Ignoring a domain restriction means giving an answer that is correct algebra but wrong for some of the inputs. It is a common reason a function question loses marks in Additional Mathematics, because the restriction is easy to lose while the expression is being simplified.
The fix is a short routine: write the restrictions first, carry them through, and state them at the end.
Why do restrictions disappear?
Most students learn functions as rules to apply: substitute, expand, simplify. Nothing in that process forces you to ask which inputs are allowed.
Simplifying makes it worse.
A square root squared, or a fraction that cancels, looks like a clean expression that applies everywhere. It does not. The original function decides the domain, and the simplified form does not erase it.
What should I check first?
Before you start any working, scan the function for the usual limits.
- A denominator: it cannot equal zero. Solve “denominator = 0” and exclude those values.
- A square root: the expression inside must be zero or positive. Solve “inside ≥ 0”.
- A stated domain: if the question gives one, such as x ≥ 0, it always applies.
Write these limits as a short first line of your working. That line is what you carry to the end.
Worked example: a composite function
Let f(x) = √(x − 1) and g(x) = x². Find fg(x) and state its domain. (Here fg(x) means f(g(x)).)
Step 1. Put g(x) inside f: fg(x) = √(x² − 1).
Step 2. The expression under the root must be zero or positive: x² − 1 ≥ 0.
Step 3. Factorise: (x − 1)(x + 1) ≥ 0. The critical values are x = −1 and x = 1. The product is positive outside them and zero at them.
Step 4. The domain is x ≤ −1 or x ≥ 1.
Check: x = 0 gives √(−1), which is undefined, so 0 is correctly excluded. x = 2 gives √3 and x = −2 gives √3, and both are allowed.
What is the plausible mistake?
A student reasons: “g(x) = x² accepts every real number, so fg accepts every real number too.” They write “domain: all real x”.
It goes wrong because the domain of a composite needs the outer function to accept what the inner one produces. g can output 0, but f(0) = √(−1) does not exist. The fix is to substitute the inner function, then test the outer function’s restriction on the whole expression, as in steps 2 to 4.
The reverse order shows the same trap. gf(x) = (√(x − 1))² simplifies to x − 1, yet it is only valid for x ≥ 1, because f cannot take anything smaller. The plain look of x − 1 hides the restriction.
Worked example: an inverse needs a domain
h(x) = x² − 4 with domain x ≥ 0. Find h⁻¹(x) and its domain.
Write y = x² − 4, so x² = y + 4 and x = √(y + 4). We take the positive root because the original domain is x ≥ 0. So h⁻¹(x) = √(x + 4).
The domain of h⁻¹ is the range of h. Since h(0) = −4 and h increases for x ≥ 0, the range is h(x) ≥ −4. So the domain of h⁻¹ is x ≥ −4.
Check: h(3) = 9 − 4 = 5 and h⁻¹(5) = √9 = 3. The two functions undo each other.
Self-check
- State the largest possible domain of f(x) = (x + 5)/(x² − 9).
- State the largest possible domain of g(x) = √(6 − 2x).
- With f(x) = √x and g(x) = x − 4, find fg(x) and its domain.
Show answer
- The denominator is zero when x² = 9, so x = 3 or x = −3. Domain: all real x except x = 3 and x = −3.
- We need 6 − 2x ≥ 0, so x ≤ 3.
- fg(x) = √(x − 4). We need x − 4 ≥ 0, so the domain is x ≥ 4.
How do I build the habit?
Use the same three-line opening every time: list the restrictions, do the algebra, compare the answer with the restrictions. You can test your own functions in the function composition and inverse explorer, and practise exact working on the arithmetic side.
The full set of lessons sits in functions and restrictions. Start with finding valid inputs and the lesson on composite functions in the correct order. Terms such as domain, range and many-to-one are explained in the Additional Mathematics terminology guide.
If this keeps happening even with the routine, online one-to-one Additional Mathematics tuition lets a teacher see exactly where your working drops the restriction. Check which topics your exam year covers on the Cambridge subject page linked below.