Functions for IGCSE Mathematics - Additional: Methods and Worked Examples

You've seen functions before, even if you don't realise it

Think about the last time you converted a temperature from Celsius to Fahrenheit. You took a number, multiplied it by 9/5, then added 32. That process - putting a number in, following a rule, getting a number out - is exactly what a function does. The IGCSE Additional Mathematics course takes that simple idea and builds an entire toolkit around it, and it's one of the topics you'll see on almost every past paper.

If you're comfortable with basic algebra, you already have the foundation. What this topic adds is a formal language for describing how inputs map to outputs, how to reverse that process, and how to chain functions together. Once you get the hang of the notation, you'll find it surprisingly satisfying.

The building blocks: function notation, domain, and range

A function is a rule that assigns each input exactly one output. You'll see it written in two main ways:

• f(x) = 2x + 3 - the most common notation, read as "f of x equals 2x plus 3"
• f: x → 2x + 3 - mapping notation, read as "f maps x to 2x plus 3"

Both say the same thing. Your exam might use either, so get comfortable reading both.

Now, every function has two important sets attached to it:

• Domain - the set of all allowed inputs. If no domain is stated, assume all real numbers unless something breaks (like dividing by zero or taking the square root of a negative number).
• Range (or image set) - the set of all possible outputs. You work this out by thinking about what values f(x) can actually produce.

One-one and many-one: why it matters

Here's a distinction that catches a lot of students off guard. A one-one function means every output comes from exactly one input. If you know the output, you can trace back to a single input with certainty. The function f(x) = 2x + 3 is one-one because different inputs always give different outputs.

A many-one function means two or more different inputs can produce the same output. The classic example is f(x) = x^2. Both x = 3 and x = -3 give f(x) = 9. You can't uniquely reverse it.

Why does this matter? Because only one-one functions have inverses. If a function is many-one, you can't "undo" it unambiguously. The exam will sometimes ask you to explain in words why a particular function doesn't have an inverse, and the answer always comes back to this: multiple inputs map to the same output, so reversing the function gives more than one answer.

Notice that third row. Restricting the domain can turn a many-one function into a one-one function. Exam questions sometimes give you f(x) = x^2 for x ≥ 0 specifically so that the inverse exists. Always check the domain before deciding whether an inverse is possible.

Composite functions: the order matters more than you think

A composite function chains two functions together. The notation fg(x) means "apply g first, then apply f to the result." Think of it like a two-step machine: your input goes into the g machine, whatever comes out goes straight into the f machine, and what emerges is your final answer.

Here's the crucial point that costs students marks every exam session: fg(x) is not the same as gf(x). The order matters.

Worked example: finding fg(x) and gf(x)

Suppose f(x) = x^2 + 1 and g(x) = 3x - 2.

Finding fg(x):

1. Start with g(x) = 3x - 2
2. Now substitute this into f, replacing every x in f with (3x - 2)
3. fg(x) = (3x - 2)^2 + 1 = 9x^2 - 12x + 4 + 1 = 9x^2 - 12x + 5

Finding gf(x):

1. Start with f(x) = x^2 + 1
2. Now substitute this into g, replacing every x in g with (x^2 + 1)
3. gf(x) = 3(x^2 + 1) - 2 = 3x^2 + 3 - 2 = 3x^2 + 1

You can see they're completely different expressions. If a question asks for fg(2), calculate g(2) first: g(2) = 3(2) - 2 = 4. Then calculate f(4): f(4) = 4^2 + 1 = 17. So fg(2) = 17.

Finding inverse functions step by step

The inverse of a function reverses what the original function did. If f takes 5 to 13, then f^(-1) takes 13 back to 5. The notation f^(-1)(x) means "the inverse of f." (It does not mean 1/f(x) - that's a different thing entirely, and confusing the two is a common mistake.)

Here's your reliable method for finding an inverse:

1. Write y = f(x)
2. Swap x and y, so now you have x = f(y)
3. Rearrange to make y the subject
4. Replace y with f^(-1)(x)

Worked example: finding the inverse

Find f^(-1)(x) given that f(x) = (2x + 5) / 3.

1. Write y = (2x + 5) / 3
2. Swap: x = (2y + 5) / 3
3. Multiply both sides by 3: 3x = 2y + 5
4. Subtract 5: 3x - 5 = 2y
5. Divide by 2: y = (3x - 5) / 2
6. So f^(-1)(x) = (3x - 5) / 2

You can verify this works. Try f(1) = (2(1) + 5) / 3 = 7/3. Now check: f^(-1)(7/3) = (3(7/3) - 5) / 2 = (7 - 5) / 2 = 1. It takes you right back where you started.

Graphs of functions and their inverses

There's a beautiful visual relationship between a function and its inverse: the graph of f^(-1)(x) is the reflection of f(x) in the line y = x. If you fold the graph paper along the line y = x, the two curves land exactly on top of each other.

This means:

• If the point (a, b) lies on the graph of f, then (b, a) lies on the graph of f^(-1)
• The two graphs always intersect on the line y = x (wherever f(x) = x)
• The domain of f becomes the range of f^(-1), and vice versa

When the exam asks you to sketch f^(-1) given the graph of f, pick three or four key points on f, swap their coordinates, plot them, and draw a smooth curve through them. Don't forget to also draw the line y = x as a dashed line to show the reflection.

Modulus functions: dealing with the absolute value

The modulus (or absolute value) of a number is its distance from zero, which is always positive. The notation |x| means "take x, and if it's negative, make it positive." So |5| = 5 and |-5| = 5.

In the IGCSE Additional Mathematics syllabus, you'll encounter two types of modulus function:

Type 1: y = |f(x)|

To sketch this, first draw y = f(x) normally. Then take any part of the graph that falls below the x-axis and reflect it upward. The sections already above the x-axis stay exactly where they are. The result is a graph that never dips below y = 0.

Type 2: y = f(|x|)

This one works differently. You only need the right-hand side of y = f(x) (where x ≥ 0). Delete the left-hand side and reflect the right-hand side across the y-axis. The result is always symmetric about the y-axis.

Worked example: solving a modulus equation

Solve |2x - 1| = 5.

The expression inside the modulus brackets can be either positive or negative, so you set up two cases:

1. Case 1: 2x - 1 = 5, so 2x = 6, giving x = 3
2. Case 2: 2x - 1 = -5, so 2x = -4, giving x = -2

Both solutions are valid. Always check by substituting back: |2(3) - 1| = |5| = 5 (correct) and |2(-2) - 1| = |-5| = 5 (correct).

Common mistakes and how to avoid them

After working through hundreds of student papers, certain errors come up again and again. Here are the ones you should actively watch for:

• Confusing fg with gf. Always identify which function acts first. The one written closest to x goes first.
• Writing f^(-1)(x) = 1/f(x). The inverse function notation uses a superscript -1, but it does not mean a reciprocal. These are completely different operations.
• Forgetting to restrict the domain. If a question defines f(x) = x^2 for x ≥ 0, the inverse only works on that domain. Don't ignore the restriction.
• Mixing up |f(x)| and f(|x|). One reflects below the x-axis upward. The other mirrors the right side of the graph to the left. Drawing a quick sketch helps you avoid the wrong transformation.
• Not verifying inverses. If you have time, check that f(f^(-1)(x)) = x for a simple value. It takes 30 seconds and can save you marks.

Self-check questions

Test your understanding with these questions before moving on. Try each one on paper, then check your method against the approach described in the relevant section above.

1. Given f(x) = 4x - 7, find f^(-1)(x) and verify that ff^(-1)(3) = 3.
2. If f(x) = x^2 + 2 and g(x) = 3x, find fg(x) and gf(x). Are they the same?
3. Explain why f(x) = x^2 for all real x does not have an inverse, but f(x) = x^2 for x ≤ 0 does.
4. Sketch y = |x^2 - 4| and state the coordinates where the graph touches the x-axis.
5. For f(x) = (x + 1) / (x - 2), x ≠ 2, find f^(-1)(x) and state any value excluded from the domain of the inverse.

Functions might look intimidating with all the notation, but once you break each problem into its steps, you'll see the same patterns repeating. Keep practising composites and inverses until the process feels automatic, and always double-check your order of operations. You've got this.