2.10 Composite and inverse functions — Learn

Prerequisites: substituting into an expression; rearranging a formula; solving a modulus equation from 2.9; the graph of $y = x$.

By the end of this section you can: use function notation and state a domain and a range; form a composite function and explain why order matters; decide whether a function has an inverse; and find an inverse by rearranging, including after restricting the domain.

1. Function language

A function is a rule that turns each allowed input into exactly one output. The set of allowed inputs is the domain, and the set of outputs is the range.

has the whole real line as its domain and the whole real line as its range.

A function is one-to-one when different inputs always give different outputs. Only a one-to-one function can be reversed, because the reverse rule must give a single answer for each output.

2. Composite functions

A composite function applies one function and then another. The notation $f(g(x))$, often written $fg(x)$, means "apply $g$ first, then $f$".

Example 1. Given $f(x) = 3x + 1$ and $g(x) = x^2$, find $f(g(x))$ and $g(f(x))$.

Substitute $g(x)$ into $f$:

Substitute $f(x)$ into $g$:

The two answers are different, so $f(g(x))$ and $g(f(x))$ are not the same. The order of the functions matters. Work from the inside out.

The domain of a composite is the set of inputs allowed by the first function applied, and then by the second. If $g(x) = \sqrt{x}$ and $f(x) = x - 3$, then $f(g(x)) = \sqrt{x} - 3$ needs $x \geq 0$.

3. Inverse functions

The inverse function $f^{-1}$ undoes $f$. If $f$ sends $a$ to $b$, then $f^{-1}$ sends $b$ back to $a$. Composing a function with its inverse returns the input:

To find an inverse, write $y = f(x)$, make $x$ the subject, then swap the letters.

Example 2. Find the inverse of $f(x) = 3x - 2$.

Write $y = 3x - 2$. Add $2$:

Divide by $3$:

Swap the letters to name the inverse as a function of $x$:

Check: $f^{-1}(f(x)) = \frac{(3x - 2) + 2}{3} = \frac{3x}{3} = x$.

Example 3. Find the inverse of $f(x) = \dfrac{x + 1}{x - 2}$, where $x \neq 2$.

Write $y = \frac{x + 1}{x - 2}$ and multiply both sides by $x - 2$:

Expand and collect the $x$ terms on one side:

Divide by $y - 1$:

Swap the letters:

4. The graph of an inverse

The graph of $y = f^{-1}(x)$ is the graph of $y = f(x)$ reflected in the line $y = x$. The reflection swaps the two coordinates, which is exactly what reversing a function does. The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$.

Example 4. The function $f(x) = x^2$ has domain $x \geq 0$. Find its inverse.

Write $y = x^2$. Then $x = \pm\sqrt{y}$. The domain $x \geq 0$ selects the positive root, so $x = \sqrt{y}$. Swap the letters:

Without the restriction, $f(x) = x^2$ has no inverse: $f(2) = f(-2) = 4$, so the reverse rule would not know whether to return $2$ or $-2$.

Watch out. Two errors appear again and again.

Treating $fg$ as a product. $fg(x)$ means $f(g(x))$, not $f(x) \times g(x)$.

Assuming every function has an inverse. A function must be one-to-one. $f(x) = x^2$ is not, so restrict its domain; a function such as $f(x) = x^3$ is one-to-one already.

Re-check: after finding an inverse, compose both ways. Both compositions must give $x$ on the allowed domain.

5. Try it: reflecting a graph in $y = x$

Predict first. The figure draws the line $y = f(x)$ and its reflection in the line $y = x$. Before you touch the sliders, choose a point on the line, read its two coordinates, and predict the coordinates of its image. Write the prediction down.

Test. Move the slider for the position of $P$ along the graph, and watch the image. Then change the gradient and the intercept of $f$, and watch both lines move together.

Explain. Write two sentences saying why the image of the point $(p, q)$ is $(q, p)$, and why the reflected graph is the graph of the inverse function.

Progress check

Return to the chapter problem. The height of the arch at a distance $x$ from the centre is

1. State the domain and the range of $h$.
2. Show that $h$ is not one-to-one by comparing $h(2)$ and $h(-2)$.
3. Restrict the domain to $0 \leq x \leq 4$ and find the inverse function $h^{-1}$.
4. Use the inverse to find the distance from the centre at which the arch is $3.2$ m high. Compare your answer with the interval you found in 2.6.

You should now be able to

• use function notation and state a domain and a range;
• form a composite function and explain why order matters;
• decide whether a function is one-to-one and so has an inverse;
• find an inverse by rearranging and swapping the letters;
• restrict a domain to make an inverse possible;
• state the domain and range of an inverse.