Functions

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Functions — Quick Facts

Definition: A function $f: A \to B$ assigns to each element of set $A$ exactly one element in set $B$.

• $A$ is the domain (input set)
• $B$ is the codomain (possible outputs)
• The range is the actual set of outputs achieved

Function Notation:

• $f(x) = y$ means the function $f$ maps $x$ to $y$
• $f: x \mapsto 2x + 1$ means “f maps x to 2x + 1”

Common Functions:

• Linear: $f(x) = ax + b$ (straight line)
• Quadratic: $f(x) = ax^2 + bx + c$ (parabola)
• Polynomial: $f(x) = a_n x^n + … + a_0$
• Modulus: $f(x) = |x|$ (V-shape, always non-negative)

Even and Odd Functions:

• Even: $f(-x) = f(x)$ (symmetric about y-axis)
• Odd: $f(-x) = -f(x)$ (symmetric about origin)

⚡ CAT Exam Tip: To find range from an expression like $f(x) = \frac{x+1}{x-2}$, find domain restrictions first, then determine what values $f(x)$ can never take.

---

🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Functions — Study Guide

Domain Restrictions:

Expression Type	Restriction
$\frac{1}{g(x)}$	$g(x) \neq 0$
$\sqrt{g(x)}$	$g(x) \geq 0$
$\log g(x)$	$g(x) > 0$
$\frac{\sqrt{g(x)}}{h(x)}$	$g(x) \geq 0$, $h(x) \neq 0$

Example: Find domain of $f(x) = \frac{1}{\sqrt{x-3}} + \frac{1}{x-5}$

Requirements:

• $x - 3 > 0 \Rightarrow x > 3$ (for square root in denominator)
• $x - 5 \neq 0 \Rightarrow x \neq 5$

Domain: $(3, 5) \cup (5, \infty)$

Composite Functions:

If $f: A \to B$ and $g: B \to C$, then $g \circ f: A \to C$ $(g \circ f)(x) = g(f(x))$

Example: If $f(x) = x + 1$ and $g(x) = x^2$, find $g(f(3))$.

$g(f(3)) = g(4) = 4^2 = 16$

Inverse Functions:

$f^{-1}(x)$ reverses the mapping of $f(x)$.

For $f(x) = 2x + 3$:

1. Let $y = 2x + 3$
2. Solve for $x$: $x = \frac{y - 3}{2}$
3. Replace $y$ with $x$: $f^{-1}(x) = \frac{x - 3}{2}$

Condition for inverse: $f$ must be one-to-one (injective)

⚡ Common Student Mistake: Thinking $f^{-1}(x) = \frac{1}{f(x)}$. This is actually the reciprocal, not the inverse function.

---

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Functions — Comprehensive Notes

Types of Functions:

1. One-to-One (Injective): Each element of domain maps to a unique element in range. $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$

2. Onto (Surjective): Every element of codomain is an output. $\forall y \in B, \exists x \in A: f(x) = y$

3. Bijective: Both one-to-one AND onto. Inverse function exists.

Piecewise Functions:

$$f(x) = \begin{cases} x^2 & \text{if } x < 0 \ 2x + 1 & \text{if } x \geq 0 \end{cases}$$

Example: Evaluate $f(-2) + f(3)$ for the above.

$f(-2) = (-2)^2 = 4$ $f(3) = 2(3) + 1 = 7$ Sum = $11$

Range of Quadratic Functions:

For $f(x) = ax^2 + bx + c$:

• If $a > 0$ (opens upward): minimum at $x = -\frac{b}{2a}$, range = $[y_{\min}, \infty)$
• If $a < 0$ (opens downward): maximum at $x = -\frac{b}{2a}$, range = $(-\infty, y_{\max}]$

Example: Find range of $f(x) = -x^2 + 4x - 3$

Vertex: $x = -\frac{4}{2(-1)} = 2$ $f(2) = -4 + 8 - 3 = 1$ Since $a < 0$, maximum = 1, range = $(-\infty, 1]$

Maximum Value Problem:

For $f(x) = -2x^2 + 8x - 3$, maximum occurs at $x = -\frac{b}{2a} = 2$, $f(2) = 5$

Modulus Function Properties:

• $|x| \geq 0$ for all $x$
• $|xy| = |x||y|$
• $|x + y| \leq |x| + |y|$ (triangle inequality)
• $|x - y| \geq ||x| - |y||$

Solving Modulus Equations:

Example: $|2x - 3| = 5$

Case 1: $2x - 3 = 5 \Rightarrow x = 4$ Case 2: $2x - 3 = -5 \Rightarrow x = -1$

JAMB Pattern Analysis (CAT 2015-2024):

• 2015: Finding domain of rational + square root functions
• 2017: Composite function evaluation
• 2019: Inverse function
• 2021: Range of quadratic (max/min)
• 2023: Piecewise function evaluation
• 2024: Modulus equation solving

⚡ Exam Strategy: When dealing with composite functions $g(f(x))$, first find the output of $f(x)$, then substitute that into $g$. When finding inverse, always swap $x$ and $y$ at the replacement step.