Functions — IB Math AA

1. Function Concepts

A function $f: A o B$ assigns each element of the domain $A$ exactly one element of the codomain $B$. The set of all output values is the range.

One-to-one (injective): Different inputs give different outputs. The horizontal line test: any horizontal line meets the graph at most once.
Onto (surjective): Every element of the codomain is achieved by at least one input.
Bijective: Both one-to-one and onto — necessary for an inverse function to exist over the full codomain.

Function Notation: $f(x)$ means "the output of $f$ when the input is $x$". The domain must be specified if it is not the natural (largest possible) domain. For example, $f(x) = \sqrt{x}$ has natural domain $x \geq 0$.

2. Transformations of Functions

Transformations map the graph of $y = f(x)$ to a new position or shape. The table below summarises the key transformations:

Transformation	New Function	Effect on Graph
Vertical shift up by $k$	$f(x) + k$	Graph moves up $k$ units
Vertical shift down by $k$	$f(x) - k$	Graph moves down $k$ units
Horizontal shift left by $k$	$f(x + k)$	Graph moves left $k$ units
Horizontal shift right by $k$	$f(x - k)$	Graph moves right $k$ units
Vertical stretch by factor $a$	$a \cdot f(x)$	$y$-values multiplied by $a$
Horizontal compression by $a$	$f(ax)$	$x$-values divided by $a$
Reflection in $x$-axis	$-f(x)$	$y$-values negated
Reflection in $y$-axis	$f(-x)$	$x$-values negated

General Transformed Form

$$y = a \cdot f(b(x - h)) + k$$

$a$: vertical stretch/reflection (if $a < 0$, reflect in $x$-axis)
$b$: horizontal compression/reflection (if $b < 0$, reflect in $y$-axis)
$h$: horizontal shift right by $h$
$k$: vertical shift up by $k$

3. Composite & Inverse Functions

Composite Functions

The composite function $(f \circ g)(x) = f(g(x))$ applies $g$ first, then $f$. Note that $f \circ g \neq g \circ f$ in general.

The domain of $f \circ g$ is the set of $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.

Inverse Functions

The inverse function $f^{-1}$ satisfies $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

Finding an Inverse Function

Write $y = f(x)$
Swap $x$ and $y$: write $x = f(y)$
Solve for $y$: this gives $y = f^{-1}(x)$

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

Domain Restriction: If $f$ is not one-to-one on its natural domain, we must restrict the domain before finding an inverse. For example, $f(x) = x^2$ has no inverse on $\mathbb{R}$, but restricting to $x \geq 0$ makes it one-to-one, and $f^{-1}(x) = \sqrt{x}$.

4. Key Function Families

You must recognise and sketch the following families of functions:

Linear: $f(x) = mx + c$ — straight line, gradient $m$, $y$-intercept $c$
Quadratic: $f(x) = a(x-h)^2 + k$ — parabola, vertex at $(h, k)$
Cubic: $f(x) = ax^3 + \cdots$ — S-shaped curve, one or three real roots
Absolute value: $f(x) = |x|$ — V-shape, vertex at origin
Reciprocal: $f(x) = \frac{1}{x}$ — hyperbola with asymptotes at $x=0$ and $y=0$
Exponential: $f(x) = a^x$ — always positive, horizontal asymptote $y=0$
Logarithm: $f(x) = \log_a x$ — defined for $x > 0$, vertical asymptote $x=0$

5. Mathematical Modelling

Choosing an appropriate function model for real-world data is a key IB skill.

Linear growth/decay → linear model $f(x) = mx + c$
Exponential growth/decay → $f(x) = A \cdot b^x$ or $f(x) = Ae^{kx}$
Periodic phenomena (tides, temperature) → sinusoidal model $f(x) = A\sin(Bx + C) + D$
Bounded growth → logistic model (HL)

Worked Examples

Worked Example 1

Transformations — Describing and Applying

The graph of $y = f(x)$ is transformed to $y = -2f(x+3) + 1$. Describe the sequence of transformations.

Horizontal shift: $f(x) o f(x+3)$: translate left by 3 units.

Vertical stretch: $f(x+3) o 2f(x+3)$: stretch vertically by factor 2 (away from $x$-axis).

Reflection: $2f(x+3) o -2f(x+3)$: reflect in the $x$-axis.

Vertical shift: $-2f(x+3) o -2f(x+3)+1$: translate up by 1 unit.

Worked Example 2

Composite Functions — Finding Domain and Range

Let $f(x) = x^2 + 1$ for $x \in \mathbb{R}$ and $g(x) = \sqrt{x-2}$ for $x \geq 2$. Find $(f \circ g)(x)$ and state its domain and range.

$(f \circ g)(x) = f(g(x)) = f(\sqrt{x-2}) = (\sqrt{x-2})^2 + 1 = (x-2) + 1 = x - 1$

Domain: $g(x)$ requires $x \geq 2$, and the output $\sqrt{x-2} \geq 0$ is in the domain of $f$ (all reals). So domain is $x \geq 2$.

Range: When $x \geq 2$, $(f \circ g)(x) = x - 1 \geq 1$. So range is $[1, \infty)$.

Worked Example 3

Inverse Functions — Finding and Verifying

Let $f(x) = \dfrac{2x+3}{x-1}$, $x \neq 1$. Find $f^{-1}(x)$ and verify that $f(f^{-1}(x)) = x$.

Write $y = \dfrac{2x+3}{x-1}$. Swap $x$ and $y$: $x = \dfrac{2y+3}{y-1}$.

Solve for $y$: $x(y-1) = 2y+3 \implies xy - x = 2y + 3 \implies y(x-2) = x+3 \implies y = \dfrac{x+3}{x-2}$

So $f^{-1}(x) = \dfrac{x+3}{x-2}$, $x \neq 2$.

Verification: $f(f^{-1}(x)) = \dfrac{2 \cdot \frac{x+3}{x-2} + 3}{\frac{x+3}{x-2} - 1} = \dfrac{\frac{2(x+3)+3(x-2)}{x-2}}{\frac{(x+3)-(x-2)}{x-2}} = \dfrac{5x}{5} = x$ ✓

Practice Problems

Q1. The function $f(x) = 3\sqrt{x-1}$ has domain $x \geq 1$. State the range of $f$ and find $f^{-1}(x)$, stating its domain.

Show Solution

Since $x \geq 1$, we have $\sqrt{x-1} \geq 0$, so $f(x) \geq 0$. Range of $f$: $[0, \infty)$.

To find $f^{-1}$: let $y = 3\sqrt{x-1}$. Swap: $x = 3\sqrt{y-1} \implies \frac{x}{3} = \sqrt{y-1} \implies \frac{x^2}{9} = y - 1 \implies y = \frac{x^2}{9} + 1$.

$f^{-1}(x) = \dfrac{x^2}{9} + 1$, domain $x \geq 0$.

Q2. Given $f(x) = 2x - 1$ and $g(x) = x^2 + 3$, find $(g \circ f)(x)$ and $(f \circ g)(x)$. Show that they are not equal.

Show Solution

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

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

Since $4x^2 - 4x + 4 \neq 2x^2 + 5$ in general (e.g., at $x=1$: $4 eq 7$), they are not equal.

Q3. The graph of $y = x^2$ is transformed to pass through the points $(1, 3)$ and $(-1, 3)$, and has vertex $(0, 0)$ shifted to vertex $(2, -1)$. Write the equation of the transformed graph.

Show Solution

The vertex form is $y = a(x-h)^2 + k$. Vertex at $(2,-1)$ gives $y = a(x-2)^2 - 1$.

The shape should pass through nearby points. If we use $a = 1$: $y = (x-2)^2 - 1$. Check: at $x=3$, $y=0$; at $x=2$, $y=-1$ ✓.

The equation is $\boxed{y = (x-2)^2 - 1}$.

Q4. A function is defined by $f(x) = e^x + 1$. Find $f^{-1}(x)$ and state its domain and range.

Show Solution

Let $y = e^x + 1$. Swap: $x = e^y + 1 \implies e^y = x - 1 \implies y = \ln(x-1)$.

$f^{-1}(x) = \ln(x-1)$.

Domain of $f^{-1}$: $x - 1 > 0 \implies x > 1$, i.e., $(1, \infty)$. Range of $f^{-1}$: all real numbers, $\mathbb{R}$.

(These are the range and domain of $f$, respectively, as expected.)