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Additional Mathematics · Functions

Solving equations with an unknown function

The question hides a constant inside the function, and you cannot tell what to equate.

Many Functions questions give you a rule with something missing, then ask you to find it. The method is to turn what the question says into an equation, then solve and check.

This is the last lesson in the Functions chapter. It uses composite functions and inverse functions.

What kinds of unknown can appear?

Type The question gives You set up
Unknown constant f(2) = 7 and f(x) = ax + 3 2a + 3 = 7
Composite equals a value fg(x) = 11 the composite, equal to 11
Function equals its inverse f(x) = f⁻¹(x) two expressions, equal

Worked example 1: an unknown constant in a composite

Given f(x) = 4x − 1 and g(x) = x + a, and fg(2) = 15. Find a.

  1. fg(2) = f(g(2)). First g(2) = 2 + a.
  2. Then f(2 + a) = 4(2 + a) − 1.
  3. Set 4(2 + a) − 1 = 15, so 4(2 + a) = 16 and 2 + a = 4.
  4. So a = 2. Check: g(2) = 4 and f(4) = 15.

Worked example 2: f equals its own inverse

Given f(x) = (x + 4) ÷ 3. Solve f(x) = f⁻¹(x).

First find the inverse: y = (x + 4) ÷ 3 gives x = 3y − 4, so f⁻¹(x) = 3x − 4.

Set (x + 4) ÷ 3 = 3x − 4. Multiply by 3: x + 4 = 9x − 12, so 16 = 8x and x = 2.

Check: f(2) = 6 ÷ 3 = 2 and f⁻¹(2) = 6 − 4 = 2. Both equal 2, so the answer stands.

Worked example 3: rejecting a value

Given f(x) = x + 2 and g(x) = x², with domain x > 0. Solve gf(x) = 25.

gf(x) = (x + 2)². Set (x + 2)² = 25, so x + 2 = 5 or x + 2 = −5. This gives x = 3 or x = −7.

The domain is x > 0, so x = −7 is rejected. The answer is x = 3. Check: f(3) = 5 and g(5) = 25.

The mistake that costs marks

The slip is to solve fg(x) when the question says gf(x), or to keep a root the domain forbids. Both come from not underlining what the question asks before starting. Write the composite you need on the first line, for example “gf(x) = g(f(x))”.

Check yourself

Given f(x) = x − 1 and g(x) = 2x². Solve gf(x) = 18.

Answer

gf(x) = g(f(x)) = 2(x − 1)². Set 2(x − 1)² = 18, so (x − 1)² = 9 and x − 1 = 3 or −3.

So x = 4 or x = −2, because no domain restriction is given.

Check x = 4: f(4) = 3 and g(3) = 18. Check x = −2: f(−2) = −3 and g(−3) = 18.

What to study next

Test the whole chapter with the functions practice set. If you are unsure how to begin an unfamiliar question, read choosing the right method in an unfamiliar Add Maths question.

If you want a teacher to work through these question types with you, see online one-to-one Additional Mathematics tuition.

Common questions

How do I find a constant hidden inside f(x)?

Use any value the question gives. If f(2) = 7 and f(x) = ax + 3, substitute x = 2 into the rule to get 2a + 3 = 7, then solve for a. One known input and output is enough for one unknown.

What does f²(x) mean?

It means ff(x), apply f twice. Do not square the answer. If f(x) = 2x − 3 then f²(x) = 2(2x − 3) − 3 = 4x − 9.

How do I solve f(x) = f⁻¹(x)?

Find f⁻¹(x) first, set the two expressions equal and solve. Then substitute the answer into both sides to confirm they match. Some questions also ask you to reject a root outside the domain.

When do I reject a solution?

When it lies outside the domain stated in the question. The algebra can give two values, and only the ones that belong to the domain are valid answers.

If the algebra runs fine but you are unsure what to equate, one-to-one Add Maths lessons let a teacher ask you what each side of the equation stands for. Tell us the form and the type of question that stalls you.

  • Online one-to-one lessons for your child with an experienced teacher.
  • Your first class is a one-hour trial, from RM50. The fee is agreed before you book.
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