Composite functions $f \circ g$ and $g \circ f$, the existence and form of inverse functions $f^{-1}$, the relationship between a function and its inverse (reflection in $y = x$, domain and range swap), and the one-to-one restriction

What this dot point is asking

VCAA wants you to construct composite functions, determine when an inverse function exists, find the inverse algebraically, and recognise the graphical relationship between a function and its inverse.

Composite functions

A composite function applies one function inside another:

Read "f of g of x". Apply $g$ first, then $f$.

The order matters: $f \circ g$ is in general not the same as $g \circ f$.

Domain considerations: the composite $f(g(x))$ requires $x$ to be in the domain of $g$, and $g(x)$ to be in the domain of $f$.

Worked example

$f(x) = \sqrt{x}$, $g(x) = x - 4$.

$(f \circ g)(x) = f(x - 4) = \sqrt{x - 4}$, defined for $x \geq 4$.

$(g \circ f)(x) = g(\sqrt{x}) = \sqrt{x} - 4$, defined for $x \geq 0$.

Different functions with different domains.

Inverse functions

The inverse of $f$ is the function $f^{-1}$ such that:

on appropriate domains.

When does $f^{-1}$ exist?

$f^{-1}$ exists if and only if $f$ is one-to-one: no two inputs map to the same output. Equivalently, $f$ passes the horizontal line test.

Functions that are not one-to-one (parabolas, $\sin$ on $\mathbb{R}$) require a domain restriction to be invertible.

Finding $f^{-1}$ algebraically

1. Write $y = f(x)$.
2. Swap $x$ and $y$.
3. Solve for $y$.
4. **Write $f^{-1}(x) =$** (the solved expression).

Domain and range swap

Under inversion:

• Domain of $f^{-1}$ = range of $f$.
• Range of $f^{-1}$ = domain of $f$.

Graphical relationship

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

If $(a, b)$ is on $f$, then $(b, a)$ is on $f^{-1}$.

Intersections of $f$ and $f^{-1}$ lie on $y = x$. To find them, solve $f(x) = x$.

Worked examples

Linear. $f(x) = 3x + 2$. Swap: $x = 3y + 2$. Solve: $y = (x - 2)/3$. So $f^{-1}(x) = (x - 2)/3$.

Quadratic with restriction. $f: [0, \infty) \to [0, \infty)$, $f(x) = x^2$. Swap: $x = y^2$. Solve (positive root, since range of $f^{-1}$ is $[0, \infty)$): $y = \sqrt{x}$. So $f^{-1}(x) = \sqrt{x}$ on $[0, \infty)$.

Exponential. $f(x) = 2^x$. The inverse is $f^{-1}(x) = \log_2(x)$ on $(0, \infty)$.

Verifying the inverse

To check $g(x) = f^{-1}(x)$, verify both $f(g(x)) = x$ and $g(f(x)) = x$ on the appropriate domains.

Common errors

Confusing $f^{-1}(x)$ with $1/f(x)$. Inverse function vs reciprocal. Different things.

Forgetting domain restriction. Asking for the inverse of $x^2$ on $\mathbb{R}$ is not well-defined. Restrict to make $f$ one-to-one.

Wrong root sign on quadratic inverse. Solving $y = x^2$ gives $x = \pm\sqrt{y}$; the correct sign depends on the restricted domain.

Composite order confusion. $(f \circ g)(x) = f(g(x))$: apply $g$ first. Many students apply $f$ first by mistake.

Domain ignored in composite. $f(g(x))$ requires $x$ in the domain of $g$ AND $g(x)$ in the domain of $f$.

In one sentence

Composite functions apply one function inside another ($(f \circ g)(x) = f(g(x))$, $g$ first then $f$), and inverse functions $f^{-1}$ undo $f$ ($f(f^{-1}(x)) = x$); $f^{-1}$ exists only when $f$ is one-to-one (passes the horizontal line test), is found algebraically by swapping $x$ and $y$ and solving for $y$, has swapped domain and range, and is graphed as the reflection of $f$ in the line $y = x$.