Hybrid functions and inverse functions: VCE Math Methods Unit 4

What this dot point is asking

VCAA wants you to define and analyse hybrid (piecewise) functions, especially their continuity and differentiability at join points, and to find inverse functions, requiring a one-to-one domain restriction when necessary. The dot point combines algebraic skill (solving for the inverse), calculus skill (matching derivatives at joins) and graphical reasoning (reflection in $y = x$).

Hybrid (piecewise) functions

A hybrid function is defined by different rules on different intervals of its domain. The general form:

Each piece $f_i$ has its own formula and its own interval $I_i$. The intervals partition the domain (no overlap, no gap).

Examples:

• $\lvert x \rvert = \begin{cases} -x & \text{if } x < 0 \\ x & \text{if } x \geq 0 \end{cases}$
• Income tax brackets (different marginal rates on different income bands).
• Heaviside step function: $H(x) = 0$ for $x < 0$ and $H(x) = 1$ for $x \geq 0$.

Continuity at a join

A hybrid function is continuous at the join point $x = a$ if the left-hand value and right-hand value of $f$ agree at $a$:

In practice: substitute $x = a$ into the left-piece formula and the right-piece formula, and require the two values to match.

Differentiability at a join

A hybrid function is differentiable at the join point $x = a$ if it is continuous at $a$ AND the left and right derivatives agree:

In practice: differentiate each piece, substitute $x = a$, and require the two derivatives to match.

Continuity alone is not enough. For example, $\lvert x \rvert$ is continuous at $x = 0$ (both pieces give 0) but not differentiable there (left derivative is $-1$, right derivative is $+1$).

Two-constant problems

Many Paper 1 hybrid questions give a hybrid function with two unknown constants in one piece, and ask to find the constants such that the function is continuous and differentiable at a specified join.

Procedure:

1. Continuity equation. Substitute the join $x$-value into both pieces; set the two expressions equal.
2. Differentiability equation. Differentiate each piece; substitute the join $x$-value into both derivatives; set the two expressions equal.
3. Solve the simultaneous equations for the two constants.

The example in the 2024 past question above follows this exactly.

Inverse functions

The inverse of a function $f$ is a function $f^{-1}$ that "undoes" $f$:

(on appropriate domains).

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

A function has an inverse if and only if it is one-to-one (no two inputs map to the same output). Geometrically, the graph passes the horizontal line test: every horizontal line meets the graph in at most one point.

Functions that are not one-to-one (parabolas, $\sin$, $\cos$) can be made invertible by restricting the domain to an interval on which they are monotonic.

Finding $f^{-1}$ algebraically

To find the inverse of $y = f(x)$:

1. Swap $x$ and $y$. (Conceptually, you are looking at the inverse from the output side.)
2. Solve for $y$ in terms of $x$.
3. 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 $f$.

If $f: A \to B$ is one-to-one, then $f^{-1}: B \to A$ (with the swapped roles).

Graphical interpretation: reflection in $y = x$

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

This means:

• If $(a, b)$ is on the graph of $f$, then $(b, a)$ is on the graph of $f^{-1}$.
• Intersection points of $f$ and $f^{-1}$ lie on $y = x$. To find them, solve $f(x) = x$.
• The asymptotes swap: a horizontal asymptote $y = c$ of $f$ becomes a vertical asymptote $x = c$ of $f^{-1}$.

Worked example. Linear function

$f(x) = 3x - 7$ on $\mathbb{R}$.

Swap. $x = 3y - 7$.

Solve. $y = (x + 7) / 3$.

So $f^{-1}(x) = (x + 7) / 3$.

Check: $f(f^{-1}(x)) = 3 \cdot (x + 7)/3 - 7 = x + 7 - 7 = x$. Correct.

Worked example. Exponential

$f(x) = e^x$ on $\mathbb{R}$.

Swap. $x = e^y$.

Solve. $y = \ln(x)$.

So $f^{-1}(x) = \ln(x)$, with domain $(0, \infty)$ and range $\mathbb{R}$.

The graphs of $e^x$ and $\ln x$ are reflections of each other in $y = x$, intersecting nowhere (asymptotic to opposite axes).

Worked example. Quadratic requiring restriction

$f(x) = x^2$ on $\mathbb{R}$ has no inverse (not one-to-one; $f(-2) = f(2) = 4$).

Restrict. $f: [0, \infty) \to [0, \infty)$, $f(x) = x^2$.

Now invertible. Swap: $x = y^2$. Solve: $y = \sqrt{x}$ (positive root, because the restricted range of $f^{-1}$ is $[0, \infty)$).

$f^{-1}(x) = \sqrt{x}$ on $[0, \infty)$.

Equivalent restriction. $f: (-\infty, 0] \to [0, \infty)$, $f(x) = x^2$. Inverse: $f^{-1}(x) = -\sqrt{x}$.

The choice of restriction determines which branch of the inverse you get.

Inverse of a hybrid function

A hybrid function is invertible only if it is one-to-one over its full domain. Each piece must be monotonic in the same direction (all increasing or all decreasing across the joins), and the joins must not produce overlapping outputs. In practice the question gives a hybrid that already satisfies these conditions.

To invert: invert each piece separately, taking care that the domain of each inverse piece is the range of the corresponding original piece.

Examples in context

Example 1. A piecewise pricing rule. A car park charges a flat $\$5$ for the first hour and then $\$3$ per hour after, so the cost is $C(t) = 5$ for $0 < t \le 1$ and $C(t) = 5 + 3(t - 1)$ for $t > 1$. At the join $t = 1$, both pieces give $C = 5$, so the cost is continuous. But the left rate is $0$ and the right rate is $3$, so $C$ is not differentiable at $t = 1$ (a kink where the charging rate jumps).

Example 2. Inverting a temperature model. A probe's reading is $f(x) = (x - 1)^2 + 2$ on the restricted domain $x \ge 1$ (so it is one-to-one and increasing). To recover $x$ from a reading $y$: $(x - 1)^2 = y - 2$, and since $x \ge 1$, $x = 1 + \sqrt{y - 2}$. Thus $f^{-1}(x) = 1 + \sqrt{x - 2}$, with domain $x \ge 2$ (the range of $f$) and range $x \ge 1$.

Try this

Q1. A hybrid function has $f(x) = 3x$ for $x \le 2$ and $f(x) = x + c$ for $x > 2$. Find $c$ so $f$ is continuous at $x = 2$. [2 marks]

• Cue. $3(2) = 2 + c \Rightarrow c = 4$.

Q2. For the same join, is $f$ differentiable at $x = 2$ when $c = 4$? Justify. [2 marks]

• Cue. Left derivative $3$, right derivative $1$; not equal, so not differentiable (a kink).

Q3. Find the inverse of $f(x) = (x + 2)^2$ on $x \ge -2$. [3 marks]

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