Chapter 1: Functions
1.1 Definition of a Function, Domain, Range, One-One Functions
Definition (formal):
A function \(f\) from set \(A\) to set \(B\) is a rule that assigns to each element \(x \in A\) exactly one element \(f(x) \in B\).
We write: \(f: A \to B\) or \(f(x) = \text{expression}\).
Key terminology:
| Term |
Symbol |
Meaning |
| Domain |
\(D_f\) |
Set of all possible input values \(x\) |
| Codomain |
\(B\) |
Set that contains all possible outputs |
| Range (image set) |
\(R_f\) |
Set of actual outputs \(\{ f(x) : x \in D_f \}\) |
| One-one (injective) |
-- |
Each output comes from exactly one input: \(f(a) = f(b) \Rightarrow a = b\) |
Proof: A quadratic function \(f(x) = x^2\) is not one-one on \(\mathbb{R}\)
Take \(a = 2\) and \(b = -2\). Then \(f(2) = 4\) and \(f(-2) = 4\). So \(f(a) = f(b)\) but \(a \neq b\). This violates the one-one condition.
Proof: A linear function \(f(x) = mx + c\) with \(m \neq 0\) is one-one
Assume \(f(a) = f(b)\): \(ma + c = mb + c\). Subtract \(c\): \(ma = mb\). Divide by \(m \neq 0\): \(a = b\). Hence \(f\) is one-one.
Worked Example 1.1 (Finding domain and range)
For \(f(x) = \sqrt{x - 2}\), find:
- Domain
- Range
Solution:
Domain: Square root requires \(x - 2 \ge 0 \Rightarrow x \ge 2\). So \(D_f = [2, \infty)\).
Range: As \(x\) increases from 2 to \(\infty\), \(\sqrt{x-2}\) increases from 0 to \(\infty\). So \(R_f = [0, \infty)\).
Worked Example 1.2 (Testing one-one)
Determine whether \(f(x) = \frac{1}{x}\) (for \(x \neq 0\)) is one-one.
Solution:
Assume \(f(a) = f(b)\): \(\frac{1}{a} = \frac{1}{b}\). Cross-multiply: \(b = a\). So \(f\) is one-one.
1.2 Composite Functions
Definition:
Given \(f: A \to B\) and \(g: B \to C\), the composite function \(f \circ g\) (written \(fg\)) is:
\[
(f \circ g)(x) = f(g(x))
\]
Read as "\(f\) of \(g\) of \(x\)". Apply \(g\) first, then \(f\).
Proof: Composition of two one-one functions is one-one
Assume \(h(a) = h(b)\): \(f(g(a)) = f(g(b))\). Since \(f\) is one-one, \(g(a) = g(b)\). Since \(g\) is one-one, \(a = b\). Thus \(h\) is one-one.
Worked Example 1.3 (Composite functions)
Given \(f(x) = 2x + 1\) and \(g(x) = x^2 - 3\), find:
- \(fg(4)\)
- \(gf(4)\)
- \(fg(x)\) in simplified form
Solution:
(a) \(g(4) = 4^2 - 3 = 13\), \(f(13) = 2(13) + 1 = 27\) \(\Rightarrow\) \(fg(4) = 27\)
(b) \(f(4) = 2(4) + 1 = 9\), \(g(9) = 9^2 - 3 = 78\) \(\Rightarrow\) \(gf(4) = 78\)
(c) \(fg(x) = f(g(x)) = f(x^2 - 3) = 2(x^2 - 3) + 1 = 2x^2 - 5\)
1.3 Inverse Functions
Definition:
If \(f\) is a one-one function with domain \(A\) and range \(B\), then its inverse function \(f^{-1}\) has:
- Domain \(B\)
- Range \(A\)
- Property: \(f^{-1}(y) = x \iff f(x) = y\)
\(f^{-1}(f(x)) = x\) for all \(x \in D_f\)
\(f(f^{-1}(x)) = x\) for all \(x \in D_{f^{-1}}\)
Proof: Only one-one functions have inverses
Suppose \(f\) is not one-one. Then there exist \(a \neq b\) with \(f(a) = f(b) = c\). If an inverse function \(f^{-1}\) existed, it would have to send \(c\) to both \(a\) and \(b\). But a function cannot send one input to two outputs. Contradiction.
Worked Example 1.5 (Finding an inverse algebraically)
Find the inverse of \(f(x) = \frac{2x - 3}{5}\).
Solution:
Step 1: Write \(y = \frac{2x - 3}{5}\)
Step 2: Swap \(x\) and \(y\): \(x = \frac{2y - 3}{5}\)
Step 3: Solve for \(y\): \(5x = 2y - 3 \Rightarrow 2y = 5x + 3 \Rightarrow y = \frac{5x + 3}{2}\)
Step 4: Write \(f^{-1}(x) = \frac{5x + 3}{2}\)
1.4 Modulus Functions
Definition:
\[
|x| = \begin{cases}
x & \text{if } x \ge 0 \\
-x & \text{if } x < 0
\end{cases}
\]
Properties of Modulus
- \(|x| \ge 0\) for all \(x\)
- \(|x| = |-x|\)
- \(|ab| = |a||b|\)
- \(|a + b| \le |a| + |b|\) (triangle inequality)
1.5 Relationship Between \(y = f(x)\) and \(y = |f(x)|\)
The graph of \(y = |f(x)|\) is obtained by:
- Sketch \(y = f(x)\) as normal
- Where \(f(x) \ge 0\), keep the graph unchanged
- Where \(f(x) < 0\), reflect that portion across the x-axis
Chapter 1 Summary
| Concept |
Key points |
| Function |
Each input \(\rightarrow\) exactly one output |
| Domain |
Allowed \(x\) values |
| Range |
All possible \(y = f(x)\) values |
| One-one |
\(f(a) = f(b) \Rightarrow a = b\) |
| Composite \(fg\) |
Apply \(g\) first, then \(f\): \(fg(x) = f(g(x))\) |
| Inverse \(f^{-1}\) |
Exists only for one-one functions |
| Modulus \(|f(x)|\) |
Reflect negative parts above x-axis |
Exercises – Chapter 1
Easy (Drill – Non-Calculator)
- For \(f(x) = 4x - 7\), find:
(a) \(f(3)\) (b) \(f(0)\) (c) \(f(-2)\)
- State the domain and range of:
(a) \(f(x) = \sqrt{x + 5}\)
(b) \(g(x) = \frac{1}{x - 3}\)
(c) \(h(x) = x^2 - 4\) with domain \(-2 \le x \le 2\)
- Determine whether each function is one-one:
(a) \(f(x) = 3x + 2\)
(b) \(f(x) = x^4\)
(c) \(f(x) = \frac{2}{x}\)
Medium (Examination Style)
- Given \(f(x) = 5x - 1\) and \(g(x) = x^2 + 2\):
(a) Find \(fg(3)\)
(b) Find \(gf(3)\)
(c) Find \(f^2(x)\) in simplest form
- Find the inverse of:
(a) \(f(x) = \frac{3x + 4}{2}\)
(b) \(f(x) = x^2 - 1\) for \(x \ge 0\)
- Sketch \(y = |2x - 6|\) for \(0 \le x \le 6\). Label the vertex and intercepts.
- Given \(f(x) = \sqrt{x}\) and \(g(x) = x - 9\), find the domain of \(fg\).
Hard (Challenge for A*)
- Prove that if \(f\) and \(g\) are both one-one, then \(f^{-1} \circ g^{-1} = (g \circ f)^{-1}\).
- The function \(f\) is defined by \(f(x) = \frac{ax + b}{cx - a}\), \(x \neq \frac{a}{c}\). Prove that \(f(f(x)) = x\) for all \(x\) in the domain. Hence find \(f^{-1}(x)\).
- Mauritian context: A mobile phone plan charges Rs 200 per month plus Rs 1.50 per minute. The cost \(C(m) = 200 + 1.5m\) for \(m \ge 0\).
(a) Is \(C(m)\) one-one? Explain.
(b) Find the inverse function and interpret its meaning in context.
(c) If a customer's bill is Rs 350, how many minutes did they use?