Form composite functions, determine when a function has an inverse, find and graph the inverse, and use restriction of domain to invert non-one-to-one functions

What this dot point is asking

NESA wants you to form composite functions $f \circ g$ and $g \circ f$, determine when a function has an inverse using the horizontal line test or one-to-one criterion, find the inverse algebraically, sketch it as the reflection in $y = x$, and restrict the domain of a non-one-to-one function so an inverse exists.

The answer

Composite functions

The composite $f \circ g$ is the function $(f \circ g)(x) = f(g(x))$: apply $g$ first, then $f$ to the result.

Domain: $x$ must be in the domain of $g$, and $g(x)$ must be in the domain of $f$. In other words, the natural domain of $f \circ g$ is

Composition is not commutative: usually $f(g(x)) \neq g(f(x))$.

Composition is associative: $(f \circ g) \circ h = f \circ (g \circ h) = f(g(h(x)))$.

When does an inverse exist

A function $f$ has an inverse (on its given domain) if and only if it is one-to-one: every value of $y$ in the range comes from exactly one $x$.

Equivalent test (the horizontal line test): every horizontal line meets the graph at most once. Strictly increasing or strictly decreasing functions are automatically one-to-one. Many "natural" functions like $x^2$, $\sin x$, $\cos x$, and $|x|$ are not one-to-one on their natural domain and need their domain restricted.

Finding an inverse algebraically

The inverse $f^{-1}$ "undoes" $f$: $f^{-1}(f(x)) = x$ for $x \in \text{dom}(f)$ and $f(f^{-1}(y)) = y$ for $y \in \text{range}(f)$.

To find $f^{-1}$ from $y = f(x)$:

1. Swap $x$ and $y$.
2. Solve for $y$ in terms of $x$.
3. State the domain of $f^{-1}$, which is the range of $f$.

The inverse graph

The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$. Point $(a, b)$ on $f$ corresponds to $(b, a)$ on $f^{-1}$. Horizontal asymptotes of $f$ become vertical asymptotes of $f^{-1}$ and vice versa.

If $f$ is increasing, so is $f^{-1}$. If $f$ is decreasing, so is $f^{-1}$.

Restricting the domain

For a non-one-to-one function $f$, choose a domain on which $f$ is one-to-one, then invert. Different restrictions give different inverses.

• IMATH_52 : restrict to $x \ge 0$ to get $f^{-1}(x) = \sqrt{x}$. Restricting to $x \le 0$ gives $f^{-1}(x) = -\sqrt{x}$.
• IMATH_57 : standard restriction is $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$. Inverse is $\arcsin x$ on $[-1, 1]$.
• IMATH_61 : standard restriction is $0 \le x \le \pi$. Inverse is $\arccos x$ on $[-1, 1]$.

Operations and inverses

The inverse of a composition reverses the order:

provided both inverses exist on appropriate domains. Think of putting on socks then shoes: to undo, take off shoes then socks.

Worked examples

Composition with domain check

Let $f(x) = \sqrt{x}$ (domain $x \ge 0$) and $g(x) = x - 4$. Find $f(g(x))$ and its domain.

$f(g(x)) = \sqrt{x - 4}$.

Domain: need $g(x) \ge 0$, that is $x \ge 4$. Domain of $f \circ g$ is $[4, \infty)$.

Inverse of a linear function

$f(x) = 5 - 2 x$. Swap and solve: $x = 5 - 2 y$, so $y = \frac{5 - x}{2}$.

$f^{-1}(x) = \frac{5 - x}{2}$. Both $f$ and $f^{-1}$ are decreasing linear functions, and their graphs reflect across $y = x$.

Inverse of an exponential

$f(x) = e^{x} + 1$ has domain $\mathbb{R}$ and range $(1, \infty)$.

Swap: $x = e^y + 1$, so $e^y = x - 1$, so $y = \ln(x - 1)$.

$f^{-1}(x) = \ln(x - 1)$, with domain $(1, \infty)$ (the range of $f$).

Domain restriction

$f(x) = (x - 1)^2$ on $\mathbb{R}$ is not one-to-one (the parabola has its vertex at $x = 1$ and is symmetric about it).

Restrict to $x \ge 1$: $f$ is increasing with range $[0, \infty)$.

Swap: $x = (y - 1)^2$, so $y - 1 = \sqrt{x}$ (positive root because $y \ge 1$), so $y = 1 + \sqrt{x}$.

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

Checking with composition

For $f(x) = 5 - 2 x$ and $f^{-1}(x) = \frac{5 - x}{2}$:

$f(f^{-1}(x)) = 5 - 2 \cdot \frac{5 - x}{2} = 5 - (5 - x) = x$. So $f^{-1}$ is correct.

Common traps

Composing in the wrong order. $f(g(x))$ applies $g$ first; $g(f(x))$ applies $f$ first. These give different functions in general.

Confusing the inverse with the reciprocal. $f^{-1}(x)$ means the inverse function, not $\frac{1}{f(x)}$. For example, if $f(x) = x + 2$, then $f^{-1}(x) = x - 2$, not $\frac{1}{x + 2}$.

Forgetting to restrict. Writing $y = \sqrt{x}$ as "the inverse of $y = x^2$" without specifying $x \ge 0$ misses half the picture: the original function must be made one-to-one first.

Wrong domain for the inverse. The domain of $f^{-1}$ equals the range of $f$. For $f(x) = e^x$ with range $(0, \infty)$, the inverse $\ln x$ has domain $(0, \infty)$.

Mixing up which side gets the $\pm$. When solving $x = y^2$ for $y$, the choice of $\sqrt{x}$ or $-\sqrt{x}$ is determined by the restricted domain of $f$, not arbitrary.

In one sentence

A function has an inverse precisely when it is one-to-one; the inverse is found by swapping $x$ and $y$ then solving, its graph is the reflection of $f$ in $y = x$, and non-one-to-one functions must first have their domain restricted before an inverse exists.