1The idea in one paragraph

A function is a rule that takes an input and gives back exactly one output. That "exactly one" is the whole definition, and it is what lets you write f(3) and mean one number. The set of inputs the rule is allowed to take is the domain. The set of outputs it actually produces is the range. A graph shows both at once: the domain is how far the graph stretches left and right, and the range is how far it stretches up and down. Some functions can be run backwards, so that each output leads you back to the one input it came from. That backwards function is the inverse, f⁻¹, and its graph is the graph of f reflected in the line y = x. It exists only when no two inputs share an output.

2What makes a rule a function

A relation is any pairing of inputs with outputs. A function is a relation with one extra promise: each input is paired with exactly one output. Figure 1 shows the difference with arrow diagrams, which the guide calls mappings.

In panel (a) the rule is "square it". Both −2 and 2 go to 4. That is allowed: two inputs may share an output, which is called many-to-one. Each input still has only one arrow leaving it. In panel (b) the rule is "take the square root, positive or negative". The input 4 has two arrows, to −2 and to 2. That is one-to-many, and it is not a function, because "the output for 4" does not name one number.

On a graph the same test becomes the vertical line test, as Figure 2 shows. Every vertical line is one value of x. If any vertical line crosses the graph more than once, that x has more than one y, and the graph is not a function. The parabola y = x² − 2 passes. The circle x² + y² = 9 fails: the line x = 1.5 meets it twice.

The same check works on a list of pairs. {(1, 4), (2, 4), (3, 5)} is a function, since each first coordinate appears once. {(1, 4), (1, 5), (2, 6)} is not, since the input 1 has two outputs.

3Function notation

Function notation names the function and its input together. f(x) is read "f of x" and means "the output of f when the input is x". The letters change with the context: v(t) is a velocity at time t, C(n) is the cost of n items. The IB also uses mapping notation, f: x ↦ 3x² − 2x, which reads "f maps x to 3x² − 2x" and means the same as f(x) = 3x² − 2x.

The input is the independent variable; you choose it. The output is the dependent variable; it depends on your choice. On a graph the input goes across and the output goes up, so y = f(x).

Take f(x) = 3x² − 2x. Evaluating means substituting the input everywhere x appears, in brackets.

Solving runs the other way: you are given an output and asked for the input or inputs.

Two inputs give the output 8. That is fine for a function, but it will matter in section 9.

Keep the two apart: f(2) asks for an output; f(x) = 2 asks for an input. Most sign and method errors in this topic come from blurring them.

4Domain: which inputs are allowed

The domain of a function is the set of inputs it is defined for. Unless a question says otherwise, the exam takes the largest possible domain, which will usually be all real numbers, x ∈ ℝ. You only have to restrict it when something would break. At standard level three things break a rule for real numbers:

• Dividing by zero. g(x) = 5 ÷ (x − 3) is undefined at x = 3. Domain: x ∈ ℝ, x ≠ 3.
• The square root (or any even root) of a negative number. h(x) = √(2x + 8) needs 2x + 8 ≥ 0, so the domain is x ≥ −4.
• The logarithm of zero or a negative number. ln(x − 1) needs x − 1 > 0, so x > 1. You meet this properly in 2.9.

Combine them when they appear together. For k(x) = 1 ÷ √(6 − x), the root needs 6 − x ≥ 0 and the division needs √(6 − x) ≠ 0, so 6 − x > 0 and the domain is x < 6, with a strict inequality.

The guide's own example is f(x) = √(2 − x). The expression under the root must not be negative: 2 − x ≥ 0, so the domain is x ≤ 2. Figure 3 shows it. The graph starts at (2, 0) and runs away to the left forever; along the x-axis, that is exactly x ≤ 2.

Write domains as inequalities in x, like x ≤ 2 or x ≠ 3, or with set notation, x ∈ ℝ, x ≠ 3.

5Range: which outputs you get

The range is the set of outputs the function actually produces from its domain. It is written in terms of the output: f(x) ≥ 0, or y ≥ 0. The guide says it plainly: a graph is helpful in visualising the range. Read it by looking at how far the graph reaches up and down.

For f(x) = √(2 − x) in Figure 3, the square root is never negative, it equals 0 when x = 2, and it grows without limit as x runs left. So the range is f(x) ≥ 0.

Three more cases, each with its own lesson.

A square pushed up. f(x) = x² + 3. Since x² ≥ 0 for every x, x² + 3 ≥ 3. The range is f(x) ≥ 3. The minimum is where the square is zero.

A line on a restricted domain. h(x) = 2x − 1 for −1 ≤ x ≤ 4. A line with positive gradient is lowest at the left end and highest at the right: h(−1) = −3 and h(4) = 7. The range is −3 ≤ h(x) ≤ 7.

A curve on a restricted domain. p(x) = (x − 1)² for −2 ≤ x ≤ 2. The end points give p(−2) = 9 and p(2) = 1. It is tempting to write 1 ≤ p(x) ≤ 9, and it is wrong. Figure 4 shows why: the curve dips to its vertex at (1, 0) inside the domain. The range is 0 ≤ p(x) ≤ 9.

Find a range from the graph, not from the end points. Sketch it, or look at it on your GDC, and read off the lowest and the highest outputs, including any turning point inside the domain.

On Paper 1 you reason as above: squares and roots are never negative, and a line is lowest and highest at its ends.

6A function as a model

The guide treats a function as a mathematical model: a rule that describes a real quantity. When it does, the context sets the domain, and that is often smaller than the domain the algebra allows.

A ball is thrown straight up from the ground. Its height in metres after t seconds is modelled by h(t) = 20t − 4.9t². The formula works for any t. The ball does not. It leaves the ground at t = 0 and lands when h(t) = 0 again.

So a sensible domain is 0 ≤ t ≤ 4.08. The highest point is at t = 20 ÷ 9.8 ≈ 2.04 s, where h ≈ 20.4 m, so the range in context is 0 ≤ h(t) ≤ 20.4. Figure 5 shows the model and the part of the curve that means nothing.

Other contexts restrict the domain differently: a cost C(n) for n tickets only makes sense for whole numbers n ≥ 0. When a Paper 2 question asks for "a suitable domain", it wants the domain the context allows, with a reason.

7The inverse: undoing a function

An inverse function reverses the effect of a function. If f takes a to b, then f⁻¹ takes b back to a.

f(a) = b means exactly the same as f⁻¹(b) = a.

Temperature conversion is the everyday example. f(x) = 1.8x + 32 turns degrees Celsius into degrees Fahrenheit: multiply by 1.8, then add 32. To undo it you do the opposite operations in the opposite order: subtract 32, then divide by 1.8. So f⁻¹(x) = (x − 32) ÷ 1.8 turns Fahrenheit back into Celsius. Body temperature of 98.6 °F is f⁻¹(98.6) = 66.6 ÷ 1.8 = 37 °C.

This gives the guide's key link: solving f(x) = k is the same as finding f⁻¹(k). You do not need a formula for f⁻¹ to find one value of it.

On Paper 2, for a function you cannot rearrange easily, the GDC does it: to find f⁻¹(20) for f(x) = x³ + 2x + 1, solve x³ + 2x + 1 = 20 with the equation solver or by intersecting y = f(x) with y = 20.

One warning about the notation. f⁻¹(x) is the inverse function, not 1 ÷ f(x). The −1 here is not a power. For f(x) = 1.8x + 32, 1 ÷ f(x) is 1 ÷ (1.8x + 32), which converts nothing into anything.

Finding a formula for f⁻¹ by algebra is the job of 2.5.

8The inverse as a reflection in y = x

If (a, b) is on the graph of f, then f(a) = b, so f⁻¹(b) = a, and the point (b, a) is on the graph of f⁻¹. Swapping the coordinates of every point is the same as reflecting the whole graph in the line y = x. So:

The graph of y = f⁻¹(x) is the reflection of the graph of y = f(x) in the line y = x.

Figure 6 does it for the guide's function f(x) = √(2 − x). Take four points on f: (−7, 3), (−2, 2), (1, 1) and (2, 0). Swap each: (3, −7), (2, −2), (1, 1) and (0, 2). Join them with the reflected shape. The point (1, 1) lies on the mirror line, so it stays where it is. The inverse here is f⁻¹(x) = 2 − x² for x ≥ 0, a result 2.5 shows you how to find.

The reflection swaps x and y, so it swaps what you read along the x-axis with what you read up the y-axis. That is why:

• the domain of f⁻¹ is the range of f, and
• the range of f⁻¹ is the domain of f.

In Figure 6, the range of f is f(x) ≥ 0, so the domain of f⁻¹ is x ≥ 0. That is where the "x ≥ 0" in f⁻¹(x) = 2 − x² comes from. Without it, 2 − x² would be a whole parabola, and half of it would not be the reflection of anything.

To sketch an inverse in the exam: draw the line y = x lightly, pick three or four clear points on f including any intercepts and end points, swap their coordinates, and join them with the mirror-image shape. An x-intercept (a, 0) on f becomes a y-intercept (0, a) on f⁻¹, and the other way round. If f has an asymptote y = k, f⁻¹ has the asymptote x = k.

9Only one-to-one functions have inverses

A function is one-to-one when every output comes from exactly one input: no two inputs share an output. Only a one-to-one function has an inverse, because the inverse has to send each output back to a single input.

Figure 7 shows what goes wrong otherwise. f(x) = x² on all real numbers sends both −1.58 and 1.58 to 2.5; the horizontal line y = 2.5 meets the graph twice. Reflect the graph in y = x and you get a sideways curve that fails the vertical line test, so the reflection is not a function. This is the horizontal line test: if any horizontal line meets the graph of f more than once, f is not one-to-one and has no inverse.

The fix is to restrict the domain. On x ≥ 0 alone, x² is one-to-one, and its inverse is √x, as panel (b) shows.

Section 3 gave a warning sign for this. f(x) = 3x² − 2x had two solutions to f(x) = 8, so two inputs share the output 8, so f is not one-to-one on ℝ and has no inverse there.

A function that only ever increases, or only ever decreases, over its domain always passes the horizontal line test. That is the quickest reason to give when a question asks you to explain why an inverse exists.