes every solution of q(x) = 0;
- an even root \displaystyle \sqrt{u(x)} or \displaystyle \sqrt[4]{u(x)} requires u(x) \ge 0;
- an odd root \displaystyle \sqrt[3]{u(x)} adds no condition of its own;
- an even root in a denominator, as in \displaystyle \dfrac{1}{\sqrt{u(x)}}, requires u(x) > 0, strictly.
The implied domain is the set of x that satisfy every condition together.
Stated and context domains. Every stated or context domain is part of the implied domain.
Why it is true
No division by zero. To say \dfrac{a}{0} = c would mean c \cdot 0 = a. If a \ne 0, no c works, since c \cdot 0 = 0. If a = 0, every c works, so there is no single answer. Either way the quotient has no value.
No even root of a negative. A square root of u is a number r with r^2 = u. Every real r has r^2 \ge 0: a positive times a positive is positive, a negative times a negative is positive, and 0^2 = 0. So when u < 0, no real r exists. The same holds for a fourth root, since r^4 = (r^2)^2 \ge 0. An odd power keeps the sign, (-2)^3 = -8, so \displaystyle \sqrt[3]{-8} = -2 and an odd root accepts every real number.
A root in a denominator. In \displaystyle \dfrac{1}{\sqrt{u}} the root needs u \ge 0. But u = 0 makes the denominator \displaystyle \sqrt{0} = 0, which the first fact forbids. What survives is u > 0.
Worked examples
A square root of a linear expression
Find the implied domain of \displaystyle f(x) = \sqrt{10 - 2x}. Write it in interval notation and in words.
SolutionThere is no denominator, so the only checkpoint is the square root. Its inside must not be negative:
Subtracting 10 from both sides keeps the inequality as it is. Dividing both sides by -2 reverses it, because dividing by a negative number reverses order, as in solving linear inequalities.
The domain is (-\infty, 5]: all real numbers less than or equal to 5. The bracket says 5 is included, and it should be, because \displaystyle f(5) = \sqrt{0} = 0.
Check. \displaystyle f(-3) = \sqrt{16} = 4 is a real number, while x = 6 gives \displaystyle \sqrt{-2}, which is not.
A denominator with two zeros
Find the implied domain of g(x) = \dfrac{x + 1}{x^2 - 4x}.
SolutionThe only checkpoint is the denominator. Factor it to find where it is zero:
A product is zero only when one of its factors is zero (the zero-product property), so the denominator is zero at x = 0 and at x = 4. Remove those two numbers and keep everything else:
In words: all real numbers except 0 and 4. Figure 2 shows the line with two holes. The set falls into three intervals, so it takes two unions to write.
The numerator restricts nothing. At x = -1 the numerator is 0, and g(-1) = \dfrac{0}{5} = 0 is a perfectly good output. Zero on top is allowed; zero on the bottom is not. This is the search for excluded values of a rational expression, now reported as a domain.
A root over a denominator
Find the implied domain of \displaystyle h(x) = \dfrac{\sqrt{x + 3}}{x - 5}.
SolutionThere are two checkpoints, and an input must pass both.
The square root needs x + 3 \ge 0, so x \ge -3. The denominator needs x - 5 \ne 0, so x \ne 5.
Together: every number from -3 upward, except 5. Because 5 lies in the stretch the root allows, removing it splits that stretch in two, as Figure 1 shows:
In words: all real numbers greater than or equal to -3, except 5. The endpoint -3 is included, since h(-3) = \dfrac{0}{-8} = 0.
A root in the denominator
Find the implied domain of \displaystyle k(x) = \dfrac{1}{\sqrt{x - 1}}, and compare it with the domain of \displaystyle \sqrt{x - 1}.
SolutionThe root needs x - 1 \ge 0. The root is also the whole denominator, so it must not be 0, which rules out x - 1 = 0. Together, the inside must be strictly positive:
The domain of k is (1, \infty): all real numbers greater than 1. For \displaystyle \sqrt{x - 1} alone the domain is [1, \infty), which includes 1 because \displaystyle \sqrt{0} = 0 is a fine output. Moving the root into a denominator costs exactly one input, the one where the root is 0, and the bracket becomes a parenthesis.
Check. \displaystyle k(5) = \dfrac{1}{\sqrt{4}} = \dfrac{1}{2}, but k(1) would be \dfrac{1}{0}, which has no value.
When nothing is excluded
Find the implied domain of \displaystyle p(x) = \dfrac{\sqrt[3]{x - 6}}{x^2 + 9}.
SolutionThe cube root is an odd root. Every real number has a real cube root, for example \displaystyle \sqrt[3]{-8} = -2, so it adds no condition.
The denominator could only cause trouble where x^2 + 9 = 0. But x^2 \ge 0 for every real x, so x^2 + 9 \ge 9 and it is never zero.
No checkpoint turns anything away. The domain is (-\infty, \infty), all real numbers. A fraction or a root in a formula is a reason to look, not a promise that something is removed.
A quadratic under the root
Find the implied domain of \displaystyle m(x) = \sqrt{15 + 2x - x^2}.
SolutionThe inside must be at least 0. Multiply both sides by -1 so the x^2 term is positive, which reverses the inequality, then factor:
The product is zero at x = -3 and x = 5. Between them the two factors have opposite signs, so the product is negative; outside them the factors share a sign and the product is positive. This is a quadratic inequality, settled by the signs of its factors.
So -3 \le x \le 5, and the domain is [-3, 5]: all real numbers from -3 to 5, both included.
Check. \displaystyle m(1) = \sqrt{16} = 4. At x = 6 the inside is 15 + 12 - 36 = -9, so m(6) does not exist, as expected.
A weather balloon: context narrows the domain
Ursicinus uses a simple made-up model, not real physics, for a weather balloon’s radius, in meters, at a height of x kilometers:
The balloon is released from the ground and bursts when its radius reaches 3 meters. Find the implied domain of r, then the domain that makes sense for the balloon.
SolutionImplied domain. The root is in the denominator, so its inside must be strictly positive: 36 - x > 0, which gives x < 36. The implied domain is (-\infty, 36).
Context domain. Heights start at the ground, so x \ge 0; at launch the radius is r(0) = \dfrac{6}{6} = 1 meter. The balloon bursts when r(x) = 3:
As x grows, 36 - x shrinks, so the denominator shrinks and r grows. The radius stays at most 3 exactly while x \le 32, so the domain for the balloon is [0, 32].
Figure 3 shows the context domain sitting inside the implied one. The formula happily gives r(-28) = \dfrac{6}{8} = 0.75 and r(35) = 6, but there is no balloon underground and none after it bursts. The implied domain says where the formula gives a number; the context says which of those numbers mean something.
Boundaries
Real numbers only. The implied domain here assumes real inputs and real outputs. In the Complex numbers, \displaystyle \sqrt{-4} = 2i exists, and the even-root condition disappears.
Read the formula as written, before simplifying. Canceling a factor or squaring a root can hide a restriction: \displaystyle \left(\sqrt{x}\right)^2 equals x only for x \ge 0, and a canceled factor still excludes its zero.
Two breaking points in this course, more later. Logarithms bring a third checkpoint: a logarithm needs a positive input. Each new kind of operation brings its own condition.
A stated domain can shrink the implied domain, never extend it. Writing \displaystyle f(x) = \sqrt{x} for -1 \le x \le 4 does not create an output at -1; that stated domain does not describe a function.
The implied domain is a fact about the formula, not the situation. It can contain inputs that mean nothing, as in Example 7, and the person building the model has to cut them out.
Where this goes next
Excluded inputs are where graphs do interesting things. Near an input removed by a denominator, a graph often shoots up or down along a vertical asymptote, and finding the domain of a rational function is the first step in graphing one. Where an even root’s condition begins, the graph has an endpoint, the starting corner of a square root graph. When one function is divided by another, the domain of the quotient gathers the conditions of both parts plus a new denominator. And when you solve a rational equation, any answer outside the domain is thrown away, which is why the domain comes first.
Exercises
One restriction at a time
Find the implied domain of each function, in interval notation.
- f(x) = 4x^3 - x + 7
- f(x) = \dfrac{5}{x + 8}
- \displaystyle g(x) = \sqrt{3x - 12}
- \displaystyle g(x) = \sqrt{6 - 4x}
- h(x) = \dfrac{x - 2}{x^2 - 25}
- \displaystyle h(x) = \sqrt[3]{2x + 9}
- \displaystyle k(x) = \sqrt[4]{x + 10}
- k(x) = \dfrac{3x}{x^2 + 1}
Combining restrictions
Find the implied domain of each function, in interval notation. For Exercises 9 and 13, also describe it in words.
- \displaystyle f(x) = \dfrac{\sqrt{x - 2}}{x - 7}
- \displaystyle f(x) = \dfrac{\sqrt{x + 6}}{x + 1}
- \displaystyle g(x) = \dfrac{x - 3}{\sqrt{x + 4}}
- \displaystyle g(x) = \dfrac{2}{\sqrt{5 - x}}
- \displaystyle h(x) = \sqrt{x} + \sqrt{8 - x}
- \displaystyle h(x) = \dfrac{\sqrt{x - 3}}{x + 2}
- \displaystyle k(x) = \dfrac{1}{x} + \sqrt{x + 1}
Quadratics under a root
Find the implied domain of each function, in interval notation.
- \displaystyle f(x) = \sqrt{x^2 - 6x}
- \displaystyle g(x) = \dfrac{1}{\sqrt{16 - x^2}}
Stated and context domains
- Zdzislaw’s weather station buys balloons for a season. The supplier charges a delivery fee of 150 dollars plus 12 dollars a balloon, so the average cost per balloon, in dollars, when the station buys n balloons is A(n) = \dfrac{150 + 12n}{n}. Find the implied domain of A, the domain that makes sense here, and A(10).
- Agrippina defines \displaystyle f(x) = \sqrt{x - 2} for 0 \le x \le 10. Explain why this stated domain does not work, and find the largest stated domain inside 0 \le x \le 10 that does.
- Vincenza defines f(x) = \dfrac{x + 5}{x - 1} for x \ge 3. What is the domain of f? Is 2 an input of f? Find f(3).
Reasoning
- Explain why \displaystyle \sqrt[3]{x} accepts every real number but \displaystyle \sqrt[4]{x} accepts only [0, \infty).
- Write a function whose implied domain is [-1, 3) \cup (3, \infty), and explain why it works.
- Tertia says \dfrac{x^2 - 16}{x - 4} and x + 4 are the same function. Is she right? Explain.
- Can an implied domain contain no numbers at all? Give an example or explain why not.
Answers
- (-\infty, \infty). A polynomial has no denominator and no root, so every real number works.
- (-\infty, -8) \cup (-8, \infty). The denominator x + 8 is zero only at x = -8.
- [4, \infty). 3x - 12 \ge 0 gives 3x \ge 12, so x \ge 4.
- \left(-\infty, \tfrac{3}{2}\right]. 6 - 4x \ge 0 gives -4x \ge -6; dividing by -4 reverses the inequality, so x \le \tfrac{3}{2}.
- (-\infty, -5) \cup (-5, 5) \cup (5, \infty). x^2 - 25 = (x - 5)(x + 5) is zero at 5 and -5. The numerator’s zero, 2, is allowed.
- (-\infty, \infty). A cube root accepts every real number.
- [-10, \infty). A fourth root is an even root, so x + 10 \ge 0.
- (-\infty, \infty). x^2 + 1 \ge 1, so the denominator is never zero.
- [2, 7) \cup (7, \infty): all real numbers greater than or equal to 2, except 7. The root needs x \ge 2 and the denominator needs x \ne 7.
- [-6, -1) \cup (-1, \infty). The root needs x \ge -6, and -1 lies in that stretch, so removing it splits the stretch in two.
- (-4, \infty). The root is the denominator, so x + 4 > 0.
- (-\infty, 5). The root is the denominator, so 5 - x > 0, which gives x < 5.
- [0, 8]: all real numbers from 0 to 8, both included. The first root needs x \ge 0 and the second needs 8 - x \ge 0, that is x \le 8.
- [3, \infty). The root needs x \ge 3. The denominator rules out -2, but -2 already fails the root, so nothing more is removed.
- [-1, 0) \cup (0, \infty). The root needs x \ge -1, and \displaystyle \frac{1}{x} needs x \ne 0.
- (-\infty, 0] \cup [6, \infty). The inside is x(x - 6) \ge 0. The product is zero at 0 and 6, and positive outside them, where both factors share a sign.
- (-4, 4). The root is in the denominator, so 16 - x^2 > 0, that is x^2 < 16, which holds exactly when -4 < x < 4.
- The implied domain is (-\infty, 0) \cup (0, \infty), since n = 0 makes the denominator zero. In context n counts balloons bought, so the sensible domain is the positive integers 1, 2, 3, \dots. A(10) = \dfrac{150 + 120}{10} = 27, so the average is 27 dollars a balloon.
- For 0 \le x < 2 the inside x - 2 is negative, so f has no real output there, and those inputs cannot be in its domain. The implied domain is [2, \infty), and its part inside 0 \le x \le 10 is [2, 10], so the stated domain 2 \le x \le 10 works.
- The domain is [3, \infty). The implied domain is every number except 1, and x \ge 3 lies inside it, so the stated domain stands. 2 is not an input of f, even though the formula would give \displaystyle \frac{7}{1} = 7 there. \displaystyle f(3) = \frac{8}{2} = 4.
- An odd power keeps the sign, (-2)^3 = -8, so every real number has a real cube root. A fourth power is never negative, r^4 = (r^2)^2 \ge 0, so a negative number has no real fourth root, and \displaystyle \sqrt[4]{x} needs x \ge 0.
- One answer: \displaystyle f(x) = \dfrac{\sqrt{x + 1}}{x - 3}. The root gives x \ge -1, which puts the bracket at -1, and the denominator removes 3, which lies inside that stretch.
- No. Wherever both are defined they agree, since \displaystyle \frac{(x - 4)(x + 4)}{x - 4} = x + 4 for x \ne 4. But at x = 4 the fraction is \displaystyle \frac{0}{0} and has no value, while x + 4 gives 8. Their domains differ, so they are different functions.
- Yes. In \displaystyle f(x) = \sqrt{-1 - x^2} the inside is -1 - x^2 \le -1 for every real x, so it is always negative. No input works, and the implied domain is empty: the formula gives no output for any real number.