Revision for domain and range, one-one and many-one functions, composite functions, inverse functions and their graphs, and the modulus of a function. Cambridge 0606 only as a topic of its own. Edexcel 4PM1 has no separate functions section, but uses function notation throughout.
A function gives exactly one output for each input. The domain is the set of inputs; the range is the set of outputs.
fg(x) means apply g first, then f. fg and gf are usually different, so check the order every time.
Only a one-one function has an inverse. If two inputs give the same output (a many-one function), restrict the domain first.
To find f⁻¹(x): write y = f(x), rearrange to make x the subject, then swap the letters. The domain of f⁻¹ is the range of f.
The graphs of f and f⁻¹ are reflections of each other in the line y = x.
For y = |f(x)|, reflect any part of y = f(x) below the x-axis in the x-axis.
Watch out: Write domains and ranges with the right letter: the domain uses x, the range uses f(x) (or y). Writing 'x ≥ 2' for a range loses the mark.
Worked example· 4 marks
The functions f and g are defined by \(f(x) = 3x - 5\) and \(g(x) = x^2 + 1\) for \(x \in \mathbb{R}\).
(a) Find \(fg(2)\).
(b) Find an expression for \(gf(x)\), simplifying your answer.
M1 (a) g(2) = 5 then f(5)
A1 (a) 10
M1 (b) (3x − 5)² + 1
A1 (b) 9x² − 30x + 26
fg(2) means apply g first: g(2) = 2² + 1 = 5.
Then f(5) = 3 × 5 − 5 = 10.
gf(x) = g(3x − 5) = (3x − 5)² + 1.
Expand: 9x² − 30x + 25 + 1 = 9x² − 30x + 26.
Answer: (a) \(10\) (b) \(9x^2 - 30x + 26\)
Functions questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the course whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 1 · 4 marks0606no calculator
The function f is defined by \(f(x) = \frac{2x + 3}{x - 1}\) for \(x \neq 1\).
Find an expression for \(f^{-1}(x)\) and state the value of \(x\) for which it is not defined.
Show the mark scheme and worked solution
M1 write y = (2x + 3)/(x − 1) and multiply out: y(x − 1) = 2x + 3
M1 collect the x terms: x(y − 2) = y + 3
A1 f⁻¹(x) = (x + 3)/(x − 2)
B1 x ≠ 2
Let y = (2x + 3)/(x − 1), so y(x − 1) = 2x + 3.
yx − y = 2x + 3, so yx − 2x = y + 3 and x(y − 2) = y + 3.
x = (y + 3)/(y − 2), so f⁻¹(x) = (x + 3)/(x − 2).
The denominator is zero when x = 2, so f⁻¹(x) is not defined there.
Answer: \(f^{-1}(x) = \frac{x + 3}{x - 2}\), not defined for \(x = 2\)
Question 2 · 5 marks0606no calculator
The function f is defined by \(f(x) = x^2 - 6x + 11\) for \(x \geq 3\).
(a) Write \(f(x)\) in the form \((x - a)^2 + b\) and hence state the range of f.
IGCSE Math Revision