Functions and Equations

Function Notation

A function $f$ maps each element of a set (the domain) to exactly one element of another set (the codomain).

F: X \to Y

We write $f(x) = y$ where $x$ is the input (independent variable) and $y$ is the output (dependent Variable).

Key Terminology

Term	Definition
Domain	Set of all valid inputs
Codomain	Set of possible outputs
Range	Set of actual outputs (subset of codomain)
Argument	The input value, e.g., $x$ in $f(x)$
Image	The output for a given input

Vertical Line Test

A relation is a function if and only if every vertical line intersects the graph at most once.

Domain and Range

Finding the Domain

The domain of a real-valued function is restricted by:

Denominators must be non-zero: $g(x) \neq 0$
Square roots must have non-negative arguments: $g(x) \ge 0$
Logarithms must have positive arguments: $g(x) \gt 0$

Given $f(x) = \sqrt{x}$ and $g(x) = x - 5$ Find the domain of $f \circ g$ .

(f \circ g)(x) = f(x - 5) = \sqrt\{x - 5\}

We need $x - 5 \ge 0$ So $x \ge 5$ .

Domain of $f \circ g$ : $[5, \infty)$ .

Inverse Functions

Definition

The inverse function $f^{-1}$ of $f$ satisfies:

F^\{-1\}(f(x)) = x \quad \mathrm\{and\} \quad f(f^\{-1\}(x)) = x

Existence of Inverses

A function has an inverse if and only if it is one-to-one (injective), meaning each output comes from exactly one input. This is verified by the horizontal line test: no horizontal line Intersects the graph more than once.

Finding the Inverse

Write $y = f(x)$ .
Swap $x$ and $y$ .
Solve for $y$ .
Replace $y$ with $f^{-1}(x)$ .

Find the inverse of $f(x) = \dfrac{2x + 3}{x - 1}$ .

Y = \frac\{2x + 3\}\{x - 1\}

Swap $x$ and $y$ :

X = \frac\{2y + 3\}\{y - 1\}

X(y - 1) = 2y + 3

Xy - x = 2y + 3

Xy - 2y = x + 3

Y(x - 2) = x + 3

F^\{-1\}(x) = \frac\{x + 3\}\{x - 2\}

Domain and Range of Inverses

The domain of $f^{-1}$ equals the range of $f$ And the range of $f^{-1}$ equals the domain of $f$ .

Graph of Inverse Functions

The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ .

Restricting Domains

Functions that are not one-to-one on their natural domain can have inverses if their domain is Restricted.

$f(x) = x^2$ is not one-to-one on $\mathbb{R}$ But $f: [0, \infty) \to [0, \infty)$ defined by $f(x) = x^2$ has inverse $f^{-1}(x) = \sqrt{x}$ .

Function Transformations

Summary of Transformations

Given $y = f(x)$ :

Transformation	Effect on Graph	Equation
Vertical translation up by $k$	Moves up $k$ units	$y = f(x) + k$
Vertical translation down by $k$	Moves down $k$ units	$y = f(x) - k$
Horizontal translation right by $h$	Moves right $h$ units	$y = f(x - h)$
Horizontal translation left by $h$	Moves left $h$ units	$y = f(x + h)$
Vertical stretch by factor $a$	Stretches vertically by $a$	$y = af(x)$
Vertical compression by factor $a$	Compresses by $\dfrac{1}{a}$	$y = af(x)$ where $0 \lt a \lt 1$
Horizontal stretch by factor $b$	Stretches horizontally by $b$	$y = f\!\left(\dfrac{x}{b}\right)$
Reflection in $x$ -axis	Flips vertically	$y = -f(x)$
Reflection in $y$ -axis	Flips horizontally	$y = f(-x)$
Reflection in $y = x$	Swaps $x$ and $y$	$y = f^{-1}(x)$

Describe the sequence of transformations that maps $f(x) = x^2$ to $g(x) = 2(x-3)^2 + 1$ .

Starting from $f(x) = x^2$ :

Translate right by 3: $(x-3)^2$
Vertical stretch by factor 2: $2(x-3)^2$
Translate up by 1: $2(x-3)^2 + 1$

The vertex moves from $(0, 0)$ to $(3, 1)$ And the parabola is narrower.

Effect on Key Points

Point on $y = f(x)$	Point on $y = f(x-h)+k$
$(x, y)$	$(x+h, y+k)$

Graphing Functions

Key Features to Identify

Domain and range
Intercepts: $x$ -intercepts (zeros) and $y$ -intercept
Symmetry: even ( $f(-x) = f(x)$ ), odd ( $f(-x) = -f(x)$ ), periodic
Asymptotes: vertical, horizontal, oblique
Stationary points: local maxima and minima
End behaviour: as $x \to \pm\infty$

Asymptotes

Vertical asymptotes occur at values of $x$ where the function is undefined and the function Approaches $\pm\infty$ .

Horizontal asymptotes describe the behaviour as $x \to \pm\infty$ .

For rational functions $\dfrac{P(x)}{Q(x)}$ :

If $\deg P \lt \deg Q$ : horizontal asymptote at $y = 0$ .
If $\deg P = \deg Q$ : horizontal asymptote at $y = \dfrac{\mathrm{leading coefficient of } P}{\mathrm{leading coefficient of } Q}$ .
If $\deg P = \deg Q + 1$ : oblique asymptote (found by polynomial division).

Divide $x^3 + 2x^2 - 5x - 6$ by $(x + 1)$ using synthetic division.

-1 | 1   2   -5   -6
   |     -1   -1    6
   |----------------
     1   1   -6    0

Result: $x^2 + x - 6 = (x+3)(x-2)$ .

So $x^3 + 2x^2 - 5x - 6 = (x+1)(x+3)(x-2)$ .

Sum and Product of Roots

For $ax^n + bx^{n-1} + \cdots = 0$ with roots $\alpha, \beta, \gamma, \ldots$ :

Quadratic ( $ax^2 + bx + c = 0$ ):

\alpha + \beta = -\frac\{b\}\{a\}, \quad \alpha\beta = \frac\{c\}\{a\}

Cubic ( $ax^3 + bx^2 + cx + d = 0$ ):

\alpha + \beta + \gamma = -\frac\{b\}\{a\}, \quad \alpha\beta + \beta\gamma + \gamma\alpha = \frac\{c\}\{a\}, \quad \alpha\beta\gamma = -\frac\{d\}\{a\}

Inequalities

Linear Inequalities

Ax + b \gt 0 \implies x \gt -\frac\{b\}\{a\} \quad (\mathrm\{if \} a \gt 0)

Solve simultaneously: $y = x^2 - 4x + 3$ and $y = 2x - 3$ .

Substitute: $2x - 3 = x^2 - 4x + 3$ .

X^2 - 6x + 6 = 0

X = \frac\{6 \pm \sqrt\{36 - 24\}\}\{2\} = \frac\{6 \pm 2\sqrt\{3\}\}\{2\} = 3 \pm \sqrt\{3\}

When $x = 3 + \sqrt{3}$ : $y = 2(3 + \sqrt{3}) - 3 = 3 + 2\sqrt{3}$ .

When $x = 3 - \sqrt{3}$ : $y = 2(3 - \sqrt{3}) - 3 = 3 - 2\sqrt{3}$ .

Modulus Functions

Definition

|x| = \begin`\{cases}` x & x \ge 0 \\ -x & x \lt 0 \end`\{cases}`

Graph

The graph of $y = |x|$ is V-shaped, with the vertex at the origin.

Solving Modulus Equations

Square both sides or use the definition casewise.

Solve $|x - 2| = 3x - 1$ .

Since $|x - 2| \ge 0$ We need $3x - 1 \ge 0 \implies x \ge \dfrac{1}{3}$ .

Case 1 ( $x \ge 2$ ): $x - 2 = 3x - 1 \implies -2x = 1 \implies x = -\dfrac{1}{2}$ . Rejected ( $x \ge 2$ ).

Case 2 ( $x \lt 2$ ): $-(x - 2) = 3x - 1 \implies -x + 2 = 3x - 1 \implies 4x = 3 \implies x = \dfrac{3}{4}$ .

Check: $|3/4 - 2| = 5/4$ and $3(3/4) - 1 = 5/4$ . Valid.

Solution: $x = \dfrac{3}{4}$ .

IB Exam-Style Questions

Question 1 (Paper 1 style)

Given $f(x) = \dfrac{x}{x + 2}$ and $g(x) = 2x - 1$ :

(a) Find $(f \circ g)(x)$ and state its domain.

(f \circ g)(x) = f(2x - 1) = \frac\{2x - 1\}\{2x - 1 + 2\} = \frac\{2x - 1\}\{2x + 1\}

Domain: $2x + 1 \neq 0 \implies x \neq -\dfrac{1}{2}$ .

(b) Find $f^{-1}(x)$ .

Y = \frac\{x\}\{x+2\} \implies y(x+2) = x \implies xy + 2y = x \implies x(1-y) = 2y

F^\{-1\}(x) = \frac\{2x\}\{1 - x\}, \quad x \neq 1

(c) Verify that $f^{-1} \circ f$ is the identity function.

(f^\{-1\} \circ f)(x) = f^\{-1\}\!\left(\frac\{x\}\{x+2\}\right) = \frac\{2 \cdot \frac\{x\}\{x+2\}\}\{1 - \frac\{x\}\{x+2\}\} = \frac\{\frac\{2x\}\{x+2\}\}\{\frac\{2\}\{x+2\}\} = x

Question 2 (Paper 2 style)

The function $f$ is defined by $f(x) = 2x^2 - 12x + 13$ for $x \ge 3$ .

(a) Express $f(x)$ in the form $a(x - h)^2 + k$ .

F(x) = 2(x^2 - 6x) + 13 = 2(x - 3)^2 - 18 + 13 = 2(x - 3)^2 - 5

(b) Find the range of $f$ .

Since $x \ge 3$ and $(x-3)^2 \ge 0$ : $f(x) \ge -5$ .

Range: $[-5, \infty)$ .

(c) Find $f^{-1}(x)$ and state its domain.

Y = 2(x-3)^2 - 5 \implies y + 5 = 2(x-3)^2 \implies (x-3)^2 = \frac\{y+5\}\{2\}

Since $x \ge 3$$x - 3 \ge 0$ :

X = 3 + \sqrt\{\frac\{y+5\}\{2\}\}

F^\{-1\}(x) = 3 + \sqrt\{\frac\{x+5\}\{2\}\}

Domain of $f^{-1}$ = range of $f$ : $[-5, \infty)$ .

Question 3 (Paper 1 style)

Solve the inequality $x^2 - 2x - 8 \gt 0$ .

(x - 4)(x + 2) \gt 0

The product is positive when both factors are positive or both are negative:

$x \gt 4$ or $x \lt -2$

Solution: $x \in (-\infty, -2) \cup (4, \infty)$ .

Question 4 (Paper 2 style)

The function $f$ is defined as $f(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \neq 3$ .

(a) Simplify $f(x)$ .

F(x) = \frac\{(x-3)(x+3)\}\{x-3\} = x + 3 \quad \mathrm\{for \} x \neq 3

(b) Find the equations of any asymptotes of $f$ .

There is a hole at $x = 3$ (removable discontinuity), not a vertical asymptote.

No horizontal asymptote (it behaves like $y = x + 3$ for large $x$ ).

(c) Sketch the graph of $f$ .

The graph is the line $y = x + 3$ with a hole at $(3, 6)$ .

Question 5 (Paper 1 style)

The cubic $P(x) = x^3 + ax^2 + bx - 12$ has a factor of $(x + 3)$ and leaves a remainder of $-20$ when divided by $(x - 1)$ . Find $a$ and $b$ .

Since $(x + 3)$ is a factor: $P(-3) = 0$ .

-27 + 9a - 3b - 12 = 0 \implies 9a - 3b = 39 \implies 3a - b = 13 \quad \mathrm\{(1)\}

Since $P(1) = -20$ :

1 + a + b - 12 = -20 \implies a + b = -9 \quad \mathrm\{(2)\}

Adding (1) and (2): $4a = 4 \implies a = 1$ .

From (2): $b = -10$ .

Summary

Concept	Key Point
Composite function	$(f \circ g)(x) = f(g(x))$ ; order matters
Inverse function	Reflect in $y = x$ ; swap domain/range
Vertical shift	$y = f(x) + k$ moves up by $k$
Horizontal shift	$y = f(x - h)$ moves right by $h$
Factor theorem	$(x-a)$ factor $\iff P(a) = 0$
Remainder theorem	Remainder of $P(x) \div (x-a)$ is $P(a)$
Even function	$f(-x) = f(x)$ Symmetric about $y$ -axis
Odd function	$f(-x) = -f(x)$ Rotational symmetry about origin

For function questions, always check the domain. When finding inverses, state the domain of the Inverse explicitly. For transformation questions, identify each transformation step by step from the inside out.

Reciprocal Functions

Definition

The reciprocal function of $f$ is $\dfrac{1}{f(x)}$ .

Graphing Reciprocal Functions

Key features of the graph of $y = \dfrac{1}{f(x)}$ :

Where $f(x) = 1$ The reciprocal also equals $1$ .
Where $f(x) = -1$ The reciprocal also equals $-1$ .
Where $f(x) \gt 0$ The reciprocal is positive.
Where $f(x) \lt 0$ The reciprocal is negative.
Where $f(x) = 0$ The reciprocal has a vertical asymptote.
Horizontal asymptotes of $f$ become horizontal asymptotes of $\dfrac{1}{f}$ .
Local maxima of $f$ become local minima of $\dfrac{1}{f}$ and vice versa.

Reciprocal of $f(x) = ax + b$

Y = \frac\{1\}\{ax + b\}

This is a rectangular hyperbola with vertical asymptote at $x = -\dfrac{b}{a}$ and horizontal Asymptote at $y = 0$ .

Reciprocal of Quadratic Functions

Sketch the graph of $y = \dfrac{1}{x^2 - 4}$ .

Vertical asymptotes at $x = 2$ and $x = -2$ (zeros of denominator).

Horizontal asymptote at $y = 0$ .

For $x \lt -2$ : denominator positive, so $y \gt 0$ .

For $-2 \lt x \lt 2$ : denominator negative, so $y \lt 0$ .

For $x \gt 2$ : denominator positive, so $y \gt 0$ .

Local minimum at $x = 0$ : $y = -\dfrac{1}{4}$ .

Rational Functions

Definition

A rational function is a ratio of two polynomials:

F(x) = \frac\{P(x)\}\{Q(x)\}

Features to Identify

Domain: values of $x$ where $Q(x) \neq 0$ .
Intercepts: $y$ -intercept (set $x = 0$ ), $x$ -intercepts (set $P(x) = 0$ ).
Asymptotes: vertical (zeros of $Q$ ), horizontal (compare degrees), oblique.
Behaviour near asymptotes: test values on each side.

Oblique Asymptotes

When $\deg P = \deg Q + 1$ Divide $P$ by $Q$ using polynomial division. The quotient (without Remainder) gives the oblique asymptote.

Additional Exam-Style Questions

Question 6 (Paper 2 style)

The function $f$ is defined as $f(x) = \dfrac{2x + 3}{x - 1}$ for $x \in \mathbb{R}$$x \neq 1$ .

(a) Find the inverse function $f^{-1}$ .

Y = \frac\{2x + 3\}\{x - 1\} \implies y(x-1) = 2x + 3 \implies yx - y = 2x + 3

X(y - 2) = y + 3 \implies x = \frac\{y + 3\}\{y - 2\}

F^\{-1\}(x) = \frac\{x + 3\}\{x - 2\}, \quad x \neq 2

(b) State the domain and range of $f^{-1}$ .

Domain of $f^{-1}$ : $x \neq 2$ .

Range of $f^{-1}$ : $y \neq 1$ (which equals the domain of $f$ ).

(c) Find the value of $x$ such that $f(x) = f^{-1}(x)$ .

\frac\{2x+3\}\{x-1\} = \frac\{x+3\}\{x-2\}

(2x+3)(x-2) = (x+3)(x-1)

2x^2 - x - 6 = x^2 + 2x - 3

X^2 - 3x - 3 = 0

X = \frac\{3 \pm \sqrt\{9+12\}\}\{2\} = \frac\{3 \pm \sqrt\{21\}\}\{2\}

Question 7 (Paper 2 style)

Given $f(x) = x^2 - 4x + 3$ and $g(x) = 2x - 1$ :

(a) Find $(g \circ f)(x)$ and its range.

(g \circ f)(x) = g(x^2 - 4x + 3) = 2(x^2 - 4x + 3) - 1 = 2x^2 - 8x + 5

Completing the square: $2(x - 2)^2 - 3$ .

Range: $[-3, \infty)$ .

(b) Find the set of values of $x$ for which $(f \circ g)(x) \lt 0$ .

(f \circ g)(x) = (2x-1)^2 - 4(2x-1) + 3 = 4x^2 - 4x + 1 - 8x + 4 + 3 = 4x^2 - 12x + 8

4x^2 - 12x + 8 \lt 0 \implies x^2 - 3x + 2 \lt 0 \implies (x-1)(x-2) \lt 0

Solution: $1 \lt x \lt 2$ .

Question 8 (Paper 1 style)

The function $h$ is defined by $h(x) = |2x - 5| + |x + 1|$ for all real $x$ .

Find the minimum value of $h(x)$ .

Critical points at $x = 2.5$ and $x = -1$ .

Case 1 ( $x \lt -1$ ): $h(x) = -(2x-5) + -(x+1) = -3x + 4$ . Minimum at $x = -1$ : $h(-1) = 7$ .

Case 2 ( $-1 \le x \le 2.5$ ): $h(x) = -(2x-5) + (x+1) = -x + 6$ . Minimum at $x = 2.5$ : $h(2.5) = 3.5$ .

Case 3 ( $x \gt 2.5$ ): $h(x) = (2x-5) + (x+1) = 3x - 4$ . Minimum at $x = 2.5$ : $h(2.5) = 3.5$ .

Minimum value is $3.5$ at $x = 2.5$ .

For the A-Level treatment of this topic, see Functions.

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Unit tests probe edge cases and common misconceptions. Integration tests combine Functions and Equations with other IB mathematics topics to test synthesis under exam conditions.

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Intuition

Functions are machines that take an input and produce an output — like a coffee grinder that turns beans into grounds. Composition is chaining machines together: the output of one becomes the input of the next. Inverses are reverse machines that undo the original: if f turns beans into grounds, f-inverse turns grounds back into beans. Transformations are like moving, stretching, or flipping a photograph — translations slide it, reflections flip it, stretches resize it. The vertical line test is a simple rule: if any vertical line hits the graph more than once, it is not a function because one input produces multiple outputs.

Common Pitfalls

Dropping negative signs during algebraic manipulation. Substitute back to verify your answer.
Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).
Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).

Cross-References

Topic	Site	Link
[Functions]	A-Level	View
[Functions]	IB	View
[Functions]	DSE	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Functions and Equations

Function Notation

Key Terminology

Vertical Line Test

Domain and Range

Finding the Domain

Additional Exam-Style Questions

Question 6 (Paper 2 style)

Question 7 (Paper 2 style)

Question 8 (Paper 1 style)

See Also