O Level · International Mathematics 0607

Functions

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Cambridge IGCSE · International Maths Extended

Functions

The Big Idea A function is just a reliable machine: you feed it a number (the input), it does the same fixed set of operations every time, and it spits out exactly one number (the output) — and once you understand that machine, you can run it forwards (find outputs), backwards (find inputs, i.e. the inverse), or chain two machines together (composite functions).

Summary — Everything In This Chapter

  • A function turns an input x into exactly one output f(x) using fixed operations.
  • Function notation: f(x) = ... lets you substitute numbers or expressions in for x.
  • The domain is the set of allowed inputs; the range is the set of resulting outputs.
  • Certain inputs must be excluded from a domain — usually values that cause division by zero or square-rooting a negative number.
  • A composite function like gf(x) means "do f first, then g" — read right to left, the letter closest to x goes first.
  • An inverse function f⁻¹(x) undoes what f(x) does, using opposite operations in reverse order.
  • The domain of f⁻¹ equals the range of f, and the range of f⁻¹ equals the domain of f.

1. Introduction to Functions

What is a function?

Think of a function as a little machine on a conveyor belt. You drop a number in one end (that's the input), the machine does the exact same set of mathematical operations to it every single time, and a number pops out the other end (that's the output).

For example, imagine a machine labelled "double the number and add 1." Two operations happen inside: multiply by 2, then add 1.

Seeing the machine in action
Put in 3 → 2 × 3 + 1 = 7
Put in −4 → 2 × (−4) + 1 = −7
Put in the letter x (an unknown number) → the output is 2x + 1

That last line is the important one: a function isn't really about one specific number, it's a rule that works for any input you throw at it — including letters and even whole expressions, as you'll see later.

Function notation

Instead of writing y = 2x + 1, we can give the rule a name — usually the letter f — and write f(x) = 2x + 1. Here, x is the input and f(x) is the output. You'll also see other letters used, like g, h, or j — typically a new letter signals a brand new rule/function.

Notation
f(x) = 3x − 4
Read as: "the function f, applied to input x, gives 3x − 4 as output." It means exactly the same thing as writing y = 3x − 4, just with a name attached.
Don't mix these up!

f(x) = 15 means "an input of x produces an output of 15" — this is a statement you'd solve as an equation to find x.

f(15) means "substitute the number 15 into the function" — you calculate a number, you don't solve anything.

Students lose easy marks by treating these as the same instruction. They are opposites: one gives you a number to plug in, the other gives you a number to solve for.

How a function works with different inputs

A function has an input x and an output f(x). The magic trick is: whatever you put in place of x on the left, you put in place of every x on the right.

Substituting different things in
If the input is 2, the output is f(2)
If the input is m, the output is f(m)
If the input is t + 5, the output is f(t + 5) — you replace every x in the formula with the whole expression (t + 5)
Worked Example

A function is defined as f(x) = 3x² − 2x + 1.

(a) Find f(7). Substitute x = 7 everywhere:
f(7) = 3(7)² − 2(7) + 1 = 3(49) − 14 + 1 = 147 − 14 + 1
f(7) = 134
(b) Find f(x + 3) in the form ax² + bx + c. This time the "input" is the whole expression (x + 3), so replace every x with (x + 3):
f(x + 3) = 3(x + 3)² − 2(x + 3) + 1
Expand carefully: 3(x² + 6x + 9) − 2x − 6 + 1 = 3x² + 18x + 27 − 2x − 6 + 1
f(x + 3) = 3x² + 16x + 22
Practice Question 1
A function is defined as g(x) = 5x − 3. Find g(4) and g(2a).
Practice Question 2
The function h(x) = 4x + 1 has output h(x) = 25. Find the value of x.

2. Domain & Range

Functions and graphs

Every function can be drawn as a graph. Just replace f(x) = ... with y = ..., and the x-axis becomes your set of inputs, while the y-axis becomes your set of outputs. This connection is exactly why sketching helps you "see" a function's behaviour instead of just guessing.

What is the domain?

The domain is the set of all inputs a function is allowed to take. Not every function accepts every number — some numbers would break the machine (like trying to divide by zero), and some domains are simply restricted on purpose because a question only cares about a certain range of inputs.

Golden rule

Domains always describe values of x — never y, never f(x). You can use "≠" (not equal to) or inequality signs to describe them.

Values that must be excluded

Two classic troublemakers show up again and again in IGCSE functions:

Trouble spot 1 — Dividing by zero
f(x) = 1 / x → domain: x ≠ 0
You physically cannot divide by zero, so 0 must be excluded. For a more complicated denominator like f(x) = 1 / [(x−1)(x+7)], you exclude whichever x-values make the denominator equal zero — here, x = 1 and x = −7.
Trouble spot 2 — Square-rooting a negative
f(x) = √x → domain: x ≥ 0
You can't take the square root of a negative number and get a real answer, so negative inputs are banned. For f(x) = √(x − 3), you need x − 3 ≥ 0, so the domain excludes anything below x = 3.

Compare this to something like f(x) = x² or f(x) = 3x + 2 — these two never break, no matter what you put in, so their domain is simply "all values of x."

What is the range?

The range is the set of all outputs a function actually produces, once you know the domain. Crucially — the range depends on the domain. Change what you're allowed to put in, and you change what comes out.

Golden rule

Ranges always describe values of f(x) — never x, never y. Same tools apply: "≠" and inequality signs.

Why the range depends on the domain

Take f(x) = 3x + 2 with domain x > 0.

If x = 0, then f(0) = 3(0) + 2 = 2 (even though x = 0 isn't actually allowed, it tells us the boundary)
Since the function increases as x increases, every input greater than 0 produces an output greater than 2
Range: f(x) > 2
A squaring example — watch the shape flip

Take f(x) = x² with domain "all values of x."

Squaring destroys negative signs — a negative input still comes out positive (or zero)
The smallest possible output happens at x = 0, giving f(0) = 0
Range: f(x) ≥ 0
Study tip

Sketching a function is the single best way to "see" both its domain and range at once. If you know what the graph of y = 1/x, y = x², y = x³, or the trig graphs look like, you can read off the range almost instantly — including for transformed and restricted versions of them.

Worked Example

Two functions are given by f(x) = 10 − x and g(x) = 1 / (2x − 1).

(a) If the domain of f is 2 < x ≤ 4, find the range.
f(2) = 10 − 2 = 8, and f(4) = 10 − 4 = 6
Since f(x) = 10 − x is a straight line with negative gradient, it decreases from 8 down to 6 as x goes from 2 to 4
The domain has 2 < x (strict) but x ≤ 4 (equal to). This flips: f(x) = 6 is achieved (equal to), while f(x) = 8 is not.
Range: 6 ≤ f(x) < 8
(b) Find the value of x excluded from the domain of g.
Set the denominator equal to zero: 2x − 1 = 0, so x = 1/2
x = 1/2 must be excluded
Watch out

When a domain flips a strict inequality (<) into a "not-strict" inequality (≤) for the range, it happens because of which end of the graph is increasing vs decreasing. Always check with a quick sketch or by testing both endpoints — never assume the inequality signs "carry over" unchanged.

Practice Question 1
State the domain of f(x) = 1 / (x − 5).
Practice Question 2
f(x) = x² has the domain x ≥ 3. Find the range of f(x).

3. Composite Functions

What is a composite function?

A composite function is what happens when you feed the output of one function straight into another function as its input — like two machines bolted together on the same conveyor belt. The number goes through machine 1, whatever pops out gets fed straight into machine 2, and only then do you get your final answer.

The notation — and why it's backwards from what you'd expect

Composite notation
gf(x) = g(f(x))
This means: do f first, then apply g to that result. The order of applying the functions runs right to left — the letter closest to the bracket "(x)" is applied first. This trips up almost everyone the first time, because it reads the opposite way to how we usually read left-to-right!
Memory trick

Rewrite gf(x) as g(f(x)) in your head or on paper — literally add the brackets back in. Now it visually reads "g of f(x)," which makes it obvious f happens first (it's the inner bracket) and g happens second (it's applied to whatever f produced).

Also worth knowing: fg(x) means do g first, then f. And ff(x) — sometimes written f²(x) — means apply f to itself twice in a row. Be careful: f²(x) does not mean [f(x)]², i.e. it does not mean "square the output of f." It means "run the input through f, then run that result through f again."

Substituting numbers into composite functions

The strategy: put the number into whichever function is closest to it, get an output, then feed that output into the other function.

Worked Example — numbers

f(x) = 2x + 1 and g(x) = 1/x.

Find gf(2):
Step 1 — apply f first (it's closest to the 2): f(2) = 2(2) + 1 = 5
Step 2 — feed that 5 into g: g(5) = 1/5
gf(2) = 1/5
Find fg(2):
Step 1 — apply g first: g(2) = 1/2
Step 2 — feed that 1/2 into f: f(1/2) = 2(1/2) + 1 = 1 + 1
fg(2) = 2

Notice gf(2) ≠ fg(2) — the order genuinely matters, exactly like putting socks on before shoes gives a very different result than shoes before socks!

Composite functions algebraically

Same idea, but now you substitute the entire expression for the inner function in place of every x in the outer function.

Worked Example — algebra

f(x) = 2x − 1 and g(x) = (x + 2)².

(a) Find fg(4). g is on the inside, so apply g first:
g(4) = (4 + 2)² = 6² = 36
Now apply f to 36: f(36) = 2(36) − 1 = 72 − 1
fg(4) = 71
(b) Find gf(x). f is on the inside, so substitute the whole expression (2x − 1) into every x in g:
gf(x) = g(2x − 1) = ((2x − 1) + 2)²
Simplify inside the bracket first: (2x + 1)²
gf(x) = (2x + 1)²
Practice Question 1
f(x) = 3x − 2 and g(x) = x². Find fg(3) and gf(3).
Practice Question 2
f(x) = x + 4 and g(x) = 2x − 5. Find fg(x) as a simplified expression.

4. Inverse Functions

What is an inverse function?

An inverse function undoes exactly what the original function did — like hitting rewind on the machine. If a function "doubles the number, then adds 1," its inverse must "subtract 1, then halve the result." Notice the operations are reversed and the order is flipped — this is exactly the same logic you already use when solving equations or rearranging a formula.

Notation
f⁻¹(x)
Read as "f inverse of x." Warning: this is NOT the same as 1/f(x) — the "−1" here is a label meaning "inverse function," not an exponent.
The input/output swap

If f(a) = b, then f⁻¹(b) = a. In other words, whatever came out of f, if you put that back into f⁻¹, you get back exactly what you started with. For example, if f(3) = 7, then f⁻¹(7) = 3.

Finding an inverse function algebraically

There's a clean, reliable 4-step method — memorise this exact sequence:

The Method
Step 1 — Write the function as y = ...
Step 2 — Swap every x and every y (letters change, nothing moves position)
Step 3 — Rearrange the new equation to make y the subject again
Step 4 — Replace y with f⁻¹(x) — your final answer should never contain the letter y
Worked Example

Find the inverse of f(x) = 2x + 1.

Step 1: y = 2x + 1
Step 2: swap x and y → x = 2y + 1
Step 3: rearrange for y → x − 1 = 2y, so y = (x − 1)/2
Step 4: replace y with f⁻¹(x)
f⁻¹(x) = (x − 1) / 2

Inverse and composite functions together

Here's a beautiful shortcut: applying a function and then its inverse (in either order) always cancels out and just gives you back x, untouched.

The cancelling rule
ff⁻¹(x) = f⁻¹f(x) = x
f and f⁻¹ are opposites, so applying one straight after the other undoes everything — you're left with exactly what you started with.

This is incredibly useful when finding f⁻¹(x) algebraically would be too hard or even impossible (for example, with exponential functions before you've learned logarithms).

Worked Example — using the cancelling trick

Solve f⁻¹(x) = 5, where f(x) = 2ˣ.

Finding f⁻¹(x) directly is genuinely hard here (it needs logarithms). Instead, apply f to both sides of the equation:
f(f⁻¹(x)) = f(5)
The left side cancels completely (that's the whole point of the rule!), leaving just x:
x = f(5) = 2⁵
x = 32

Domain and range of an inverse function

This is one of the most elegant (and most-tested) ideas in the whole chapter: the inverse function literally swaps the roles of domain and range.

The Swap Rule
Domain of f⁻¹ = Range of f
Range of f⁻¹ = Domain of f

This makes total sense once you think about it: f⁻¹ takes f's outputs and turns them back into f's inputs. So whatever used to come out of f becomes what goes into f⁻¹, and vice versa.

Watch out

Always write the domain of f⁻¹ in terms of x (e.g. "x > 5"), and always write the range of f⁻¹ in terms of f⁻¹(x) (e.g. "f⁻¹(x) > 5") — even though the numbers are identical to f's range and domain, the letters used to describe them must change to match whichever function you're now describing.

Worked Example

A function f(x) = 5 − 3x has the domain −2 < x ≤ 7.

(a) Find f⁻¹(x) using algebra:
y = 5 − 3x → swap → x = 5 − 3y → rearrange: 3y = 5 − x, so y = (5 − x)/3
f⁻¹(x) = (5 − x) / 3
(b) Find the domain of f⁻¹(x):
The domain of f⁻¹ = the range of f, so first find the range of f
f(−2) = 5 − 3(−2) = 11, and f(7) = 5 − 3(7) = −16 — since gradient is negative, f decreases from 11 to −16
Because the domain has −2 < x (strict) but x ≤ 7 (equal to), the range flips to −16 ≤ f(x) < 11
Domain of f⁻¹(x): −16 ≤ x < 11
Practice Question 1
Find the inverse of f(x) = 4x − 7.
Practice Question 2
f(x) = 3/(x + 1) has range f(x) > 5. State the domain of f⁻¹(x).

What to Memorise

f(x) notation: x is the input, f(x) is the output. f(x) = 15 (solve for x) is different from f(15) (calculate the output).
Domain: allowed set of inputs (values of x). Exclude x that causes division by zero or a negative under a square root.
Range: resulting set of outputs (values of f(x)). Always depends on the domain — sketch the graph to find it.
Composite gf(x) = g(f(x)): apply f first, then g. Read right-to-left — opposite of normal reading order.
ff(x) or f²(x): apply f twice. This is NOT the same as [f(x)]² (squaring the output).
Finding f⁻¹(x): write y = f(x) → swap x and y → rearrange to make y the subject → replace y with f⁻¹(x).
Cancelling rule: ff⁻¹(x) = f⁻¹f(x) = x always. Use this to solve tricky inverse equations without finding f⁻¹ directly.
Swap rule: Domain of f⁻¹ = Range of f. Range of f⁻¹ = Domain of f.

Concepts Checklist

Exam Tips & Common Mistakes

Trap #1 — Composite order

The single most common error: students calculate fg(x) as if it means "f, then g" because that's the order the letters appear. It's the opposite! gf(x) means f happens first. Always rewrite gf(x) as g(f(x)) before you start — the brackets don't lie.

Trap #2 — Domain restrictions

When a denominator is a product like (x − 1)(x + 7), students often only spot one root and forget the other. Set the ENTIRE denominator equal to zero and solve for all values — every root must be excluded from the domain.

Trap #3 — Inequality direction in range

When converting a domain into a range, don't assume the inequality signs stay in the same place. Always test both endpoints of the domain, work out which output is bigger, and pay attention to whether each endpoint is "equal to" (≤/≥) or "strictly less/greater than" (</>) in the original domain — carry that distinction through correctly.

Trap #4 — f⁻¹(x) ≠ 1/f(x)

The "−1" in f⁻¹(x) is not an exponent — it's just notation meaning "the inverse function." Never confuse it with 1 divided by f(x). They are completely different objects.

Trap #5 — Forgetting to swap x and y

When finding an inverse, it's easy to just rearrange for x without ever swapping x and y first. Remember: Step 2 (swap) MUST happen before Step 3 (rearrange), or your final answer will be wrong even though your algebra is technically correct.

Examiner's favourite question style

Expect combined questions: "Find gf(x), then state its domain and range" — testing composite functions AND domain/range together in one question. Also very common: "Use ff⁻¹(x) = x to solve f⁻¹(x) = [number]" for functions where finding f⁻¹ directly is hard (exponentials, etc.) — this shows up specifically to test whether you understand the cancelling relationship, not just mechanical algebra.

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