IB Applications and Interpretation SL · Functions · Section 4 of 7

Inverse function: how to find it and what it means

3 worked examples · By Álvaro López

Key idea

The inverse function $f^{- 1}$ undoes what $f$ does: if $f$ takes $x$ to $y$ , the inverse takes $y$ back to $x$ . It is used to solve for the input variable of a model, for example to find the price before a surcharge that corresponds to a given final price.

One-to-one functions

A function is injective (or one-to-one) if two different values of $x$ always give different values of $f (x)$ . This is the condition for it to be undone: if two inputs gave the same output, you would not know, on going back, which one you started from.

The function $f (x) = 2 x + 1$ passes the test, because a non-horizontal line is always one-to-one. By contrast, $f (x) = x^{2}$ does not: $f (2) = f (- 2) = 4$ . However, it can be fixed by restricting the domain: if only the values $x \geq 0$ are considered, the function is one-to-one and has an inverse.

Definition of the inverse

Notation warning: $f^{- 1} (x)$ is not $\frac{1}{f ( x )}$ . The $- 1$ denotes the inverse function, not an exponent.

How to find the inverse

With $f (x) = 2 x + 1$ : from $y = 2 x + 1$ we solve $x = \frac{y - 1}{2}$ , and swapping the names gives $f^{- 1} (x) = \frac{x - 1}{2}$ . Check: $f (3) = 7$ and $f^{- 1} (7) = \frac{6}{2} = 3$ .

The graph of the inverse

If the point $(a, b)$ lies on the graph of $f$ , the point $(b, a)$ lies on the graph of $f^{- 1}$ . That is why the two graphs are symmetric about the line $y = x$ : one is the reflection of the other in that mirror. If a function has an $x$ -intercept at $(a, 0)$ , its inverse has a $y$ -intercept at $(0, a)$ .

Function f(x) = x² − 4 for x ≥ 0 and its inverse f⁻¹(x) = √(x + 4) as a reflection in the line y = x, with pairs of symmetric points marked

Composition of functions

The composition $f \circ g$ applies $g$ first and then $f$ :

$(f \circ g) (x) = f (g (x))$

It is read as " $f$ after $g$ ". In general, $f \circ g$ and $g \circ f$ are different. Composition explains the inverse: applying a function and then its inverse returns the starting value,

$(f^{- 1} \circ f) (x) = x (f \circ f^{- 1}) (x) = x$

This property is useful for checking that an inverse has been calculated correctly.

Worked examples

Example 1 Basic

Inverse of a linear function

Let $f (x) = 3 x - 7$ .

a) Find $f^{- 1} (x)$ .

b) Calculate $f^{- 1} (8)$ .

Solution

a) Write $y = 3 x - 7$ and solve for $x$ : $3 x = y + 7$ , so $x = \frac{y + 7}{3}$ . Swap the names of the variables: $f^{- 1} (x) = \frac{x + 7}{3}$ .

b) $f^{- 1} (8) = \frac{8 + 7}{3} = 5$ . Check: $f (5) = 15 - 7 = 8$ .

Example 2 Medium

Inverse with a restricted domain

Let $f (x) = x^{2} - 4$ , with domain $x \geq 0$ .

a) Find $f^{- 1} (x)$ .

b) Write down the domain of $f^{- 1}$ .

c) Calculate $f^{- 1} (5)$ .

Solution

a) From $y = x^{2} - 4$ we obtain $x^{2} = y + 4$ . As the domain of $f$ is $x \geq 0$ , only the positive root works: $x = y + 4$ . Swapping the names, $f^{- 1} (x) = x + 4$ .

b) The domain of $f^{- 1}$ is the range of $f$ . For $x \geq 0$ , $x^{2} \geq 0$ , so $f (x) \geq - 4$ . The range of $f$ is $[- 4, \infty [$ , and that is the domain of $f^{- 1}$ . It agrees with the condition for the root: $x + 4 \geq 0$ .

c) $f^{- 1} (5) = 5 + 4 = 3$ . Check: $f (3) = 9 - 4 = 5$ .

Example 3 Hard

Price with surcharge and discount

An online shop calculates the final price of an item from its base price $p$ , in euros. First it applies a $20 %$ surcharge for handling costs, using the function $g (p) = 1.2 p$ . Then it subtracts a voucher of $18$ €, using the function $h (x) = x - 18$ .

a) Find the function $C = h \circ g$ that gives the final price as a function of $p$ .

b) Find the inverse function $C^{- 1} (x)$ .

c) A customer pays $102$ € for an item. Find its base price.

d) Explain what $C^{- 1}$ represents in this context.

Solution

a) Apply $g$ first and then $h$ : $C (p) = h (g (p)) = g (p) - 18 = 1.2 p - 18$

b) Write $y = 1.2 p - 18$ and solve for $p$ : $1.2 p = y + 18 \Rightarrow p = \frac{y + 18}{1.2} = \frac{5 ( y + 18 )}{6}$ Swapping the names of the variables, $C^{- 1} (x) = \frac{5 ( x + 18 )}{6}$ .

c) The final price is $102$ , so the base price is $C^{- 1} (102) = \frac{5 \cdot 120}{6} = 100 €$ Check with the direct function: $C (100) = 120 - 18 = 102$ .

d) $C^{- 1} (x)$ takes the final price paid, $x$ , and returns the base price $p$ of the item. It undoes the two steps in reverse order: first it adds back the $18$ € voucher, then it divides by $1.2$ (removing the surcharge).

In the exam: in a composition, the order matters: $h \circ g$ means "first $g$ ". If you do it the other way round, you get $1.2 (p - 18)$ , which is a different function. Always check the inverse with a value, as in part c).

Common mistakes

Summary

A function has an inverse if it is one-to-one: no horizontal line cuts its graph more than once.
$f^{- 1}$ undoes $f$ : $f (x) = y ⟺ f^{- 1} (y) = x$ . It is not $\frac{1}{f}$ .
To find it: write $y = f (x)$ , solve for $x$ and swap the names of the variables.
Domain of $f^{- 1}$ = range of $f$ , and range of $f^{- 1}$ = domain of $f$ .
The graphs of $f$ and $f^{- 1}$ are symmetric about $y = x$ .
$(f \circ g) (x) = f (g (x))$ applies $g$ first; also $(f \circ f^{- 1}) (x) = (f^{- 1} \circ f) (x) = x$ .

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