Domain and Range of Square Root Functions

A student stares at f(x) = √(3x − 12) and types x ≥ 0 into the answer box. The real domain of square root function starts with one inequality: set the radicand ≥ 0.Domain is [4, ∞). Every square root with an even index (square root, fourth root, sixth root) follows this rule. OpenStax Precalculus 2e Section 3.2 defines domain as "the set of possible input values." For square roots, that set is restricted by the radicand.

The failure case: a student forgets the inequality entirely and writes "all real numbers." That answer gets zero credit. Another student writes the inequality but solves it backwards, giving x ≤ 4. Check your work by plugging in a test value. For x = 3, 3(3)−12 = −3, and √(−3) is not a real number. The calculator confirms the domain step by step, showing the inequality before the interval.

Range follows from the domain and the direction of the function. For the parent function f(x) = √x with domain [0, ∞), the smallest output is √0 = 0. As x increases, the principal root increases without bound. Range is [0, ∞). Every square root function has a range that starts at the function's minimum y-value and goes to ∞ or, ∞, depending on the sign in front of the radical.

For f(x) =, √x, the range becomes (, ∞, 0]. The negative sign flips the outputs downward. The starting point is still at x = 0, but now y = 0 is the maximum, not the minimum. Paul's Online Math Notes Algebra section defines range as "the set of all values that the function assumes." For square roots, the range is always bounded on one side by the principal root value.

Set The Radicand Restriction

The general radical function is f(x) = a√[n]{b(x, h)} + k, where n is the index. For even n (n = 2, 4, 6...): set b(x, h) ≥ 0. Solve that inequality for the domain. The range depends on a. If a > 0, range is [k, ∞). If a < 0, range is (, ∞, k]. The point (h, k) is the starting point of the graph, also called the "origin" of the transformed radical.

Work Through An Example

For f(x) = 2√(x, 3) + 1: domain from x, 3 ≥ 0 → x ≥ 3. Range starts at y = 1 (when x = 3, 2√0 + 1 = 1) and increases. Range is [1, ∞). The amplitude a = 2 stretches the outputs vertically, but the range still starts at k = 1.

Rational Functions Follow A Different Rule

Domain and range of rational functions follow a separate logic: domain excludes denominator zeros, and range may exclude the horizontal asymptote value but does not always. That topic needs its own treatment. For radical functions, the rule stays on radicand restrictions.

When a square root appears in a denominator, the radicand must be strictly greater than 0, not ≥ 0. A denominator of zero makes the function undefined. So for f(x) = 1/√(x, 5), solve x, 5 > 0 → x > 5. Domain is (5, ∞). The endpoint x = 5 is excluded because √0 = 0, and division by zero is not allowed.

Students commonly write x ≥ 5 for this function. That is wrong. Test x = 5: √(5-5) = 0, so f(5) = 1/0, undefined. The domain is (5, ∞), and x = 5 is an excluded value.

Cube roots have an odd index. Odd-index roots (cube root, fifth root, seventh root) allow any real radicand. Negative radicands produce negative outputs. So the domain of cube root functions is all real numbers, (, ∞, ∞). The range is also all real numbers.

For f(x) = ∛(x + 2), any x works: x =, 3 gives ∛(, 1) =, 1; x =, 2 gives ∛0 = 0; x = 1 gives ∛3 ≈ 1.44. No inequality to solve. Students who memorize "radicand ≥ 0" for all radicals make errors here. Cube root domain is unrestricted.

Example 1: Standard Square Root

f(x) = √(2x + 8). Domain: 2x + 8 ≥ 0 → 2x ≥ -8 → x ≥ -4.Range: a = 1 > 0, so range starts at f(-4) = √0 = 0 and increases. Range: [0, ∞).

Example 2: Square Root With Negative Coefficient

g(x) = -3√(x-1) + 2.Range: a = -3 < 0, so range goes downward from k = 2. At x = 1, g(1) = -3√0 + 2 = 2. Range: (-∞, 2].

Example 3: Square Root in Denominator

h(x) = 5/√(4-x). Domain: 4-x > 0 → -x > -4 → x < 4. Domain: (-∞, 4). Range: as x approaches 4 from the left, h(x) → +∞; as x → -∞, h(x) → 0. Range: (0, ∞).

Example 4: Cube Root Function

j(x) = ∛(3x-9)-1. Domain: all real numbers, (-∞, ∞). Range: all real numbers, (-∞, ∞). No inequality. The cube root function's range is unbounded in both directions.

Find the domain and range for each function. Write answers in interval notation. Check against the solutions below.

1. f(x) = √(5x, 15)
2. g(x) =, √(x + 4)-3
3. h(x) = 2/√(6-2x)
4. j(x) = ∛(4x + 12) + 5
5. k(x) = √(x² + 1) (hint: x² + 1 is always positive)

Solutions: 1) Domain [3, ∞), range [0, ∞). 2) Domain [−4, ∞), range [−3, ∞). 3) Domain (−∞, 3), range (0, ∞). 4) Domain (−∞, ∞), range (−∞, ∞). 5) Domain (−∞, ∞), range [1, ∞). The fifth example shows that a radicand of x² + 1 is always ≥ 1, so no restriction on x.

Function Type	Domain Rule	Range Rule	Example Domain	Example Range
Square Root, f(x)=√(ax+b)	ax+b ≥ 0	Starts at principal root, goes to ∞	[–b/a, ∞) for a>0	[0, ∞) for a>0
Negative Square Root, f(x)=–√(ax+b)+k	ax+b ≥ 0	Starts at k, goes to –∞	[–b/a, ∞) for a>0	(–∞, k] for a<0
Square Root in Denominator, f(x)=1/√(ax+b)	ax+b > 0 (strict)	Positive outputs approach 0 and ∞	(–b/a, ∞) for a>0	(0, ∞)
Cube Root, f(x)=∛(ax+b)	All real numbers	All real numbers	(–∞, ∞)	(–∞, ∞)

Students treat every square root the same: set radicand ≥ 0. For a square root in a denominator, that gives the wrong domain. The strict > 0 rule is the single most common error. Check your function for a denominator before writing the inequality. If the radicand is in a denominator, use > 0, not ≥ 0. The calculator flags this distinction automatically.