Functions • Topic 1 of 4

Domain & Range

The domain is the set of inputs the rule is allowed to accept; the range is the set of outputs it actually produces. CAT domain questions reduce to a short checklist: a square root forces the inside to be ≥ 0, a denominator forces it to be ≠ 0, and a logarithm forces its argument to be strictly > 0. Combine the conditions with AND — the domain is their intersection. Range is the harder half. The fast methods are: complete the square for a quadratic (a parabola has a clean minimum or maximum), bound expressions using the fact that a perfect square is ≥ 0, or solve y = f(x) for x and ask which y keep x real. For f(x) = 1/(x − a), every real except one value of y is reachable. Never plot points blindly — reason about what the rule can and cannot output.

✅ Solved examples

1. Find the domain of f(x) = √(x − 3) + 1/(x − 7).
Need x − 3 ≥ 0 ⇒ x ≥ 3, and x − 7 ≠ 0 ⇒ x ≠ 7. Domain = [3, 7) ∪ (7, ∞).
2. Find the range of f(x) = x² − 4x + 9.
Complete the square: (x − 2)² + 5. Minimum is 5 at x = 2, and it grows without bound. Range = [5, ∞).
3. Find the domain of f(x) = log(x² − 5x + 6).
Argument > 0: x² − 5x + 6 = (x − 2)(x − 3) > 0 ⇒ x < 2 or x > 3. Domain = (−∞, 2) ∪ (3, ∞).
4. Find the range of f(x) = (2x + 1)/(x − 3).
Solve for x: y(x − 3) = 2x + 1 ⇒ x(y − 2) = 3y + 1 ⇒ x = (3y + 1)/(y − 2). Real x exists for every y except y = 2. Range = ℝ − {2}.

✏️ Practice — try these, take hints as needed

1. Domain of f(x) = √(4 − x²).
Inside the root must be ≥ 0.
4 − x² ≥ 0 ⇒ x² ≤ 4.
So −2 ≤ x ≤ 2.
[−2, 2]
2. Range of f(x) = 7 − (x − 1)².
A square is always ≥ 0.
(x − 1)² ≥ 0, so subtract it from 7.
Maximum is 7, decreasing without bound.
(−∞, 7]
3. Domain of f(x) = 1/√(x − 2).
Root needs ≥ 0 and denominator needs ≠ 0.
Together: x − 2 > 0 (strict).
So x > 2.
(2, ∞)
4. Range of f(x) = x/(x² + 1) for real x.
Set y = x/(x² + 1) and clear: yx² − x + y = 0.
For real x the discriminant 1 − 4y² ≥ 0.
So y² ≤ 1/4.
[−1/2, 1/2]
5. Domain of f(x) = √(x − 1)/(x − 4).
Numerator root: x − 1 ≥ 0.
Denominator: x − 4 ≠ 0.
Intersect x ≥ 1 with x ≠ 4.
[1, 4) ∪ (4, ∞)

📝 Topic test — 8 questions

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