Restrict Domain to Obtain Inverse Function

The inverse of a function is also a function only if the original function is one-to-one.

In other words, for some function $f$f, the inverse function $f^{-1}$f−1 exists if $f$f is one-to-one. We can then say that $f$f is invertible.

Some functions are already one-to-one over their natural domain, so their inverse exists and is also one-to-one. These are invertible functions and include linear functions like $f(x)=ax+b$f(x)=ax+b, and hyperbolic functions like $f(x)=\frac{k}{x}$f(x)=kx​.

Only one-to-one functions are invertible.

But for the type of function we call many-to-one, its inverse is not a function. Quadratics and other even-power polynomials, trigonometric functions, and semicircles are all examples of many-to-one functions.

To be able to get an inverse function from a many-to-one function we need to restrict the domain of the original function so that it becomes one-to-one.

Testing invertibility

For a function described as a set of ordered pairs, we can test for invertibility by checking that for every pair of values the $y$y-values appear only once in the set. If one of the $x$x-values has a $y$y-value that matches another $x$x-value then the function is many-to-one and will not be invertible.

 

 

We can see with the function on the left-hand side that both the second and third pairs have a $y$y-value of $3$3, so this is a many-to-one function and is not invertible. The function on the right hand side has unique $y$y-values for each $x$x-value and so it is invertible.

Horizontal line test

We can use the horizontal line test to determine whether a function is one-to-one or many-to-one, and hence whether it is invertible.

The key property of a many-to-one function is that there are two or more values of $x$x from the domain that correspond to the same value of $f(x)$f(x) in the range.

Let's say that $f(a)=c$f(a)=c and $f(b)=c$f(b)=c, so that the numbers $a$a and $b$b from the domain correspond to the same number $c$c in the range. Graphically, we can think of the points $\left(a,c\right)$(a,c) and $\left(b,c\right)$(b,c) as being the points of intersection of the graph $y=f(x)$y=f(x) and the horizontal line $y=c$y=c.

$f\left(x\right)$f(x) does not pass the horizontal line test.

If we can draw a horizontal line that intersects the graph of a function in at least two places, then the function is many-to-one. If, however, there is no horizontal line that we can draw that passes through the graph of the function more than once, then the function "passes the horizontal line test", and we know it is one-to-one and invertible.

Worked example

Determine whether the function $f(x)=2x-3$f(x)=2x−3 has an inverse.

Think: This is a linear function, and its graph is a straight line that increases at a constant rate. The natural domain of $f\left(x\right)$f(x) is $\left(-\infty,\infty\right)$(−∞,∞).

Do: Using the horizontal line test, we can see that any horizontal line we draw will intersect the graph of $f(x)$f(x) at only one point.

$f\left(x\right)$f(x) does pass the horizontal line test.

The function passes the horizontal line test, so it is a one-to-one function. This means that it does have an inverse.

Reflect: Find the inverse of a function amounts to reflecting the graph of the function across the line $y=x$y=x. Algebraically, this is equivalent to swapping the $x$x and $y$y variables in the equation. Let's use this idea to find $f^{-1}$f−1, the inverse of $f(x)$f(x).

$f(x)$f(x)	$=$=	$2x-3$2x−3
$y$y	$=$=	$2x-3$2x−3	(Substitute $f(x)=y$f(x)=y for notational convenience)
$x$x	$=$=	$2y-3$2y−3	(Swap $x$x and $y$y in the equation)
$x+3$x+3	$=$=	$2y$2y
$\frac{x+3}{2}$x+32​	$=$=	$y$y	(Rearrange to solve for $y$y)
$\frac{x+3}{2}$x+32​	$=$=	$f^{-1}\left(x\right)$f−1(x)	(Identify $y$y with $f^{-1}\left(x\right)$f−1(x))

The function $f(x)=2x-3$f(x)=2x−3 has an inverse function $f^{-1}\left(x\right)=\frac{x+3}{2}$f−1(x)=x+32​. We have determined the inverse without needing to restrict the natural domain of $f\left(x\right)$f(x).

Making a function invertible

If a function is not one-to-one over a particular domain, we can restrict the domain in such a way that the graph of the function passes the horizontal line test, and the function becomes invertible.

Notice that in general the actual algebraic expression for the function will not change in this process. We are only reducing the domain, the set of possible input values that we can feed into the function.

Worked example

Choose a domain restriction on the function $f\left(x\right)=\left(x-4\right)^2+1$f(x)=(x−4)2+1 that makes $f\left(x\right)$f(x) invertible.

Think: This function is a quadratic, so there are two values of $x$x in the natural domain $\left(-\infty,\infty\right)$(−∞,∞) that correspond to each value in the range $\left[1,\infty\right)$[1,∞), except for the point representing the vertex (or turning point) at $\left(4,1\right)$(4,1).

Do: Looking at the graph of $f\left(x\right)$f(x), if we select any interval that includes the vertex of the parabola then we will still have a many-to-one function. We need to choose a domain that represents at most one half of the parabola.

$f\left(x\right)$f(x) is not one-to-one over its natural domain.

If we restrict the domain of $f\left(x\right)$f(x) to any of the following intervals, then the resulting function will be one-to-one and hence invertible.

$\left[4,\infty\right)$[4,∞), $\left(4,\infty\right)$(4,∞), $\left[6,11\right)$[6,11), $\left(-23,1\right)$(−23,1)

These are just a few valid intervals. There are infinitely many we could choose.

$f\left(x\right)$f(x) is one-to-one over the restricted domain $\left[4,\infty\right)$[4,∞).
The inverse function $f^{-1}\left(x\right)$f−1(x) is a reflection across the line $y=x$y=x.

In contrast, if we restrict the domain of $f\left(x\right)$f(x) to any of the following intervals, then the resulting function will not be one-to-one, since each interval still contains the vertex.

$\left[3,\infty\right)$[3,∞), $\left(0,8\right)$(0,8), $\left(-23,4.01\right)$(−23,4.01)

Reflect: We could select any of the valid restricted domains to make $f\left(x\right)$f(x) invertible, but for this function there happens to be two domains in particular that retain the most of the original domain. These are $\left(-\infty,4\right]$(−∞,4] and $\left[4,\infty\right)$[4,∞), and they correspond to the left and right half of the parabola respectively. We can think of these domains as the "least restricted" or the "largest" domains for which $f\left(x\right)$f(x) is invertible.

Practice questions

Question 1

Question 2

Question 3