Transformations from $y = f(x)$ to $y = A f(n(x - b)) + c$ (dilation, reflection, translation), composite functions $(f \circ g)(x) = f(g(x))$ and the conditions for their existence, and inverse functions $f^{-1}$ with the link to one-to-one functions

What this dot point is asking

VCAA wants you to manipulate functions in three standard ways: transform $y = f(x)$ to $y = A f(n(x - b)) + c$, build composites $(f \circ g)(x) = f(g(x))$ checking that they are defined, and find inverses $f^{-1}$ after restricting to a one-to-one domain. These three operations underpin most function-family Paper 1 and Paper 2 questions.

Transformations

The standard form is $y = A f(n(x - b)) + c$ with four parameters:

• **$A$** is a vertical dilation by factor $|A|$ from the x-axis (reflection in the x-axis if $A < 0$). It multiplies y-values.
• **$n$** is the reciprocal of a horizontal dilation: $n = 2$ compresses x-values horizontally by factor 2 (reflection in the y-axis if $n < 0$). Period and key x-values become $\frac{1}{|n|}$ times their original value.
• **$b$** is a horizontal translation right by $b$ (left if $b < 0$).
• **$c$** is a vertical translation up by $c$ (down if $c < 0$).

Order of operations

When described in words, the conventional order is dilation, then reflection, then translation. VCAA marking guides accept any consistent order if the algebra produces the same final rule.

Effect on key features

For each function family, transformations move key features predictably:

• Asymptotes. Vertical asymptote of $\ln(x)$ at $x = 0$ moves to $x = b$. Horizontal asymptote of $e^x$ at $y = 0$ moves to $y = c$.
• Intercepts. Apply the new rule and solve.
• Maxima, minima, midline. For $A \sin(n(x - b)) + c$: amplitude $|A|$, midline $y = c$, period $2\pi/|n|$.

Composite functions

$(f \circ g)(x) = f(g(x))$. The composite takes $x$, applies $g$ first, then applies $f$.

Existence condition

For $f \circ g$ to be defined on a set $S$, the range of $g$ restricted to $S$ must be a subset of the domain of $f$. If $g$ produces outputs outside the domain of $f$, the composite is undefined there.

Three exam patterns

VCAA examines composites in three patterns.

1. Direct evaluation. Compute $f(g(2))$ by substituting $g(2)$ into $f$.
2. Rule and domain. Determine $(f \circ g)(x)$ as a simplified rule, and state the maximal domain where it is defined.
3. Parameter conditions. Given a parameter (e.g. "for which values of $k$ is $f \circ g$ defined on all of $\mathbb{R}$?"), find the condition on the parameter so the existence condition holds.

Worked example

Let $f(x) = \sqrt{x}$ (domain $[0, \infty)$) and $g(x) = x - 4$ (domain $\mathbb{R}$).

$f \circ g(x) = \sqrt{x - 4}$. Domain: $x \geq 4$, i.e. $[4, \infty)$.

$g \circ f(x) = \sqrt{x} - 4$. Domain: $x \geq 0$, i.e. $[0, \infty)$.

The two composites differ in rule, domain and range.

Inverse functions

The inverse $f^{-1}$ undoes $f$: $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$, and $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$.

When does an inverse exist?

$f^{-1}$ exists if and only if $f$ is one-to-one (each y-value is hit by exactly one x-value, the horizontal line test).

Many standard functions are not one-to-one on their natural domain. For example, $f(x) = x^2$ on $\mathbb{R}$ is two-to-one (every positive $y$ comes from two $x$ values, $\pm \sqrt{y}$). To take its inverse, restrict the domain to $[0, \infty)$ or $(-\infty, 0]$.

Finding the inverse

Three steps.

1. Start with $y = f(x)$.
2. Swap $x$ and $y$: $x = f(y)$.
3. Solve for $y$.

The result is $y = f^{-1}(x)$.

Domain and range swap

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

Graphical property

The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y = x$. So if $f$ has y-intercept at $(0, c)$, then $f^{-1}$ has x-intercept at $(c, 0)$.

Worked example

Let $f(x) = e^{2x} + 3$ for $x \in \mathbb{R}$. Find $f^{-1}$.

Range of $f$: $(3, \infty)$ (since $e^{2x} > 0$).

Swap: $x = e^{2y} + 3$. Rearrange: $e^{2y} = x - 3$, then $2y = \ln(x - 3)$, so $y = \frac{1}{2} \ln(x - 3)$.

$f^{-1}(x) = \frac{1}{2} \ln(x - 3)$. Domain: $(3, \infty)$. Range: $\mathbb{R}$.

Common Paper 1 traps

Transformation order confusion. "Translate then dilate" produces a different result than "dilate then translate" when described in words. Always rewrite into the standard form $A f(n(x - b)) + c$ and identify the four parameters first.

Forgetting the reciprocal in horizontal dilation. $n = 3$ compresses by factor $\frac{1}{3}$, not $3$.

Skipping the existence condition for composites. Writing $f \circ g(x) = \sqrt{x - 4}$ without stating that $x \geq 4$ loses a domain mark.

Inverting a non-one-to-one function without restricting the domain. $f(x) = x^2$ on $\mathbb{R}$ has no inverse. You must restrict (e.g. to $x \geq 0$) before swapping.

Forgetting that domain and range swap for the inverse. Markers expect both stated explicitly.

In one sentence

Transformations, composites and inverses are the three standard ways to build new functions in VCE Math Methods: the four-parameter form $A f(n(x - b)) + c$ stretches and shifts a graph, composites $(f \circ g)(x) = f(g(x))$ chain functions when the range of $g$ lies in the domain of $f$, and inverses $f^{-1}$ exist only when $f$ is one-to-one (so often after a domain restriction) and reflect across $y = x$ with domain and range swapped.