Functions, inverses and the modulus

What you’ll learn

• Find domains, ranges, composites and inverse functions.
• Simplify rational expressions with their restrictions.
• Interpret modulus graphs and combined transformations.

Understand the idea

A function assigns exactly one output to each input in its domain. Its range is the set of outputs actually reached. The composite $fg(x)$ means $f(g(x))$, so the output of $g$ must lie in the domain of $f$. A one-to-one function has an inverse; a many-to-one function may first need a restricted domain.

Find an inverse by writing $y=f(x)$, solving for $x$, and exchanging the variable names. Its graph reflects in $y=x$ and its domain and range exchange roles. A square root's sign must match the chosen original domain. Cancelling factors in a rational expression simplifies its formula but does not restore originally excluded inputs.

The modulus $|f(x)|$ reflects the portions of a graph below the x-axis upwards. In contrast, $f(|x|)$ keeps the right-hand part and reflects it into the left half-plane. Solve a modulus equation by using sign cases and checking each candidate against its case conditions.

Key formulae

$$(fg)(x)=f(g(x))$$

$$f^{-1}(f(x))=x\quad\text{on the domain of }f$$

$$|u|=\begin{cases}u&u\geq0\\-u&u<0\end{cases}$$

$$\frac{x^2-a^2}{x-a}=x+a\quad(x\ne a)$$

Worked example

Let $f(x)=(x-2)^2+3$ for $x\geq2$. Find $f^{-1}$ and its domain.

1. Write $y=(x-2)^2+3$, so $(x-2)^2=y-3$.
2. Since $x\geq2$, choose $x-2=\sqrt{y-3}$ rather than the negative root.
3. Hence $x=2+\sqrt{y-3}$. The range of $f$ is $y\geq3$, becoming the inverse domain.

Answer: $f^{-1}(x)=2+\sqrt{x-3}$ for $x\geq3$.