4.05 Inverse functions (Enrichment)

Inverse functions

We call $+7$+7 and $-7$−7 inverse operations, because they do the opposite of each other. 

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

$f(x)=x^3$f(x)=x3 and $g(x)=\sqrt[3]{x}$g(x)=³√x because finding the cube root is the opposite function to cubing a number.

For a function $f$f, the notation $f^{-1}$f−1 is used for the inverse function.

Exploration

If $f(x)=2x$f(x)=2x and $f^{-1}(x)=\frac{x}{2}$f−1(x)=x2​ determine the following:

 $f(-2),f(0),f(3),f(10)$f(−2),f(0),f(3),f(10) and  $f^{-1}(-4),f^{-1}(0),f^{-1}(6),f^{-1}(20)$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(−2)=−4,f(0)=0,f(3)=6,f(10)=20

       $f^{-1}(-4)=-2,f^{-1}(0)=0,f^{-1}(6)=3,f^{-1}(20)=10$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:

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

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

Reflect: Look at the order of the $x$x and $y$y-values. Can you see that they are swapped for $f(x)$f(x) and $f^{-1}(x)$f−1(x)? This is because $f(x)$f(x) and $f^{-1}(x)$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 $x$x and $y$y-values 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.

 

Worked example

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.

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

Do: Swap the coordinates in each 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).

Find the inverse function algebraically

The steps to finding an inverse function are as follows:

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

 

Worked examples

EXAMPLE 2

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

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​. 

 

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 whether the inverse is a function or relation.

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 input values (domain) of the original function would need to be restricted, for example to only positive $x$x-values. 

 

Practice questions

QUESTION 1

QUESTION 2

Question 3

Question 4

Does the inverse function exist?

Remember that we define a function as a rule that assigns each $x$x-value 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 4

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 a graphics calculator or other technology 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 5