The valid inputs of a rational expression are every value of x except those that make the denominator equal to zero. In Additional Mathematics this appears whenever a question gives a function such as f(x) = (3x + 1)/(x² − x − 6) and asks for its domain, or asks you to explain why a value is not allowed.
It is the first skill in functions and restrictions, and it comes back in composite and inverse functions, where a restriction from one step carries into the next.
How do you find the excluded values?
A fraction is undefined when its denominator is zero. So the method is short: set the denominator equal to zero, solve, and exclude those solutions.
- Write the denominator on its own.
- Set it equal to zero and solve. Factorise if it is quadratic.
- State each solution as x ≠ a. Do not list values from the numerator.
- Write the domain in words or symbols, for example “all real x except x = 3 and x = −2”.
If the denominator can never be zero, such as x² + 9, then every real number is valid. You should say so, not leave the answer blank.
Worked example
State the largest possible domain of f(x) = (3x + 1)/(x² − x − 6).
Step 1, denominator: x² − x − 6.
Step 2, set to zero and factorise: x² − x − 6 = 0 gives (x − 3)(x + 2) = 0.
Step 3, solve: x = 3 or x = −2.
Step 4, check both: at x = 3, 9 − 3 − 6 = 0. At x = −2, 4 + 2 − 6 = 0. Both make the denominator zero.
Answer: the domain is all real x with x ≠ 3 and x ≠ −2.
Notice that the numerator 3x + 1 played no part. It would be zero at x = −1/3, which is a perfectly valid input.
The mistake to watch for
A common slip is to simplify first and read the restriction from the simplified form. Take g(x) = (x² − 4)/(x − 2).
Mistaken working: g(x) = (x − 2)(x + 2)/(x − 2) = x + 2, so the domain is all real numbers.
The simplified line x + 2 accepts x = 2, but the original g does not. At x = 2 the original denominator is 0.
The correction is to read the restriction from the original denominator: x − 2 = 0 gives x ≠ 2. The function equals x + 2 everywhere except at that one excluded input. Make it a habit to write the restriction before you cancel anything.
Check yourself
Give the largest possible domain in each case, then open the answer.
1. f(x) = 5/(2x − 7)
Show answer
2x − 7 = 0 gives x = 7/2. Domain: all real x with x ≠ 7/2 (that is, x ≠ 3.5).
2. h(x) = (x + 1)/(x² − 4x)
Show answer
x² − 4x = x(x − 4) = 0 gives x = 0 or x = 4. The numerator x + 1 is ignored. Domain: all real x with x ≠ 0 and x ≠ 4.
3. k(x) = x/(x² + 9). Is there any restriction?
Show answer
x² is never negative, so x² + 9 is at least 9 and is never zero. There is no restriction: the domain is all real x.
Where this leads next
Once excluded values feel routine, move on to determining a range from a restricted domain. When you are ready to mix the skills, try the functions and restrictions practice set. The function composition and inverse explorer lets you test inputs and see where an expression stops making sense.
Some students can do every step here but lose marks when a restriction has to be written in a longer answer. That is the kind of pattern our teachers look for in online one-to-one Additional Mathematics tuition.