Domain and Range
Every function has two fundamental properties that describe what it can and cannot do: its domain and its range. The domain tells you which inputs the function will accept. The range tells you which outputs it can produce. On the CSCA exam, domain questions are among the most frequently tested — you will see them on almost every paper, and they appear in several different forms.
The core skill is this: certain mathematical operations place restrictions on which values of x are allowed. A fraction cannot have a denominator of zero. A square root cannot take a negative input. A logarithm requires a strictly positive input. When a function combines several of these operations at once, you must find all the restrictions, solve each one, and then take the intersection — the set of x values that satisfy every restriction simultaneously.
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What domain and range mean
Think of a function as a machine. You put a number in, the machine processes it, and a number comes out. The domain is the set of all numbers you are allowed to put in. The range is the set of all numbers that can possibly come out.
Range: the complete set of output values (y values) that the function can actually produce.
When a function is written as y = f(x), the domain lives on the x-axis and the range lives on the y-axis.
For many simple functions, the domain is all real numbers — you can plug in anything. For example, f(x) = 2x + 1 works for any x. But as soon as a function involves a fraction, a square root, or a logarithm, certain values of x become forbidden, and you must identify exactly which ones.
The correct answer is B. x² is always ≥ 0 for any real x — squaring a number never gives a negative result. The smallest value x² can take is 0 (when x = 0). Adding 1 shifts everything up by 1, so the smallest possible output is 0 + 1 = 1. The function can produce any value ≥ 1, but nothing below 1. Range: [1, +∞).
Option A gives all real numbers, which would be the range of a linear function — x² + 1 can never go below 1. Option C forgets the +1 shift — that would be the range of y = x². Option D points downward, which would require a negative leading coefficient like y = −x² + 1.
The main domain restrictions
Four types of expressions create most domain restrictions on the CSCA exam (a fifth, the zero power, comes later on this page). Each one has a specific rule, and you must apply all relevant rules at the same time when a function contains more than one of these expressions.
The method for finding a domain is always the same: identify every restriction, write each one as an inequality or exclusion, then find the intersection — the values of x that satisfy all restrictions at once.
Single restrictions: building the habit
Before combining restrictions, make sure you are confident with each type individually.
The argument of the logarithm must be strictly positive: x − 2 > 0. Solving: x > 2. Domain: (2, +∞).
Notice the round bracket — the logarithm requires strictly greater than zero, so x = 2 itself is excluded.
The expression under the root must be ≥ 0: 1 − x ≥ 0. Solving: x ≤ 1. Domain: (−∞, 1].
Notice the square bracket — the root allows the expression to equal exactly zero (when x = 1, the root equals zero, which is valid). So x = 1 is included.
The denominator must not equal zero: x ≠ 0. Domain: (−∞, 0) ∪ (0, +∞), often written as {x | x ≠ 0}.
A fraction usually makes a hole in the domain (one missing point), not a cut-off at one end.
The correct answer is D. The only restriction is the denominator: 1 − x ≠ 0, so x ≠ 1. Domain: (−∞, 1) ∪ (1, +∞).
Option B comes from cancelling first: (1 − x²)/(1 − x) = (1 − x)(1 + x)/(1 − x) = 1 + x, and 1 + x accepts every x. But the domain belongs to the function as it is written. At x = 1 the original fraction is 0/0, which is undefined. Always find the domain before you simplify. Option A copies the rule for 1/x and removes x = 0. Here the denominator is 1 − x, and at x = 0 it equals 1, which is fine. Option C removes x = −1, which makes the numerator zero. A zero numerator is allowed: at x = −1 the fraction is 0/2 = 0.
Multiple restrictions: the intersection method
The harder domain questions on the exam involve functions with two or more restriction types at once. The key principle is that every restriction must be satisfied simultaneously. You find the domain of each restriction separately, then take the intersection — the overlap — of all of them.
From √(1 − x): 1 − x ≥ 0 → x ≤ 1. This gives (−∞, 1].
The correct answer is A. Square root: x + 3 ≥ 0, so x ≥ −3. Fraction: x² − 4 ≠ 0. Since x² − 4 = (x − 2)(x + 2), this means x ≠ 2 and x ≠ −2. Both holes lie inside [−3, +∞), so both must be removed: [−3, −2) ∪ (−2, 2) ∪ (2, +∞).
Option B solves x² = 4 as x = 2 only and forgets x = −2. Option C uses only the fraction and forgets the square root, so it keeps values like x = −5, where √(−2) is not real. Option D uses a round bracket at −3. But at x = −3 the root is √0 = 0, which is allowed.
Logarithm and square root combined
One of the trickier domain questions on the exam (from the October 2025 paper) combines a logarithm and a square root in the same function. The logarithm and the square root each impose their own restriction, and both must be satisfied at once.
Restriction 1 — logarithm: ln x requires x > 0.
Restriction 2 — square root: √(1 − x) requires 1 − x ≥ 0, so x ≤ 1.
Intersection: we need x > 0 AND x ≤ 1 simultaneously. The overlap is 0 < x ≤ 1.
Domain: (0, 1]. Open bracket at 0 (logarithm requires strictly positive), closed bracket at 1 (square root allows equality).
Notice how the two boundary types differ: the logarithm forces a strict inequality (open bracket at 0), while the square root allows equality (closed bracket at 1). Getting both boundary types right is what distinguishes a correct answer from a close-but-wrong one on this question type.
The correct answer is C. Logarithm: x + 2 > 0, so x > −2 (open at −2). Square root: 3 − x ≥ 0, so x ≤ 3 (closed at 3). Domain: (−2, 3]. The integers in it are −1, 0, 1, 2, 3. That is 5 integers.
Option A (4) uses an open bracket at 3 as well, so it drops 3. But √(3 − 3) = √0 = 0 is allowed. Option B (6) uses a closed bracket at −2 as well, so it counts −2. But lg(0) is undefined. Option D uses only one of the two restrictions. Each one alone leaves an endless interval, but together they cut the domain down to (−2, 3].
The zero-power trap: a special domain case
One question type that appears on the exam involves an expression raised to the power of zero — something like (3x − 7)⁰. Any non-zero number raised to the power of zero equals 1, which seems harmless. But 0⁰ is undefined. This means the base of a zero-power expression must be excluded from the domain, even though the expression looks like it always equals 1.
Restriction 1 — square root: (x − 2)^(1/2) = √(x − 2) requires x − 2 ≥ 0 → x ≥ 2.
Restriction 2 — zero power: (3x − 7)⁰ requires 3x − 7 ≠ 0 → x ≠ 7/3.
Intersection: x ≥ 2 AND x ≠ 7/3. Since 7/3 ≈ 2.33 is within [2, +∞), we must exclude it. Domain: [2, 7/3) ∪ (7/3, +∞).
The correct answer is A. Square root: x ≥ 0. Zero power: x² − 4 ≠ 0, so x ≠ 2 and x ≠ −2. The value −2 is already outside [0, +∞), so only x = 2 makes a new hole. Domain: [0, 2) ∪ (2, +∞).
Option B misses the zero-power rule: at x = 2 the first term is 0⁰, which is undefined. Option C uses a round bracket at 0. But √0 = 0 is allowed, and (0 − 4)⁰ = 1, so x = 0 is in the domain. Option D uses only the zero-power rule and forgets the square root, so it keeps negative x, where √x is not real.
Fractional exponents and their combined restrictions
Expressions like (1 − 2x)^(−3/4) look intimidating but follow a clear logic once you break them apart. A negative exponent means there is a denominator — the expression is 1/(1 − 2x)^(3/4). A fractional exponent with an even denominator means there is a root — (1 − 2x)^(1/4) requires 1 − 2x ≥ 0. Combining: the base must be strictly positive (non-negative for the root, and non-zero for the denominator).
The bottom number of the exponent tells you which root it is: ^(1/2) is a square root, ^(1/3) a cube root, ^(3/4) a 4th root. Only an even root needs a base ≥ 0. An odd root accepts negative numbers, for example ∛(−8) = −2. So read two things from the exponent: is it negative (then base ≠ 0), and is the root even (then base ≥ 0)?
The correct answer is B. (x − 2)^(−1/2) = 1/√(x − 2). The square root needs x − 2 ≥ 0, and the denominator needs x − 2 ≠ 0. Together: x − 2 > 0, so x > 2. Domain: (2, +∞).
Option A is √(x − 2). It has no denominator, so x = 2 is allowed (√0 = 0). Its domain is [2, +∞), with a square bracket. Option C is 1/(x − 2). There is no root, so negative bases are fine and only x = 2 is removed: its domain is (−∞, 2) ∪ (2, +∞). Option D is ∛(x − 2), an odd root. It accepts every real number, so its domain is ℝ. Choosing D treats every fractional exponent as a square root.
Finding the domain of a composite expression
Some domain questions involve a logarithm whose argument contains a further restriction — for example, a logarithm applied to a fraction, or a square root under a logarithm. In these cases you must apply the outer restriction (logarithm requires > 0) to the inner expression, and also check whether the inner expression itself has restrictions.
Outer restriction — square root: the expression inside must be ≥ 0: 1 − log₀.₅(x) ≥ 0 → log₀.₅(x) ≤ 1.
Solving log₀.₅(x) ≤ 1: write 1 as a log: 1 = log₀.₅(0.5). The base 0.5 is less than 1, so log₀.₅ is decreasing: a smaller log value comes from a larger x. So log₀.₅(x) ≤ log₀.₅(0.5) means x ≥ 0.5. Thus x ≥ ½.
Inner restriction — logarithm: log₀.₅(x) requires x > 0. Every x ≥ ½ already satisfies this, so it adds nothing new.
Domain: [½, +∞). There is no upper limit: as x grows, log₀.₅(x) becomes more and more negative, so 1 − log₀.₅(x) stays positive.
Your turn!
Practice questions
The correct answer is D. Logarithm: x − 1 > 0, so x > 1. Fraction: x − 3 ≠ 0, so x ≠ 3. The hole at 3 lies inside (1, +∞), so the domain is (1, 3) ∪ (3, +∞).
Option A uses ≥ for the logarithm and keeps x = 1. But ln(0) is undefined. Option B forgets the denominator: at x = 3 we would divide by zero. Option C forgets the logarithm and keeps values like x = 0, where ln(−1) is not real.
The correct answer is C. Three restrictions. Square root: x + 1 ≥ 0, so x ≥ −1. Zero power: x − 2 ≠ 0, so x ≠ 2. Logarithm: 4 − x > 0, so x < 4. Intersect: from −1 (included) up to 4 (not included), with a hole at 2. Domain: [−1, 2) ∪ (2, 4).
Option A misses the zero-power rule: at x = 2 the middle term is 0⁰, which is undefined. Option B uses ≥ for the logarithm and keeps x = 4. But lg(0) is undefined. Option D uses a round bracket at −1. But √0 = 0 is allowed, so x = −1 is in the domain.
Things to remember
| Concept | What to remember |
|---|---|
| Domain | The set of all x values for which the function produces a valid real output. Find it by identifying and solving all restrictions. |
| Range | The set of all possible output values. For y = x² + 1, the minimum output is 1 (when x = 0), so range is [1, +∞). |
| Fraction restriction | Denominator ≠ 0. Produces a hole in the domain at the forbidden value. |
| Square root restriction | Expression under root ≥ 0. Closed bracket at the boundary (equality allowed). |
| Logarithm restriction | Argument > 0 (strictly). Open bracket at the boundary (equality not allowed). |
| Zero-power restriction | Base of a zero-power expression ≠ 0. Even though A⁰ = 1, we need A ≠ 0 because 0⁰ is undefined. |
| Negative fractional exponent | Combines a root restriction (base ≥ 0) with a denominator restriction (base ≠ 0). Net result: base must be strictly > 0. |
| Multiple restrictions | Apply every restriction independently, then intersect. All conditions must be satisfied simultaneously. Never union — always intersect. |
| Log base < 1 | Decreasing function — inequality direction flips when converting logₐ(x) ≤ k to x ≥ aᵏ. |