03.6inverses

# Relations and their Inverses

Recall:

• A relation is a set of ordered pairs that can be listed, graphed or described by a rule.
• The inverse of a rule is the reverse operation that "undoes" whatever the rule has done.

Examples of inverses you have already encountered are

• $y = x + 3$ and $y = x - 3$
• $y = e^x$ and $y = \log_e(x)$
• $y = \sin(x)$ and $y = \sin^{–1}(x)$
• $y = x^2$ and $y = \pm\sqrt{x}$

.

The inverse of a relation can be found in 3 ways:

• swap the x and y coordinates of each ordered pair …. (or)
• reflect the graph of the relation across the line $y = x$ …. (or) .
• interchange x and y in the rule and rearrange to make y the subject

.

The domain and range of a relation are also swapped to form the range and domain of the inverse.

.

#### Example 1

Find the inverse of the graph shown here:

Solution:

• Swap the x and y coordinates of individual points
• Reflect the shape of the graph across the line $y = x$

.

# Inverses and Functions

Recall that a function is a relation that passes the vertical line test.

• for any x value there is no more than one y-value.

The inverse of a function is not necessarily a function.

Only one-to-one functions will have an inverse which is also a function

.

#### Example 2

A function is described by the rule: $f: R \rightarrow R, \;\; f(x) = (x + 2)^2$,

Find the inverse of $f(x)$

Solution:

… … Let $y = \big( x + 2 \big)^2$

… … domain: $x \in R$

… … range: $\big\{ y : y \geqslant 0 \big\}$

• Swap the x and y in the rule then rearrange to make y the subject
• The range of f will become the domain of the inverse of f

… … Inverse:

… … $x = \big( y + 2 \big)^2$

… … $\big( y + 2 \big)^2 = x$

… … $y + 2 = \pm \sqrt{x}$

… … $y = \pm \sqrt{x} - 2$

… … domain: $\big\{ x : x \geqslant 0 \big\}$

so the inverse of $f(x)$ is

… … $y = \pm \sqrt{x} - 2, \quad \text{ for } \big\{ x : x \geqslant 0 \big\}$

… … notice that the inverse is not a function because f(x) is not one-to-one

… … the inverse fails the vertical line test
.

Note:

• The graph of $f(x)$ has been reflected across the line $y = x$
• The coordinates of the turning point have been swapped
• $(-2, \; 0)$ has become $(0, \; -2)$

.

## The Inverse using the calculator

• But if you swap the x and the y in the rule manually
• The calculator can do the rearranging for you to make y the subject

For example 1 above, enter the following:

Enter:$\text{solve}\big( x = (y + 2)^2 , y \big)$

• Notice the two solutions.

.

## The Inverse using Matrices

The matrix operation that produces a reflection across the line $y = x$ is:

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} 0&1 \\ 1&0 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$

For example, the inverse of the point $(4, \; 2)$ can be found by:

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} 0&1 \\ 1&0 \\ \end{matrix} \right] \left[ \begin{matrix} 4 \\ 2 \\ \end{matrix} \right]$

.

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} 2 \\ 4 \\ \end{matrix} \right]$

Since finding the inverse by swapping x and y values is trivial, we usually wouldn't use this method but finding the inverse can be done using matrices.

.

## Graphs of a Relation and its Inverse

If we start with the graph of any relation, we can sketch its inverse by reflecting it across the line $y = x$.
Any particular points can be identified by swapping x and y coordinates.

.

### Activity

For each of the following graphs,

• draw in the line y = x and hence sketch the inverse relation.
• Label significant points
• state domain and range of both the graph and its inverse.
• For each graph and its inverse, state if the graph is a function.

#### Example 3

Solution

Original Inverse
Domain $x \in [-3, \; 1]$ $x \in [1, \; 3]$
Range $y \in [1, \; 3]$ $y \in [-3, \; 1]$
Function no no

.

#### Example 4

Solution

Original Inverse
Domain $x \in [-2, \; 2]$ $x \in [0, \; 2]$
Range $y \in [0, \; 2]$ $y \in [-2, \; 2]$
Function no no

.

#### Example 5

Solution

Original Inverse
Domain $x \in R$ $x \in R$
Range $y \in R$ $y \in R$
Function yes no

.

#### Example 6

Solution

Original Inverse
Domain $x \in R^+$ $x \in R$
Range $y \in R$ $y \in R^+$
Function yes yes

.

.