Functions

A-Level Maths 9709 · Topic 1.2 revision notes

Everything from the 3 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.

Functions: domain, range and composition
Inverse functions
Transformations of graphs

1Functions: domain, range and composition

Syllabus 1.2

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

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.

Answer: (a) \(-5 \le f(x) \le 5\) (b) \(0 < g(x) \le 1\)

Sketch for the range, inside-out for composites.

Range: sketch over the domain; check both ends and any vertex inside it.
Write the range in terms of \(f(x)\), with \(<\) or \(\le\) chosen carefully.
One-one: no horizontal line meets the graph twice; restrict to one side of a vertex.
Composite: \(fg(x) = f\big(g(x)\big)\), so \(g\) acts first.
Existence: \(fg\) needs the range of \(g\) inside the domain of \(f\).

Watch out!

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.

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 \(-8\).

\(fg(x)\) is NOT \(f(x) \times g(x)\). With \(f(x) = 2x + 1\) and \(g(x) = x^{2}\), the product is \(2x^{3} + x^{2}\), but \(fg(x) = 2x^{2} + 1\).

Revisit it: Functions: domain, range and composition lesson · worked practice

2Inverse functions

Syllabus 1.2

To find \(f^{-1}\): write \(y = f(x)\), make \(x\) the subject, then swap the letters. The domain of \(f^{-1}\) is the range of \(f\), the range of \(f^{-1}\) is the domain of \(f\), and the graph of \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in the line \(y = x\).

Fig. 1: f(x) = x² - 2 for x ≥ 0 and its inverse f⁻¹(x) = √(x + 2) are mirror images in y = x; they meet on that line at (2, 2)

Worked example: Linear

Find \(f^{-1}(x)\) for \(f(x) = 3x - 5\), \(x \in \mathbb{R}\).

Write \(y = 3x - 5\).
Make \(x\) the subject: \(3x = y + 5 \Rightarrow x = \dfrac{y + 5}{3}\).
Swap letters to write the inverse as a function of \(x\).

Answer: \(f^{-1}(x) = \dfrac{x + 5}{3}\), for \(x \in \mathbb{R}\)

Make \(x\) the subject, then swap.

Check the function is one-one on its domain.
Write \(y = f(x)\) and make \(x\) the subject (complete the square for a quadratic).
Choose the sign of any square root to match the domain of \(f\).
Swap letters to write \(f^{-1}(x)\).
State the domain of \(f^{-1}\): it is the range of \(f\).

Watch out!

\(f^{-1}(x)\) is NOT \(\dfrac{1}{f(x)}\). For \(f(x) = 3x - 5\), \(\dfrac{1}{3x - 5}\) is a reciprocal; the inverse is \(\dfrac{x + 5}{3}\).

Writing \(\pm\sqrt{\ }\) in the final answer is wrong: an inverse function gives ONE output. Test your choice: \(g^{-1}\) must output values \(\le \frac{1}{3}\), and \(\frac{1 - \sqrt{x + 5}}{3}\) does.

Sketch \(f^{-1}\) only over the RANGE of \(f\). In Fig. 1 the inverse starts at \((-2, 0)\); drawing \(\sqrt{x + 2}\) to the left of \(-2\), or reflecting the whole parabola, earns no marks.

Revisit it: Inverse functions lesson · worked practice

3Transformations of graphs

Syllabus 1.2

Changes OUTSIDE the function act on \(y\) as you would expect: \(f(x) + a\) moves up \(a\), \(af(x)\) stretches by factor \(a\) parallel to the \(y\)-axis. Changes INSIDE act on \(x\) the opposite way: \(f(x + a)\) moves LEFT \(a\), \(f(ax)\) stretches by factor \(\dfrac{1}{a}\) parallel to the \(x\)-axis.

Fig. 1: y = (x - 3)² + 1 is y = x² translated 3 units right and 1 unit up; every point, such as (0, 0) and (-2, 4), moves by the same vector

Worked example: Describe a translation

Describe the transformation that maps \(y = x^{2}\) onto \(y = (x - 3)^{2} + 1\) (Fig. 1).

\(x\) is replaced by \(x - 3\): the graph moves 3 units in the positive \(x\)-direction.
\(+1\) is added outside: the graph moves 1 unit up.
Check with the vertex: \((0, 0)\) moves to \((3, 1)\), the vertex of the new curve.

Answer: A translation by \(\begin{pmatrix}3\\1\end{pmatrix}\)

Outside acts on \(y\); inside acts on \(x\), backwards.

Translation \(\begin{pmatrix}p\\q\end{pmatrix}\): \(y = f(x - p) + q\).
Reflection: \(-f(x)\) in the \(x\)-axis; \(f(-x)\) in the \(y\)-axis.
Stretch: \(af(x)\) parallel to the \(y\)-axis, factor \(a\); \(f(ax)\) parallel to the \(x\)-axis, factor \(\frac{1}{a}\).
Describe fully: name the transformation, the axis or direction, and the vector or factor.
Combine one step at a time, rewriting the equation after each, to check the order.

Watch out!

\(y = f(x + 4)\) moves the graph 4 units LEFT, not right. A point \((2, 5)\) on \(y = f(x)\) goes to \((-2, 5)\): you need \(x + 4 = 2\) to get the same output.

\(y = f(2x)\) is a stretch by factor \(\frac{1}{2}\), not 2: the graph is squeezed towards the \(y\)-axis. And always say 'parallel to the \(x\)-axis' or 'parallel to the \(y\)-axis'; 'a stretch of factor 2' alone scores nothing.

Inside the bracket, translations and stretches interact: \(f(2x - 6)\) is NOT 'translate 6 right, stretch factor \(\frac{1}{2}\)' unless you do them in that order. Test any sequence by building the equation.

Revisit it: Transformations of graphs lesson · worked practice