Define inverse and composite functions, identify when an inverse function exists (one-to-one), find inverse functions algebraically, and graph inverse and composite functions

What this dot point is asking

VCAA wants you to define inverse and composite functions, find inverses algebraically when one exists, restrict domains when necessary, and connect inverse functions graphically (reflection in $y = x$).

What is an inverse function

A function $f^{-1}$ is the inverse of $f$ if:

The inverse undoes what the function does.

When does an inverse exist

A function has an inverse only if it is one-to-one: each $y$-value comes from exactly one $x$-value. Equivalently, the function passes the horizontal line test.

Most polynomials of even degree are not one-to-one over their full domain. To define an inverse, restrict the domain so the function becomes one-to-one (e.g. $x \ge 2$ for $f(x) = (x-2)^2$).

Finding $f^{-1}$ algebraically

1. Write $y = f(x)$.
2. Swap $x$ and $y$.
3. Solve for $y$ in terms of $x$.
4. Choose the correct branch (using the domain restriction).
5. The result is $f^{-1}(x)$.

Graphical relationship

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

Composite functions

$(f \circ g)(x) = f(g(x))$. Apply $g$ first, then $f$.

For $f(x) = x^2$ and $g(x) = x + 3$:

• IMATH_24 .
• IMATH_25 .

Composition is not commutative in general: $f \circ g \neq g \circ f$.

Domain of a composite

For $f \circ g$, the domain is $\{x \in \text{dom}(g) : g(x) \in \text{dom}(f)\}$. Both conditions must hold.

Worked example

Let $f(x) = 2x + 1$. Find $f^{-1}(x)$.

$y = 2x + 1 \Rightarrow x = 2y + 1 \Rightarrow y = (x - 1)/2$.

$f^{-1}(x) = (x - 1)/2$.

Check: $f(f^{-1}(x)) = 2 \cdot (x - 1)/2 + 1 = x - 1 + 1 = x$. ✓

Common traps

Treating $f^{-1}$ as $1/f$. $f^{-1}$ is the inverse function (different concept from reciprocal). $(2x + 1)^{-1}$ in inverse-function notation gives $(x - 1)/2$, not $1/(2x+1)$.

Forgetting to restrict the domain. $f(x) = x^2$ has no inverse over all real $x$, but $f$ restricted to $x \ge 0$ has inverse $f^{-1}(x) = \sqrt{x}$.

Confusing composite functions with multiplication. $f \circ g$ is not $fg$.

Domain of composite. The output of $g$ must lie in the domain of $f$.

In one sentence

A function has an inverse $f^{-1}$ only when it is one-to-one (passes the horizontal line test); $f^{-1}$ is found by swapping $x$ and $y$ in $y = f(x)$ and solving, has domain and range swapped from $f$, and is graphed as the reflection of $f$ in the line $y = x$; composite functions $(f \circ g)(x) = f(g(x))$ apply $g$ first then $f$ and are generally non-commutative.