Functions: Revision Notes

Condensed for the final weeks. For worked examples, use the Functions study guide.

Notation

f(x) = 2x + 3 means “the function f applied to x”.

f(4) means substitute 4 for x: f(4) = 2(4) + 3 = 11.

f(x) = 11 means solve 2x + 3 = 11 → x = 4.

Reading which of these is being asked is half the topic — the letter x appears in all three, so it’s the position of the equals sign and what’s already known that tells you which task is being asked for.

Domain and range

• Domain — the set of inputs (x values) allowed.
• Range — the set of outputs (y values) produced.

Restrictions arise where a function would be undefined:

• Denominator cannot be zero → exclude that x.
• Square root cannot be negative → the expression inside must be ≥ 0.

Worked example. g(x) = 1/(x − 2). State the value excluded from the domain.

x - 2 = 0  ->  x = 2 must be excluded

The denominator is zero at x = 2, so g(2) is undefined — every other real number is a valid input.

A mapping diagram shows the same relationship visually: arrows run from each domain value on the left to its corresponding range value on the right. A one-to-one function has exactly one arrow arriving at each range value; a many-to-one function can have two or more arrows arriving at the same value.

Composite functions

fg(x) means do g first, then f. Work from the inside out — the order catches nearly everyone.

f(x) = 2x + 1        g(x) = x^2

fg(x) = f(g(x)) = f(x^2)   = 2x^2 + 1
gf(x) = g(f(x)) = g(2x+1)  = (2x+1)^2

fg(x) ≠ gf(x) in general. If asked to show they differ, compute both.

Worked example. f(x) = 3/(x + 2) and g(x) = (3x + 5)². Find fg(x).

fg(x) = f(g(x)) = f((3x+5)^2) = 3 / ((3x+5)^2 + 2)

(3x + 5)² + 2 does not factorise or cancel with the 3 on top, so this is already in its simplest form — don’t force a cancellation that isn’t there.

Inverse functions

f⁻¹(x) undoes f. Method:

1. Write  y = f(x)
2. SWAP x and y
3. Rearrange to make y the subject
4. Replace y with f^-1(x)

Worked: f(x) = 3x − 5

y = 3x - 5
x = 3y - 5          (swap)
x + 5 = 3y
y = (x + 5) / 3     ->  f^-1(x) = (x + 5)/3

Check: f(f⁻¹(x)) should give x. Here f((x+5)/3) = 3·(x+5)/3 − 5 = x ✓

Graphically, y = f⁻¹(x) is the reflection of y = f(x) in the line y = x.

A function has an inverse only if it is one-to-one over its domain — each output must come from exactly one input.

Worked example. f(x) = (x - 3)^2 + 2 for x >= 3. Why must the domain be restricted for an inverse to exist? Without the restriction, f is not one-to-one — for instance f(2) = 3 and f(4) = 3, so two different inputs give the same output, and there is no single input to send 3 back to. Restricting to x >= 3 keeps only the right-hand half of the parabola, from the vertex onwards, which rises strictly and so is one-to-one — an inverse can then be defined.

Exam traps

• f⁻¹(x) is the inverse, not 1/f(x). This is the most common error.
• In fg(x), apply g first.
• Don’t forget to swap x and y when finding an inverse.
• State restrictions on the domain when a denominator or square root demands them.
• f(4) and f(x) = 4 ask opposite things.
• Forgetting that a composite function fraction must be checked for simplification before it’s left as a final answer.
• Claiming a function always has an inverse — it only does if it is one-to-one over the given domain; a full parabola is not, but half of one (with a domain restriction) can be.
• Stating the range when a question asks for the domain, or vice versa — read the question twice if the two get mixed up under time pressure.

For fuller worked examples of every case above, see the Functions study guide; for exam-style questions with full mark schemes, see the Functions practice questions.

Self-test

1. f(x) = 5x − 2. Find f(3) and solve f(x) = 18.
2. f(x) = x + 4, g(x) = 3x. Find fg(2) and gf(2).
3. Find the inverse of f(x) = (x − 1)/2.
4. Why can f(x) = 1/(x − 3) not take x = 3?
5. What is the geometric relationship between f and f⁻¹?
6. h(x) = 2x² + 3 and f(x) = 3x − 5. Find fh(x).
7. g(x) = (x + 4)/3. Verify that gg⁻¹(2) = 2.

Answers: 1. f(3) = 13; 5x − 2 = 18 → x = 4. 2. fg(2) = f(6) = 10; gf(2) = g(6) = 18. 3. y = (x−1)/2 → swap: x = (y−1)/2 → y = 2x + 1, so f⁻¹(x) = 2x + 1. 4. It would make the denominator zero, and division by zero is undefined. 5. Their graphs are reflections of each other in the line y = x. 6. fh(x) = f(2x² + 3) = 3(2x² + 3) − 5 = 6x² + 4. 7. g⁻¹(x) = 3x − 4, so g⁻¹(2) = 2; g(2) = 6/3 = 2 ✓.