Find the inverse of a one–one function using correct notation

Functions – Inverses of One‑One Functions (IGCSE / A‑Level)

Learning Objective

By the end of this unit students will be able to:

• state the precise definitions required by the Cambridge syllabus,
• determine the domain and range of a wide variety of functions,
• recognise when a function is one‑one (injective) or many‑one,
• explain verbally and diagrammatically why a non‑injective function has no inverse,
• find the inverse of a one‑one function using the correct notation,
• form and evaluate composite functions, noting that the order of composition matters.

1.1 – Key Definitions (Outcome 1.1)

• Function: a relation that assigns to each element x in a set called the domain exactly one element y in a set called the range. Written y = f(x).
• Domain: the set of all admissible inputs for the function.
• Range: the set of all possible outputs produced by the function.
• One‑one (injective) function: f is one‑one if f(a)=f(b) implies a=b. Graphically, no horizontal line meets the graph more than once (horizontal‑line test).
• Many‑one function: a function that is not one‑one; at least two different inputs give the same output.
  Example: k(x)=x^{2} on the unrestricted domain ℝ. The horizontal line y=4 meets the graph at x=-2 and x=2, so k is many‑one.
• Inverse function: for a one‑one function f, the inverse f^‑1 satisfies
  f^‑1(f(x)) = x for every x in the domain of f, and
  f(f^‑1(y)) = y for every y in the range of f. The graph of f^‑1 is the reflection of the graph of f in the line y = x.
• Composite function: given functions f and g, the composition (f ∘ g)(x) means “apply g first, then f”. In general f ∘ g ≠ g ∘ f.
• Function notation:
  • f(x) – value of f at x.
  • f^{-1}(x) – inverse function, not the reciprocal of f(x).
  • f\circ g – composition.
  • f: x \mapsto \lg x \;(x>0) – mapping description; translates to f(x)=\lg x with domain x>0.

1.2 – Finding Domain and Range (Outcome 1.2)

Three typical function types are shown below, followed by a note on composites and rational expressions.

Function	Domain (how to find it)	Resulting Domain	Range (how to find it)	Resulting Range
\(f(x)=\dfrac{1}{x-2}\)	Denominator ≠ 0 → \(x-2eq0\)	\(\mathbb{R}\setminus\{2\}\)	Vertical asymptote at \(x=2\); all real values except 0 are taken.	\(\mathbb{R}\setminus\{0\}\)
\(g(x)=\sqrt{x-3}\)	Radicand ≥ 0 → \(x-3\ge0\)	\([3,\infty)\)	Square‑root outputs are ≥ 0.	\([0,\infty)\)
\(h(x)=\sin x\)	All real numbers are allowed for the sine function.	\(\mathbb{R}\)	\(-1\le\sin x\le1\)	\([-1,1]\)

Composite functions: to find the domain of \( (f\circ g)(x)=f(g(x))\) first determine the domain of \(g\); then restrict it further so that every value produced by \(g\) lies in the domain of \(f\).

Rational expressions with quadratic denominators: for \(\displaystyle r(x)=\frac{x+1}{x^{2}-4}\) the denominator must be non‑zero, i.e. \(xeq\pm2\). Any additional restrictions (e.g. square‑root in the numerator) are applied afterwards.

1.3 – Recognising and Using Function Notation (Outcome 1.3)

Translate the following mapping description into function notation and state the domain:

• \(f: x \mapsto \lg x\) for \(x>0\) → \(f(x)=\lg x,\; \text{Domain } (0,\infty)\).

Conversely, write the function \(g(x)=\dfrac{1}{x-5}\) as a mapping description:

• \(g: x \mapsto \dfrac{1}{x-5},\; xeq5\).

1.4 – Effect of Absolute Value on Injectivity (Outcome 1.4)

Absolute value can change a one‑one function into a many‑one function. The table below summarises the effect for common types, followed by a worked algebraic example.

Function type	Original \(f(x)\)	Is \(f\) one‑one?	\(|f(x)|\) – is it one‑one?
Linear	\(2x-3\)	Yes	No (symmetry about the zero of \(f\))
Quadratic	\(x^{2}\)	No	No (unchanged)
Cubic	\(x^{3}\)	Yes	No (e.g. \(|1^{3}|=|(-1)^{3}|=1\))
Trigonometric (restricted)	\(\sin x\) on \([-\pi/2,\pi/2]\)	Yes	No (both \(x\) and \(-x\) give the same positive value)

Worked example: Solve \(|f(x)|=4\) where \(f(x)=3x-1\).

1. Remove the absolute value: \(3x-1=4\)  or  \(3x-1=-4\).
2. First equation → \(3x=5\) → \(x=\dfrac{5}{3}\).
   Second equation → \(3x=-3\) → \(x=-1\).
3. Both solutions are valid because the original function is one‑one, but after applying \(|\;|\) we obtain two pre‑images for the same output, showing that \(|f|\) is many‑one and therefore does not possess an inverse on \(\mathbb{R}\).

1.5 – Why a Function Must Be One‑One to Have an Inverse (Outcome 1.5)

If a function fails the horizontal‑line test, at least one output has more than one pre‑image. Consequently the “undo” operation is not unique, so no inverse function can exist.

Checklist for a verbal justification:

• State that the horizontal‑line test fails (or give a specific \(y\) with two different \(x\) values).
• Explain that the same output would have to be mapped back to two different inputs.
• Conclude that a single‑valued inverse function cannot be defined.

Example: \(k(x)=x^{2}\) on \(\mathbb{R}\). The line \(y=9\) meets the graph at \(x=3\) and \(x=-3\). Because one output (\(9\)) corresponds to two inputs, there is no function \(k^{-1}\) with domain \(\mathbb{R}\).

1.6 – Procedure for Finding the Inverse of a One‑One Function (Outcome 1.6)

1. Write the equation \(y = f(x)\).
2. Interchange the symbols \(x\) and \(y\) (swap the roles of input and output).
3. Solve the new equation for \(y\). The solution is \(y = f^{-1}(x)\).
4. State the domain of the inverse (the range of the original) and the range of the inverse (the domain of the original).
5. Check the result by composition for at least one convenient value.

Worked Example A – Linear function

\[ \begin{aligned} y &= 3x+2\\ x &= 3y+2\\ 3y &= x-2\\ y &= \frac{x-2}{3} \end{aligned} \]

Therefore \(\displaystyle f^{-1}(x)=\frac{x-2}{3}\). Both domain and range are \(\mathbb{R}\).

Worked Example B – Restricted quadratic

Let \(f(x)=x^{2}\) with domain \(x\ge0\).

\[ \begin{aligned} y &= x^{2},\; x\ge0\\ x &= y^{2}\\ y &= \sqrt{x} \end{aligned} \]

Hence \(\displaystyle f^{-1}(x)=\sqrt{x}\) with domain \(x\ge0\) and range \(y\ge0\).

Worked Example C – Reciprocal (self‑inverse)

\[ \begin{aligned} y &= \frac{2}{x}\\ x &= \frac{2}{y}\\ y &= \frac{2}{x} \end{aligned} \]

Thus \(\displaystyle g^{-1}(x)=\frac{2}{x}\). Domain and range are both \(\mathbb{R}\setminus\{0\}\).

Worked Example D – Trigonometric with restricted domain

Let \(p(x)=\sin x\) with domain \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\). This interval makes \(\sin\) one‑one.

\[ \begin{aligned} y &= \sin x,\; -\frac{\pi}{2}\le x\le\frac{\pi}{2}\\ x &= \arcsin y \quad\text{(by definition of the inverse sine)}\\ \end{aligned} \]

Therefore \(\displaystyle p^{-1}(x)=\sin^{-1}x\) with domain \([-1,1]\) and range \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\).

1.7 – Forming and Using Composite Functions (Outcome 1.7)

Given two functions, always check that the inner function’s range lies inside the outer function’s domain.

Example 1 – Simple linear & quadratic

\[ \begin{aligned} (f\circ g)(x) &= f(g(x)) = 2(x^{2})+1 = 2x^{2}+1,\\ (g\circ f)(x) &= g(f(x)) = (2x+1)^{2}=4x^{2}+4x+1. \end{aligned} \]

Example 2 – Composite leading to a rational function

Let \(u(x)=\dfrac{1}{x}\) (domain \(xeq0\)) and \(v(x)=x+3\) (domain \(\mathbb{R}\)).

\[ \begin{aligned} (u\circ v)(x) &= u(v(x)) = \frac{1}{x+3},\qquad xeq-3,\\ (v\circ u)(x) &= v(u(x)) = \frac{1}{x}+3,\qquad xeq0. \end{aligned} \]

The two composites are different because the order of applying the operations matters.

Common Mistakes (and How to Avoid Them)

• Not restricting the domain of a non‑linear function before seeking an inverse.
• Leaving the placeholder variable x in the final answer instead of writing f^{-1}(x).
• Skipping the swap of x and y in step 2 of the procedure.
• Omitting the domain and range of the inverse (they are the range and domain of the original).
• Confusing \((f(x))^{-1}\) (the reciprocal) with \(f^{-1}(x)\) (the inverse function).
• Assuming \(|f(x)|\) is always one‑one because \(f\) is one‑one; the absolute value can destroy injectivity.

Practice Questions

1. Find the inverse of \(f(x)=5x-7\) and state its domain and range.
2. Given \(g(x)=\dfrac{2}{x}\) with domain \(xeq0\), determine \(g^{-1}(x)\).
3. For \(h(x)=\sqrt{x+4}\) (domain \(x\ge-4\)), find \(h^{-1}(x)\) and verify that \(h^{-1}(h(a))=a\) for a convenient value \(a\).
4. Explain why \(k(x)=x^{2}\) (no domain restriction) does not have an inverse function, using the checklist from Outcome 1.5.
5. Let \(p(x)=\sin x\) with domain \([-\pi/2,\pi/2]\). Find \(p^{-1}(x)\) and state its domain and range.
6. Given \(f(x)=2x+1\) and \(g(x)=x^{2}\), compute \((f\circ g)(x)\) and \((g\circ f)(x)\). Explain why the results differ.
7. Translate the mapping description “\(q: x \mapsto \lg x\) for \(x>0\)” into function notation and give its domain.
8. Solve \(|3x-2|=7\) and discuss whether the resulting function \(|3x-2|\) could have an inverse on \(\mathbb{R}\).

Summary Table (All Examples)

Function \(f(x)\)	Domain of \(f\)	Range of \(f\)	Inverse \(f^{-1}(x)\)	Domain of \(f^{-1}\)	Range of \(f^{-1}\)
\(3x+2\)	\(\mathbb{R}\)	\(\mathbb{R}\)	\(\dfrac{x-2}{3}\)	\(\mathbb{R}\)	\(\mathbb{R}\)
\(x^{2}\) ( \(x\ge0\) )	\([0,\infty)\)	\([0,\infty)\)	\(\sqrt{x}\)	\([0,\infty)\)	\([0,\infty)\)
\(\dfrac{2}{x}\) ( \(xeq0\) )	\(\mathbb{R}\setminus\{0\}\)	\(\mathbb{R}\setminus\{0\}\)	\(\dfrac{2}{x}\)	\(\mathbb{R}\setminus\{0\}\)	\(\mathbb{R}\setminus\{0\}\)
\(\sin x\) ( \([-\pi/2,\pi/2]\) )	\([-\pi/2,\pi/2]\)	\([-1,1]\)	\(\sin^{-1}x\)	\([-1,1]\)	\([-\pi/2,\pi/2]\)

Suggested Diagram

Sketch the graph of \(y=3x+2\) together with its inverse \(y=\dfrac{x-2}{3}\). Draw the line \(y=x\) and show that each point on the first graph reflects across this line to a point on the second graph.

Quick Checklist for the Exam (Outcome 1.6 & 1.7)

• Confirm the function is one‑one (or state the domain restriction that makes it one‑one).
• Write \(y=f(x)\); swap \(x\) and \(y\); solve for \(y\).
• Present the answer exactly as \(f^{-1}(x)=\dots\).
• State the domain of \(f^{-1}\) (original range) and the range of \(f^{-1}\) (original domain).
• Check by composition for at least one value: \(f^{-1}(f(a))=a\) (or \(f(f^{-1}(b))=b\)).
• When a composite is required, write the inner function first, verify that its range lies inside the outer function’s domain, then simplify.
• Read the question carefully: “inverse” never means “reciprocal”.