Find the Domain: Rational, Cube Root & Square Root Functions (Stewart Ch.1 Q3)

Direct answer. (a) Domain of f(x) = (2x + 1) / (x² + x − 2) is (−∞, −2) ∪ (−2, 1) ∪ (1, ∞). (b) Domain of g(x) = ³√x / (x² + 1) is (−∞, ∞). (c) Domain of h(x) = √(4 − x) + √(x² − 1) is (−∞, −1] ∪ [1, 4].

The three rules every domain problem reduces to

Almost every "find the domain" question in Stewart Chapter 1 is built from exactly three building blocks. Memorize the three rules and the rest is bookkeeping.

1. Denominators cannot be zero. Any fraction, no matter how simple or how disguised, requires you to find the zeros of the denominator and exclude them from the domain.
2. Even-index radicals require a nonnegative radicand. Square roots, fourth roots, sixth roots — any radical with an even index — needs the expression underneath to be greater than or equal to zero, because even roots of negatives are not real.
3. Odd-index radicals are unrestricted. Cube roots, fifth roots, and any radical with an odd index accept every real number. They never restrict the domain on their own.

When a function combines several of these, you find the domain of each piece and intersect — the function is defined exactly where every condition holds.

(a) f(x) = (2x + 1) / (x² + x − 2)

This is a rational function: a polynomial over a polynomial. Polynomials are defined for every real number, so the numerator never restricts the domain. The only restriction comes from the denominator, which is not allowed to be zero.

Set the denominator equal to zero:

x² + x − 2 = 0

Factor:

(x + 2)(x − 1) = 0

So x = −2 or x = 1. These two values are excluded from the domain. Every other real number is allowed.

Domain: (−∞, −2) ∪ (−2, 1) ∪ (1, ∞)

Each excluded value chops the real line into a new piece. Two exclusions means three pieces, joined with the union symbol. The parentheses (never brackets) at −2 and 1 are non-negotiable: the function blows up at those points.

(b) g(x) = ∛x / (x² + 1)

Two pieces to check: the cube root in the numerator, and the polynomial in the denominator.

Cube root in the numerator. Unlike the square root, the cube root accepts every real number. ³√−8 = −2, ³√0 = 0, ³√8 = 2. So this piece introduces no restriction.

Polynomial in the denominator. We need x² + 1 ≠ 0. But x² ≥ 0 for every real x, so x² + 1 ≥ 1 > 0 always. The denominator is never zero. No restriction from this piece either.

Domain: (−∞, ∞)

Stewart includes this part as a trap. Students who see a radical and a fraction default to writing complicated domain restrictions, when in fact this function is defined everywhere. The discipline is to check each rule honestly rather than pattern-match on the shape of the formula.

(c) h(x) = √(4 − x) + √(x² − 1)

This is the part that separates students who learned the rules from students who memorized examples. Two square roots are added together. Both have to be real for the sum to be real, so both radicands must be nonnegative simultaneously.

First square root. Require 4 − x ≥ 0, which gives x ≤ 4.

Second square root. Require x² − 1 ≥ 0, which gives x² ≥ 1. Taking square roots (and tracking the absolute value) gives |x| ≥ 1, which splits into two cases: x ≤ −1 or x ≥ 1.

Intersect. Combine x ≤ 4 with (x ≤ −1 or x ≥ 1):

• x ≤ 4 AND x ≤ −1 simplifies to x ≤ −1 (the stricter condition wins).
• x ≤ 4 AND x ≥ 1 simplifies to 1 ≤ x ≤ 4.

Union the two cases:

Domain: (−∞, −1] ∪ [1, 4]

Brackets at −1, 1, and 4 because the function is defined at all three (the radicands are zero there, and √0 = 0 is a real number). Parenthesis at −∞ because infinity is never an attainable value.

The three mistakes worth memorizing

• Forgetting that cube roots accept negatives. Students who learned the square-root rule first reflexively write x ≥ 0 for any radical. Cube root, fifth root, and every odd-index root are defined for all reals — no restriction. The rule lives in the index, not in the radical symbol.
• Unioning two conditions instead of intersecting. When a function adds two radicals, the domain is the intersection of where each is real, not the union. Writing the domain of part (c) as (−∞, 4] ∪ (−∞, −1] ∪ [1, ∞) "covers all cases" but admits values like x = 5, where √(4 − 5) = √−1 is not real.
• Bracket-vs-parenthesis at zero radicands. When a square root's radicand equals zero, the radical equals zero, which is a real number. The endpoint is included, so the bracket is square. Students who write (−∞, 4) instead of (−∞, 4] for √(4 − x) are excluding a perfectly legal input.

How SOMATH teaches this problem

We teach Stewart Ch.1 Q3 as a three-station drill: rational, cube root, double square root. Same picture every time — a function with one or more pieces, and a flowchart that says "denominator? exclude zeros. even radical? require ≥ 0. odd radical? skip." Students cycle through five functions per station, with progressively trickier denominators and radicands, until each rule is automatic. By the end of the lesson a student can find the domain of any Chapter 1 function in under a minute.

Our small-group classroom at 226 W 79th St (Upper West Side, between Broadway and Amsterdam) keeps groups of three to five students at whiteboards on every wall. Each rule gets the why, the how, and three increasingly hostile examples — including ones designed to trick students into pattern-matching. This problem is the gate every pre-calc and AP Calc student walks through at SOMATH before they touch limits.

FAQ

What is the domain of a function?

The domain is the set of all input values (typically x) for which the function produces a real output. For algebraic functions in Chapter 1 of Stewart, finding the domain is equivalent to spotting every operation that has a restriction (division by zero, even roots of negatives, log of nonpositives later in the textbook) and excluding the inputs that would violate those restrictions.

Why does the cube root not restrict the domain when the square root does?

Because the cube root has an odd index, and odd roots of negative numbers are well-defined real numbers. ³√−8 = −2 because (−2)³ = −8. The square root has an even index, and there is no real number whose square is negative, so the square root of a negative input is not real. The same logic applies to every even root versus every odd root.

How do I know when to write a bracket versus a parenthesis?

Square bracket if the endpoint is in the domain — that is, the function is defined at exactly that input. Parenthesis if the endpoint is excluded, or if the endpoint is ∞ or −∞ (which are not real numbers and cannot be attained). For Stewart Ch.1 Q3 part (c), the radicands are zero at x = −1, x = 1, and x = 4, and √0 = 0 is real, so those endpoints get brackets.

Will domain questions show up on the AP Calculus exam?

Yes, directly and indirectly. AP Calc multiple-choice often asks for the domain of a function explicitly, but more often the domain appears as a setup step in a longer problem — defining the interval where a derivative is taken, where the Mean Value Theorem applies, or where a definite integral is being evaluated. A wrong domain at the start propagates into a wrong answer five steps later. Fluent domain work is one of the highest-leverage skills in the entire course.

What is the next Stewart problem after this one?

Stewart Ch.1 Q4 asks you to find the domain and sketch the graph for a piecewise-defined function. Same three rules, applied to each piece separately, and then the pieces are joined. We have walkthroughs of the surrounding problems linked below.

Related posts

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• Stewart Ch.1 Q1: Reading a Function from Its Graph (f(−1), f(2), Zeros, Domain & Range)
• Stewart Q4: Two Points — Line, Midpoint, and Circle with AB as Diameter
• Functions and Their Domains in AP Pre-Calculus
• AP Pre-Calc Rational Function: Where Is h(x) > 0?