Inverse and composite functions (VCE Maths Methods Unit 1)

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 $g(x) = x + 3$:

• $(f \circ g)(x) = f(g(x)) = (x + 3)^2$.
• $(g \circ f)(x) = g(f(x)) = x^2 + 3$.

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.

Restricting the domain to create an inverse

Many functions are not one-to-one over their natural domain, so VCAA requires you to restrict the domain to the largest interval on which $f$ is one-to-one before an inverse exists. For a parabola $f(x) = (x - h)^2 + k$, the turning point at $x = h$ is the boundary: restricting to $x \ge h$ (the increasing branch) or $x \le h$ (the decreasing branch) makes $f$ one-to-one. The branch you choose determines the sign taken when you solve for $y$, so always carry the restriction through to the final answer and use it to select between the $+$ and $-$ square root.

The same idea governs $\sqrt{\,}$ and reciprocal functions: state the maximal domain that keeps the function one-to-one, then the inverse exists on the corresponding range. Examiners frequently penalise answers that produce a "$\pm$" inverse, because an inverse must itself be a function (one output per input).

The graphical link and self-inverse functions

Because reflecting in $y = x$ swaps the roles of $x$ and $y$, the graph of $f^{-1}$ is the mirror image of $f$ in that line. A consequence worth remembering for the exam: where $f$ and $f^{-1}$ intersect, they often meet on the line $y = x$ (so solving $f(x) = x$ can locate intersection points more quickly than solving $f(x) = f^{-1}(x)$ directly). A function that is its own inverse, such as $f(x) = \dfrac{1}{x}$ or $f(x) = a - x$, is symmetric in the line $y = x$, so $f(f(x)) = x$ for all valid $x$.

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.

Examples in context

Example 1. Converting a unit-conversion rule. A currency desk converts Australian dollars to a foreign amount with $f(x) = 1.4x - 2$ (a $1.4$ rate minus a $\$2$ fixed fee). To recover the dollars paid from the foreign amount received, invert: $y = 1.4x - 2$, swap to $x = 1.4y - 2$, solve $y = \frac{x + 2}{1.4}$. So $f^{-1}(x) = \frac{x + 2}{1.4}$. A foreign amount of $19.6$ came from $f^{-1}(19.6) = \frac{21.6}{1.4} = 15.43$ dollars, and indeed $f(15.43) \approx 19.6$.

Example 2. Composing two processes. A factory applies a markup $g(x) = 1.1x$ then a flat handling charge $f(x) = x + 5$. The total cost is the composite $(f \circ g)(x) = f(g(x)) = 1.1x + 5$. Applied to a $\$40$ item: $1.1(40) + 5 = \$49$. Reversing the order, $(g \circ f)(x) = 1.1(x + 5) = 1.1x + 5.5$, gives $\$49.50$, confirming composition is not commutative.

Try this

Q1. Find the inverse of $f(x) = 3x - 7$. [2 marks]

• Cue. Swap and solve: $f^{-1}(x) = \frac{x + 7}{3}$.

Q2. For $f(x) = x^2 + 1$ with domain $x \le 0$, find $f^{-1}(x)$ and state its domain. [3 marks]

• Cue. $x = y^2 + 1 \Rightarrow y = -\sqrt{x - 1}$ (negative branch from $x \le 0$); domain $x \ge 1$.

Q3. Let $f(x) = 2x$ and $g(x) = x - 3$. Find $(f \circ g)(x)$ and $(g \circ f)(x)$. [2+2 marks]

• Cue. $(f \circ g)(x) = 2(x - 3) = 2x - 6$; $(g \circ f)(x) = 2x - 3$.