A function is a rule together with the set of inputs it accepts. Learn to state the domain, find the range, decide whether a function is one-one, and build composite functions such as \(fg(x)\), including when they can exist at all.
The domain is the set of inputs; the range is the set of outputs it produces. \(fg(x) = f\big(g(x)\big)\): apply \(g\) FIRST. The composite \(fg\) exists only when the range of \(g\) lies within the domain of \(f\).
Fig. 1: f(x) = (x - 2)² + 3 for 0 ≤ x ≤ 5. The least output is at the vertex, not at an end, so the range is 3 ≤ f(x) ≤ 12
Section 01
Domain and range.
In the lab, choose a function and follow the mapping diagram as each input in the domain is sent to its output: the set of outputs is the range. A function is written \(f : x \mapsto 2x - 3\) or \(f(x) = 2x - 3\), always with its domain, e.g. 'for \(x \in \mathbb{R}\)' or 'for \(x \ge 1\)'.
Domain: the inputs allowed, given in the question.
Range: the outputs \(f(x)\) produced from that domain. Write it in terms of \(f(x)\), e.g. \(f(x) \ge 1\), never in terms of \(x\).
Find it from a sketch: check the ends of the domain AND any turning point inside it.
Notation: \(\mathbb{R}\) is the set of all real numbers.
Worked example Linear and reciprocal
Find the range of (a) \(f(x) = 2x - 3\) for \(-1 \le x \le 4\) (b) \(g : x \mapsto \dfrac{1}{x}\) for \(x \ge 1\).
(a) \(f\) is increasing, so the ends give the extremes: \(f(-1) = -5\) and \(f(4) = 5\).
(b) \(g(1) = 1\). As \(x\) increases, \(\dfrac{1}{x}\) decreases towards 0 but never reaches it.
\(x = 2\) is inside the domain, so the least value is \(3\).
Ends: \(f(0) = 7\) and \(f(5) = 12\); the greater is 12.
Answer: \(3 \le f(x) \le 12\)
Examiner trap
Do not just substitute the two ends of the domain. For Fig. 1 that gives \(7 \le f(x) \le 12\), missing every output between 3 and 7. Always look for a vertex inside the domain.
Section 02
One-one or many-one?
A function is one-one if every output comes from exactly one input. On a graph: no horizontal line meets the curve more than once. Only one-one functions have inverses, so restricting the domain to make a function one-one is a common exam step.
Function
Type
Why
\(x^{3}\), \(x \in \mathbb{R}\)
one-one
always increasing
\(x^{2}\), \(x \in \mathbb{R}\)
many-one
\(f(-2) = f(2) = 4\)
\(x^{2}\), \(x \ge 0\)
one-one
only the right half of the parabola
\(\sin x\), \(0^{\circ} \le x \le 180^{\circ}\)
many-one
\(\sin 30^{\circ} = \sin 150^{\circ}\)
Worked example Restricting a domain
\(f(x) = x^{2} - 6x + 1\) for \(x \ge k\). Find the least value of \(k\) for which \(f\) is one-one.
Complete the square: \(f(x) = (x - 3)^{2} - 8\), so the vertex is at \(x = 3\).
To the right of the vertex the parabola only increases, so it is one-one there.
Any domain starting before 3 includes points on both sides of the vertex with equal outputs.
Answer: Least \(k = 3\)
Examiner trap
The least \(k\) is the \(x\)-coordinate of the vertex, not its \(y\)-coordinate. For \((x - 3)^{2} - 8\) the answer is \(k = 3\), not \(-8\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Find the range of \(f(x) = 3x + 1\) for \(0 \le x \le 2\).
Find the range of \(g : x \mapsto x^{2} + 1\) for \(x \in \mathbb{R}\).
Section 03
Composite functions.
In the lab, type in \(f(x)\) and \(g(x)\) and compare \(fg(x)\), where \(x\) goes into \(g\) first, with \(gf(x)\). Notice that \(fg\) and \(gf\) are usually different. Work from the inside out: the function written nearest \(x\) acts first.
Worked example Building fg and gf
\(f(x) = 2x + 1\) and \(g(x) = x^{2}\), both for \(x \in \mathbb{R}\). Find \(fg(x)\) and \(gf(x)\), and solve \(fg(x) = 19\).