4.09 Inverse functions

Relations and their inverses

Recall from relations and functions that a relation can be described using a table, a rule, a graph, a set of ordered pairs or a diagram mapping elements of two different sets. A function is a special type of relation, where each input only has one output.

Finding the inverse of a relation involves swapping the order of the ordinates in the ordered pairs. This can be done algebraically by swapping $x$x and $y$y values, or graphically by a reflection over the line $y=x$y=x. This graphical method works because a reflection over the line $y=x$y=x swaps all $x$x and $y$y values, except those sitting on the "mirror line" of $y=x$y=x.

If the inverse relation is also a function, it is referred to as the inverse function. For a function $f$f, the notation $f^{-1}$f−1 is used for the inverse function.

Remember that a relation is a function if each input has only one output or, graphically, if it passes the vertical line test.

 

Worked examples

EXAMPLE 1

Consider the set of coordinates$(-1,-5),(0,-3),(1,-1),(2,1)$(−1,−5),(0,−3),(1,−1),(2,1). Graph this set, find the inverse relation and sketch the graph of the inverse on the same axes, stating its domain and range.

Think: This question requires swapping $x$x and $y$y ordinates.

Do: Swap the ordinates in each coordinate pair to give the new set: $(-5,-1),(-3,0),(-1,1),(1,2)$(−5,−1),(−3,0),(−1,1),(1,2). This is the set of points of the inverse relation, which is graphed below:

The dotted line $y=x$y=x is included here to illustrate the reflection of the original relation (red) and the inverse (blue).

For this set, we can see that the domain = $\left\{-5,-3,-1,1\right\}${−5,−3,−1,1} and the range = $\left\{-1,0,1,2\right\}${−1,0,1,2}.

 

EXAMPLE 2

Find the inverse for the function $f(x)=3x+1$f(x)=3x+1. Graph $f$f and its inverse on the same set of axes and state the domain and range of each.

Think: Let $f(x)=y$f(x)=y. This question requires swapping $x$x and $y$y and solving the resulting equation for $y$y.

Do: Swapping $x$x and $y$y gives $x=3y+1$x=3y+1. Solving for $y$y leads to the inverse $y=\frac{x-1}{3}$y=x−13​. The original function $y$y and its inverse are plotted below:

The intersection between $y$y and its inverse is found by setting $3x+1=\frac{x-1}{3}$3x+1=x−13​ and solving for $x=\frac{-1}{2}$x=−12​. Since the intersection will always sit on the line $y=x$y=x (why?), the coordinate is $(\frac{-1}{2},\frac{-1}{2})$(−12​,−12​). Note that the inverse relation passes the vertical line test, which means that the function $f(x)=3x+1$f(x)=3x+1 has an inverse function $f^{-1}(x)=\frac{x-1}{3}$f−1(x)=x−13​. 

The domain and range for both $f(x)$f(x) and its inverse are all real numbers in $x$x and $y$y.

 

EXAMPLE 3

Find the inverse for the function $y=x^2+4$y=x2+4. Graph $y$y and its inverse on the same set of axes and state the domain and range of each.

Think: This question again requires swapping $x$x and $y$y and solving the equation for $y$y.

Do: Swapping $x$x and $y$y gives $x=y^2+4$x=y2+4. Solving for $y$y leads to the inverse $y=\pm\sqrt{x-4}$y=±√x−4. The original function $y$y and its inverse are plotted below:

Note that the inverse here is a relation and not a function as it fails the vertical line test. For the inverse to be a function, the range would need to be restricted. 

The domain of $y=x^2+4$y=x2+4 is all real $x$x-values, The range is $y\ge4$y≥4. The domain of the inverse $y=\pm\sqrt{x-4}$y=±√x−4 is $x\ge4$x≥4. The range of the inverse is all real $y$y-values.

 

In the examples above, notice that the domain of the original function becomes the range of the inverse. Similarly, the range of the original function becomes the domain of the inverse. 

 

Practice questions

QUESTION 1

Question 2

 

Inverse functions and domain restrictions

As the previous worked example demonstrated, the inverse of a function may not always be a function itself. By restricting the domain of the original function $f$f, the inverse can be guaranteed to be a function if $f$f is restricted to be one-to-one. This means that for a function $y=f(x)$y=f(x), there is only one $x$x-value for any $y$y-value and vice versa. 

A function is one-to-one if it passes both the horizontal and vertical line tests:

Examples of one-to-one functions could be:

Note that any horizontal or vertical lines drawn through these graphs would only cut through them once.

 

Worked example

Example 4

Restrict the domain of the function $f(x)=(x-1)^2+4$f(x)=(x−1)2+4 to find the inverse function.

Think: What is the largest domain this function can be restricted to so that it is one-to-one?

Do: There are two different restrictions that could be made to the domain of $f(x)$f(x): the left side of the parabola or the right side, including the $x$x-value of the turning point.

From graphing quadratics previously, we know that the turning point of $f(x)=(x-1)^2+4$f(x)=(x−1)2+4 is at $(1,4)$(1,4). So, we will restrict the domain to $x\ge1$x≥1. This implies that the range for $f$f given this restriction is $y\ge4$y≥4.

To find the inverse function, let $f(x)=y$f(x)=y and swap $x$x and $y$y to give $x=(y-1)^2+4$x=(y−1)2+4. Solving for $y$y, we have the inverse function $y=\sqrt{x-4}+1$y=√x−4+1. Note that only the positive branch is chosen here due to the domain and range restrictions.

Remember that the domain and range from $f$f swap for the inverse function, which means that the inverse $f^{-1}(x)=\sqrt{x-4}+1$f−1(x)=√x−4+1 has domain $x\ge4$x≥4 and range $y\ge1$y≥1.

The functions $f$f and $f^{-1}$f−1, with their restricted domains, are both graphed on the same axes below. 

Notice that the graphs are reflected over the dashed mirror line $y=x$y=x, as expected.

Practice questions

QUESTION 3

Question 4