Functions: notation, domain and range, composite and inverse functions, sketching graphs

Functions – Notation, Domain & Range, One‑to‑One & Inverse, Composite, Transformations, Sketching Graphs

1. Definition and Standard Notation

• A **function** assigns to each element \(x\) in a set called the domain a unique element \(y\) in a set called the range.
• Typical notation:
  • \(f:D\to R\) – emphasises the mapping.
  • \(y=f(x)\) – the dependent variable.
  • \(f(x)=\text{expression in }x\) – defines the rule.

2. Determining Domain and Range

For an algebraic expression the domain consists of all real numbers that do **not** violate any of the following restrictions.

Source of restriction	Condition to satisfy	Typical example
Denominator ≠ 0	\(xeq\) any root of the denominator	\(f(x)=\dfrac{x+1}{x-3}\) → \(xeq3\)
Even root (√, ⁴√, …)	Radicand ≥ 0	\(g(x)=\sqrt{4-x}\) → \(4-x\ge0\Rightarrow x\le4\)
Logarithm	Argument > 0	\(h(x)=\ln(x-2)\) → \(x-2>0\Rightarrow x>2\)
Inverse trigonometric functions	Argument must lie in the principal domain	\(\arcsin x\) → \(|x|\le1\)

Once the domain \(D\) is known, the range \(R\) is obtained by analysing the behaviour of \(f(x)\) on \(D\). For IGCSE the usual tools are:

• Algebraic manipulation (e.g. completing the square for quadratics).
• Limits for asymptotic behaviour (A‑Level).
• First‑derivative test for monotonicity (A‑Level).

3. One‑to‑One (Injective) Functions & Inverses

• Injective definition: \(f\) is one‑to‑one if \(f(x_1)=f(x_2)\) implies \(x_1=x_2\). Only such functions possess an inverse on the whole domain.
• Horizontal‑line test: a horizontal line must intersect the graph at most once. (See Figure 1.)
• Graphical relationship: the graph of the inverse \(f^{-1}\) is the reflection of the graph of \(f\) in the line \(y=x\). (Figure 2 illustrates the mirror image.)

Algebraic method for finding an inverse

1. Write the relation as \(y = f(x)\).
2. Interchange the symbols \(x\) and \(y\): \(x = f(y)\).
3. Solve the resulting equation for \(y\); the solution is \(f^{-1}(x)\).

Examples

Linear function \(f(x)=3x-4\)

\[ \begin{aligned} y &= 3x-4 \\ x &= 3y-4 \;\Longrightarrow\; y = \frac{x+4}{3} \end{aligned} \qquad\Rightarrow\qquad f^{-1}(x)=\frac{x+4}{3} \]

Quadratic with restricted domain \(f(x)=x^{2}\) (originally not injective)

• Restrict to \(x\ge0\) (or \(x\le0\)).
• Inverse: \(y = \sqrt{x}\) (or \(-\sqrt{x}\)).

Inverse trigonometric example \(y=\sin^{-1}x\)

• Domain: \(-1\le x\le 1\).
• Range (principal values): \(-\dfrac{\pi}{2}\le y\le\dfrac{\pi}{2}\).

4. Composite Functions

Given two functions \(f\) and \(g\), the composite \((f\circ g)\) is defined by

\[ (f\circ g)(x)=f\bigl(g(x)\bigr). \]

Domain of a composite – take all \(x\) in the domain of \(g\) for which \(g(x)\) lies in the domain of \(f\).

Worked composition with domain checking

\[ f(x)=\sqrt{x}, \qquad g(x)=\frac{1}{x-1}. \]

• Domain of \(g\): \(xeq1\). (Denominator cannot be zero.)
• Range of \(g\): all real numbers except \(0\) (since \(\frac{1}{x-1}=0\) has no solution).
• For \(f\bigl(g(x)\bigr)\) we need \(g(x)\ge0\). Solve \(\frac{1}{x-1}\ge0\):
  • \(x-1>0\) → \(x>1\) gives \(\frac{1}{x-1}>0\).
    
  • \(x-1<0\) → \(x<1\) gives \(\frac{1}{x-1}<0\) (reject).
• Hence \(\displaystyle (f\circ g)(x)=\sqrt{\frac{1}{x-1}}\) with domain \((1,\infty)\).

Note that in general \(f\circ geq g\circ f\); the reverse composition would be \((g\circ f)(x)=\dfrac{1}{\sqrt{x}-1}\) with a different domain.

5. Function Transformations (Shifts, Stretches & Reflections)

Transformations allow us to obtain the graph of a new function from a known “parent’’ graph.

Transformation	Resulting expression	Effect on the graph
Horizontal shift by \(a\)	\(y=f(x-a)\)	Right if \(a>0\); left if \(a<0\).
Vertical shift by \(b\)	\(y=f(x)+b\)	Up if \(b>0\); down if \(b<0\).
Horizontal stretch/compression by factor \(keq0\)	\(y=f(kx)\)	\(|k|>1\) → compression; \(0<|k|<1\) → stretch.
Vertical stretch/compression by factor \(aeq0\)	\(y=a\,f(x)\)	\(|a|>1\) → stretch; \(0<|a|<1\) → compression; negative \(a\) also reflects in the \(x\)-axis.
Reflection in the \(y\)-axis	\(y=f(-x)\)	Mirror about the \(y\)-axis.
Reflection in the \(x\)-axis	\(y=-f(x)\)	Mirror about the \(x\)-axis.

Worked example – Transforming \(y=x^{2}\)

Apply the following operations in order:

1. Shift three units left → \(y=(x+3)^{2}\).
2. Shift two units up → \(y=(x+3)^{2}+2\).
3. Vertical stretch by factor 2 → \(y=2\bigl[(x+3)^{2}+2\bigr]\).
4. Reflection in the \(x\)-axis → \(\displaystyle y=-2\bigl[(x+3)^{2}+2\bigr]\).

Final equation: \(\displaystyle y=-2\bigl[(x+3)^{2}+2\bigr]\).
Vertex: \((-3,\,-4)\); the parabola opens downwards and is twice as steep as the parent curve.

6. Systematic Checklist for Sketching a Graph

1. Domain & Range – list all restrictions (denominators, radicands, logs, trig arguments).
2. Intercepts
   • \(x\)-intercepts: solve \(f(x)=0\).
   • \(y\)-intercept: evaluate \(f(0)\) (if 0 lies in the domain).
3. Symmetry
   • Even: \(f(-x)=f(x)\) → symmetry about the \(y\)-axis.
   • Odd: \(f(-x)=-f(x)\) → symmetry about the origin.
4. Asymptotes
   • Vertical: values that make the function blow up (typically denominator = 0).
   • Horizontal/Oblique: limits \(\displaystyle\lim_{x\to\pm\infty}f(x)\).
5. Critical points – compute \(f'(x)\) (A‑Level) and set \(f'(x)=0\).
6. Monotonicity – sign of \(f'(x)\) tells where the function is increasing or decreasing.
7. Concavity & Points of Inflection – use the second derivative \(f''(x)\).
8. Horizontal‑line tests – solve \(f(x)=k\) for a few convenient values of \(k\) to obtain extra points.
9. Plot key points – intercepts, critical points, points near asymptotes, and any points from step 8.
10. Draw the curve – join the points smoothly, respecting asymptotes, symmetry and monotonicity.

7. Piece‑wise Functions

These are common in exam questions. Treat each piece separately, then check the behaviour at the break points.

\[ f(x)=\begin{cases} f_{1}(x), & a_{1}\le x < a_{2},\\[2mm] f_{2}(x), & a_{2}\le x < a_{3},\\[2mm] \;\vdots \end{cases} \]

Example

\[ f(x)=\begin{cases} x+2, & x\le 1,\\[2mm] 3x-4, & x>1. \end{cases} \]

• Domain: \(\mathbb{R}\).
• Continuity at \(x=1\): left‑hand value \(1+2=3\); right‑hand value \(3(1)-4=-1\). Discontinuous at \(x=1\).
• Derivative: \(f'(x)=1\) for \(x<1\) and \(f'(x)=3\) for \(x>1\). No derivative at the jump.

8. Inverse Trigonometric Functions (Principal Values)

In the Cambridge A‑Level syllabus the symbols \(\sin^{-1},\;\cos^{-1},\;\tan^{-1}\) denote the principal (restricted) inverses, not powers.

• \(\sin^{-1}x\) (or \(\arcsin x\)) Range: \(\displaystyle\bigl[-\frac{\pi}{2},\frac{\pi}{2}\bigr]\).
• \(\cos^{-1}x\) (or \(\arccos x\)) Range: \([0,\pi]\).
• \(\tan^{-1}x\) (or \(\arctan x\)) Range: \(\displaystyle\bigl(-\frac{\pi}{2},\frac{\pi}{2}\bigr)\).

These restrictions make the trigonometric functions one‑to‑one, allowing the inverses to exist.

9. Worked Example – Sketching \(\displaystyle f(x)=\frac{x}{x^{2}-1}\)

1. Domain: denominator \(eq0\) → \(xeq\pm1\). \(D=\mathbb{R}\setminus\{-1,1\}\).
2. Intercepts
   • \(y\)-intercept: \(f(0)=0\).
   • \(x\)-intercept: numerator = 0 → \(x=0\).
3. Asymptotes
   • Vertical: \(x=1\) and \(x=-1\).
   • Horizontal: \(\displaystyle\lim_{x\to\pm\infty}\frac{x}{x^{2}-1}=0\) → \(y=0\).
4. First derivative \[ f'(x)=\frac{(x^{2}-1)-x(2x)}{(x^{2}-1)^{2}} =\frac{-x^{2}-1}{(x^{2}-1)^{2}}<0\quad\forall x\in D. \] Hence the function is decreasing on each interval of its domain.
5. Second derivative \[ f''(x)=\frac{2x(x^{2}+3)}{(x^{2}-1)^{3}}. \] Sign changes at \(x=0\); therefore an inflection point at \((0,0)\).
6. Additional points for the sketch \[ \begin{aligned} f(-2)&=\frac{-2}{3}\approx-0.67,\\ f(-0.5)&=\frac{-0.5}{-0.75}\approx0.67,\\ f(0.5)&=\frac{0.5}{-0.75}\approx-0.67,\\ f(2)&=\frac{2}{3}\approx0.67. \end{aligned} \]

Connecting these points with the asymptotes gives the characteristic “S‑shaped’’ curve that lies in the four separate intervals \((-\infty,-1),\;(-1,0),\;(0,1),\;(1,\infty)\).

10. Exam‑style Tips & Links to Assessment

• Read the question carefully – the syllabus often asks for domain, range, intercepts, asymptotes and a sketch. Use the checklist in section 6.
• When an inverse is required, first check the function is one‑to‑one (horizontal‑line test). If not, state the required domain restriction.
• For composite‑function questions, always write down the domain of the inner function and then test whether the outer function is defined for those outputs.
• Mark clearly any algebraic steps used to find a domain or an inverse – examiners award marks for method as well as final answer.
• Past papers (Cambridge 9709) contain several “sketch the graph’’ questions; practising them with the systematic checklist will improve speed and accuracy.