2.05 Inverse functions

We define a function as a rule that assigns each $x$x-value in our domain to only one $y$y-value. We use the notation $y=f\left(x\right)$y=f(x) where $f$f is the function, $x$x is the input value from the domain and $y$y is the output value in the range.

What if we want to find the rule that relates each of these $y$y-values only one $x$x-value? We call this rule the inverse function, since it reverses the rule of $f$f. There's no guarantee that a function always has an inverse. Functions like $y=x^2$y=x2, takes a value of $x$x, and squares it–giving only one value. However for every positive value of $y$y, we know that there are two values of $x$x (each with opposite sign), that when squared, give $y$y.

Hence, a function must have one $x$x-value for every $y$y-value in order for an inverse to exist. We call these functions one-to-one or injective.

To tell if a function is one-to-one, we draw a horizontal line. If this line line cuts the function only once for all horizontal lines then it's one-to-one. As in the example above, $y=x^2$y=x2 is not one-to-one since the horizontal line $y=4$y=4 cuts the function twice.

Worked example

example 1

Which of the following functions have an inverse?

$y=x^3,y=x+5,y=\sin x,y=e^x$y=x3,y=x+5,y=sinx,y=ex

Think: We can imagine drawing each function and passing a horizontal line through each one. If the function is only cut once, then it is one-to-one and has an inverse function.

Do: Each of the functions are shown below. We can see they all have inverses except for $y=\sin x$y=sinx, where the function is cut more than once by the horizontal line.

Practice question

qUESTION 1

 

Restricting the domain

If a function doesn't have an inverse, since it's not one-to-one, we can always reduce the domain of a function so that it guarantees an inverse function exists.

A trivial way to reduce the domain is to reduce it to a single value. We saw above that $y=x^2$y=x2 doesn't have an inverse over all real values of $x$x. But, if we define $y=x^2$y=x2 only for $x=2$x=2, then of course the function is now one-to-one and the inverse function assigns $y=4$y=4 the value $x=2$x=2.

A more meaningful way of reducing the domain is to consider the domain $x\ge0$x≥0. Now each $y$y-value has only has one $x$x-value and the inverse function exists!

You may already be able to tell, but in this case the inverse function is the positive square root. It takes each $y$y-value, takes the positive square root, and returns the one $x$x-value.

Worked example

EXAMPLE 2

Consider the following function defined on all real values of $x$x drawn below. For which of the following values of $x$x does the function have an inverse?

$x\ge2,x\ge-2,-2\le x\le2,x\le2$x≥2,x≥−2,−2≤x≤2,x≤2

Think: The restricted domain should make the resulting function one-to-one.

Do: For $x\ge2$x≥2 and $-2\le x\le2$−2≤x≤2, the function becomes one-to-one so the function has an inverse over these domains.

Practice question

question 2

 

Finding the inverse function

Up to this point, we've only talked about the conditions for when an inverse function exists. Now we'll look at how to actually find the rule for an inverse function. The steps to finding the inverse function is as follows:

1. Make $x$x the subject in the equation of the function.
2. Replace $x$x with $y$y and $y$y with $x$x.

By making $x$x the subject, we are effectively creating a rule that assigns each $y$y-value of the original function, an $x$x-value. Swapping the instances of $x$x and $y$y is just how we conventionally describe functions–an $x$x-value goes in to a function, and a $y$y-value comes out. The inverse function is typically referred to as $f^{-1}$f−1 in function notation.

Notice that the domain and range for a function $f$f swaps for the inverse function $f^{-1}$f−1.

Worked example

example 3

Find the inverse function of $f\left(x\right)=\frac{1}{x^2}$f(x)=1x2​ where $f$f is defined over $x\ge0$x≥0.

Think: Let's use $y=f\left(x\right)$y=f(x), and then make $x$x the subject of the equation.

Do:

$y$y	$=$=	$\frac{1}{x^2}$1x2​
$x^2$x2	$=$=	$\frac{1}{y}$1y​	Multiplying by $x^2$x2 and dividing by $y$y
$x$x	$=$=	$\sqrt{\frac{1}{y}}$√1y​	Taking the positive square root since $x\ge0$x≥0
$x$x	$=$=	$\frac{1}{\sqrt{y}}$1√y​	Simplifying
$y$y	$=$=	$\frac{1}{\sqrt{x}}$1√x​	Swapping $x$x and $y$y

 

So the inverse function is $y=\frac{1}{\sqrt{x}}$y=1√x​ where $f^{-1}\left(x\right)=\frac{1}{\sqrt{x}}$f−1(x)=1√x​.

Reflect: It actually doesn't matter what order we do steps (1) and (2). Step (2) is just a way of relabelling our variables, so we can do this at any point. Notice that the range of values is $y\ge0$y≥0, which is exactly the domain of the original function.

EXAMPLE 4

The volume of a sphere is given by $V=\frac{4}{3}\pi r^3$V=43​πr3, where $r$r is the radius. Find the inverse of $r$r as a function of $V$V, and state the domain and range.

Think: We want to rearrange the equation so that $r$r is the subject.

Do:

$V$V	$=$=	$\frac{4}{3}\pi r^3$43​πr3
$r^3$r3	$=$=	$\frac{V}{\frac{4}{3}\pi}$V43​π​	Dividing by the coefficient of $r^3$r3
$r^3$r3	$=$=	$\frac{3V}{4\pi}$3V4π​	Tidying things up
$r$r	$=$=	$\sqrt[3]{\frac{3V}{4\pi}}$3√3V4π​	Taking the cube root on both sides

 

So the inverse function is $r=\sqrt[3]{\frac{3V}{4\pi}}$r=³√3V4π​. We don't really need to swap the variables $r$r and $V$V, since otherwise we would confuse the radius for the volume.

The domain and range of the inverse function is $V\ge0$V≥0 and $r\ge0$r≥0 since these represent the physical volume and radius of a sphere.

Reflect: Since we've kept the same variables, it's easy to see that the domain and range swap places when going from the original function to the inverse function.

Practice question

question 3