# 5.07 Inverses and their graphs

Lesson

When we have an equation in two variables then there is a relation between the two variables. For example, $y=x+1$y=x+1 describes a relation between $x$x and $y$y (that is, $y$y is $1$1 more than $x$x). Consequently there is also an inverse relation between $y$y and $x$x, $y=x-1$y=x1 (so $y$y is one less than $x$x).

If, for any value of $x$x there is only one value for $y$y then the relation is a function. $y=x+1$y=x+1 describes a function, but $x^2+y^2=4$x2+y2=4 does not. We can see this because if $x=0$x=0 then $y$y could be either $2$2 or $-2$2.

Some, but not all, functions have inverse functions. This means that the inverse relation is also a function. Since $y=x-1$y=x1 describes a function, it is the inverse function of $y=x+1$y=x+1. However, for example, $y=x^2$y=x2 describes a function, but its inverse relation $y^2=x$y2=x is not a function.

When we say that $y$y is a function of $x$x, we call $x$x the independent variable and $y$y the dependent variable. We sometimes write functions using function notation. For example, $f\left(x\right)$f(x) means $f$f is a function of $x$x. We can then write the inverse function as $f^{-1}\left(x\right)$f1(x).

### Finding inverse functions

To find an inverse function swap the dependent and independent variables. We can then rearrange the equation to make the new dependent variable the subject if necessary.

#### Worked example

Find the inverse function of $f\left(x\right)$f(x) defined by $y=4\left(x-3\right)^3+7$y=4(x3)3+7.

Think: $x$x is the independent variable and $y$y is the dependent variable, so first we swap $x$x and $y$y and then we can rearrange the equation to make $y$y the subject.

Do:

 $x$x $=$= $4\left(y-3\right)^3+7$4(y−3)3+7 Swapping $x$x and $y$y in $f\left(x\right)$f(x) $x-7$x−7 $=$= $4\left(y-3\right)^3$4(y−3)3 Subtracting $7$7 from both sides $\frac{x-7}{4}$x−74​ $=$= $\left(y-3\right)^3$(y−3)3 Dividing both sides by $4$4 $\sqrt[3]{\frac{x-7}{4}}$3√x−74​ $=$= $y-3$y−3 Taking the cube root of both sides $\sqrt[3]{\frac{x-7}{4}}+3$3√x−74​+3 $=$= $y$y Adding $3$3 to both sides

So the inverse function $f^{-1}\left(x\right)$f1(x) is defined by the equation $y=\sqrt[3]{\frac{x-7}{4}}+3$y=3x74+3

Reflect: Notice that each operation we did was reversing an operation in the original function. This will always work with addition, subtraction, multiplication and division (except division by $0$0). It also worked with taking the cube root, but it would not work if we wanted to take the square root instead.

### Graphs of functions and inverses

Graphs represent relations in the same way that equations do. Each point on the graph has an $x$x-value and a $y$y-value which satisfies the rules of the relation.

If a graph represents a function then each $x$x-value has only one corresponding $y$y-value. This means that any vertical line will intersect the graph at at most one point. This gives us the vertical line test: If we can draw a vertical line which intersects the graph more than once then the graph does not represent a function.

The graph of $x=1$x=1 intersects the graphs of $y=x+1$y=x+1 and $y=x^2$y=x2 once each but intersects the graph of $x^2+y^2=4$x2+y2=4 twice. This implies that $x^2+y^2=4$x2+y2=4 is not a function.

When a function has an inverse each value of the dependent variable has only one possible value for the independent variable. On a graph that means that each $y$y-value has only one corresponding $x$x-value and so any horizontal line will intersect the graph at at most one point. This gives us the horizontal line test. If we can draw a horizontal line which intersects the graph more than once then the function does not have an inverse function.

The graph of $y=1$y=1 intersects the graph of $y=x+1$y=x+1 once but intersects the graphs of $y=x^2$y=x2 and $x^2+y^2=4$x2+y2=4 twice each. This implies that $y=x^2$y=x2 and $x^2+y^2=4$x2+y2=4 do not have inverse function.

When we want to find the equation of an inverse function, we swap the positions of the independent and dependent variables. Equivalently, to find the graph of an inverse function, we swap the $x$x-coordinates with $y$y-coordinates. This is the same as reflecting the graph about the line $y=x$y=x.

The graph of the function $y=x+1$y=x+1 reflected about the line $y=x$y=x gives the inverse function $y=x-1$y=x1.

Summary

When we have an equation in two variables then there is a relation between the two variables.

There is also an inverse relation which is the relation going in the opposite direction.

The variable which can take values freely is called the independent variable.

The variable which is determined by the relation and the independent variable is called the dependent variable.

Usually, we call the independent variable $x$x and the dependent variable $y$y.

A function is a relation where for any value of the independent variable there is only one value of the dependent variable.

An inverse function is an inverse relation which is also a function.

Usually the function is called $f\left(x\right)$f(x) and the inverse function is called $f^{-1}\left(x\right)$f1(x).

We can find inverse functions and relations by swapping the position of the dependent and independent variables in the equation and then rearranging the equation to make the new dependent variable the subject.

The vertical line test says that any vertical line will intersect a function at at most one point.

The horizontal line test says that if a function has an inverse function any horizontal line will intersect the function at at most one point.

The graph of an inverse function will be the graph of the original function reflected about the line $y=x$y=x

#### Practice questions

##### Question 1

Do the following graphs have inverse functions?

Yes

A

No

B

Yes

A

No

B

Yes

A

No

B

Yes

A

No

B

Yes

A

No

B

Yes

A

No

B

Yes

A

No

B

Yes

A

No

B

Yes

A

No

B

Yes

A

No

B
##### Question 2

The lines $y=5x$y=5x (labelled $B$B) and $y=x$y=x (labelled $A$A) have been plotted below.

By reflecting $y=5x$y=5x about the line $y=x$y=x, plot the graph of the inverse of $y=5x$y=5x.

Find the inverse function of $y=-8x+6$y=8x+6.