Similarly, functions that do the opposite of each other are called inverse functions. For example the function $$f(x)=2x and $$g(x)=x2 are inverse functions because if we multiply a value by $$2, then divide it by $$2 we return to the original value. Another example of a pair of inverse functions is:

$$f(x)=x3 and $$g(x)=³√x because finding the cube root is the opposite function to cubing a number.

For a function $$f, the notation $$f−1 is used for the inverse function.

Exploration

If $$f(x)=2x and $$f−1(x)=x2 determine the following:

$$f(−2),f(0),f(3),f(10) and $$f−1(−4),f−1(0),f−1(6),f−1(20)

Do: $$f(−2)=−4,f(0)=0,f(3)=6,f(10)=20

$$f−1(−4)=−2,f−1(0)=0,f−1(6)=3,f−1(20)=10

Think: If we were to write these as coordinate pairs we would get the following:

$$f(x): $$(−2,4),(0,0),(3,6),(10,20)

$$f−1(x): $$(4,−2),(0,0),(6,3),(20,10)

Reflect: Look at the order of the $$x and $$y-values. Can you see that they are swapped for $$f(x) and $$f−1(x)? This is because $$f(x) and $$f−1(x) are inverse functions. Inverse functions have opposite coordinate pairs.

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

As the $$x and $$y-values are opposites, this means the domain and range of inverse functions are also opposites.

Domain and range for $$f(x) and $$f−1(x)

The domain of a function is the range of the inverse function.

The range of a function is the domain of the inverse function.

Worked example

EXAMPLE 1

Consider the set of coordinates$$(−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 $$x and $$y values.

Do: Swap the coordinates in each pair to give the new set: $$(−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 is included here to illustrate the reflection of the original relation (red) and the inverse (blue).

For this blue inverse set, we can see that the domain = $${−5,−3,−1,1} and the range = $${−1,0,1,2}, which is the opposite of the original red set.

Find the inverse function algebraically

The steps to finding an inverse function are as follows:

Write $$f(x) as $$y.
Swap $$x with $$y and $$y with $$x.
Make $$y the subject in the equation of the function.
Replace $$y with $$f−1(x) notation.
Check that the inverse is in fact a function using the vertical line test.

Worked examples

EXAMPLE 2

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

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

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

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

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

EXAMPLE 3

Find the inverse for the function $$y=x2+4. Graph $$y and its inverse on the same set of axes and state whether the inverse is a function or relation.

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

Do: Swapping $$x and $$y gives $$x=y2+4. Solving for $$y leads to the inverse $$y=±√x−4. The original function $$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 domain of the original function would need to be restricted, for example to only positive $$x-values.

Practice questions

QUESTION 1

Examine the two curves, shown in the graph below.

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A graph of two curves from distinct functions on a Cartesian coordinate system, with both the x-axis and y-axis labeled and ranging from -4 to 4. If both curves pass the vertical line test AND are reflections of each other about the line $$y=x, then they are inverse functions of each other.

Are the curves in the graph inverse functions of each other?
Yes
A
No
B

QUESTION 2

Consider the function given by $$f(x)=x+6 defined over the interval $$[0,∞).

Plot the function $$f(x)=x+6 over its domain.

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Find the inverse $$f−1.
State the domain and range of $$f−1 in interval notation.

Domain: $$

Range: $$
Plot the function $$f−1 over its domain.

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QUESTION 3

Consider the function $$f(x)=6x3.

Find an expression for the inverse function $$f−1(x). You may let $$y=f−1(x).

Question 4

Below we have sketched the line $$y=x24+1 as defined for $$x≤0 (labelled $$B) over the line $$y=x (labelled $$A).