5.08 Inverse functions and restricted domains

Does the inverse function exist?

Remember that we define a function as a rule that assigns each $x$x-value in our domain to only one $y$y-value.

There's no guarantee that a function always has an inverse function. For example, a function like $y=x^2$y=x2, takes a value of $x$x, and squares it, so the points $(2,4)$(2,4) and $(-2,4)$(−2,4) are points on the original function . However when we swap the coordinates to form the inverse function we get $(4,2)$(4,2) and $(4,-2)$(4,−2) so we now have two $y$y-values for the same $x$x-value. Our inverse is NOT a function! 

One-to-one functions

A function must have one $x$x-value for every $y$y-value in order for an inverse function 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 sketch each function, or use our graphics calculator to draw each one and pass an imaginary 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 (AA only)

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 way of reducing the domain of the function $y=x^2$y=x2,  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 examples

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.

EXAMPLE 3

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 questions

question 2

 

question 3