Chapter 2.5 Inverse Functions
In this section, we learn about how to reflect a function across the line \(y=x\). In some cases, the reflection itself is a function, called an inverse function. In some cases, it is a not a function, but we can restrict the original function so its reflection does become a function. In either case, we will learn about some properties these new inverse function possess.
Reflection and Inverse Functions
Take \(f\) to be a real valued function on \(\mathbb R\).
Recall that the reflection of the point \((a,b)\) across the line \({\rm pow}_1\) is the point \((b,a)\).
We will define an inverse relation and what it means for a function to be invertible.
Inverse Relation and Invertible
The inverse relation \(f^{-1}\) (read: eff inverse) is the reflection of \(f\) across the line \({\rm pow}_1\).
For any point \(y\) in the range of \(f\), define \(f^{-1}(y)\) by \[f^{-1}(y) = \{x\in \mathcal D(f)\colon f(x) = y\}.\]
If for each \(y\) in the range of \(f\), \(f^{-1}(y)\) contains only one element, then \(f\) is invertible.
The function \(f\) is invertible if and only if it passes the horizontal line test, that is, \[f(x_1) = f(x_2) \quad \text{implies that}\quad x_1 = x_2.\]
Why? What happens if it does not pass this test?
See this image below:
If it does not pass the horizontal line test, then the set \(f^{-1}(y)\) contains more than one point because there are two \(x\) for which \(f(x)=y\) has a solution.
These special kinds of function that are invertible have the following associated function.
Inverse Function
If \(f\) is invertible, then for every \(y\) in the range of \(f\), there is a single \(x\) in the domain of \(f\) with \[f^{-1}(y) = \{x\}.\]
In this case, redefine \(f^{-1}\) to be the function (the inverse function) that takes \(y\) to \(x\), that is, \[f^{-1}(y) = x \quad {\rm where}\quad f(x) = y.\]
Here we will find the inverse function of a linear function.
Example 1
Solve for \(x\) as a function of \(y\) in order to determine \(f^{-1}\), where \(f\) is the function given by \[f(x) = 2x + 1.\]
There are two ways to think about this.
Reflecting a line produces another line. Find \(f^{-1}\) by reflecting two different points on \(f\) over the line \(y=x\) and then creating a line that passes through those two points. For example \((0,1)\) and \((1,3)\) are on \(f\), so the reflected points \((1,0)\) and \((3,1)\) are on \(f^{-1}\). The line that passes through \((1,0)\) and \((3,1)\) is \(y=\frac{1}{2}(x-1)+0\). Thus \(f^{-1}(x)=\frac{1}{2}(x-1)\).
The definition of the inverse is \(f^{-1}=\{(y,x)\colon y=2x+1\}=\{(x,y)\colon x=2y+1\}.\) So solve \(2y+1=x\) for \(y\), to get \(y=\frac{1}{2}(x-1)\).
Try it again with this example.
Example 2
Understanding the geometry of the problem allows you to solve the previous example without making a single calculation. Repeat the previous example, but for \(f\) given by \[f(x) = 4x + 5.\]
Reflecting a line reverses the role of \(x\) and \(y\). So then the slope will be the recipocal of the original. The slope of \(f\) is \(4\), which means the slope of \(f^{-1}\) will be \(\frac{1}{4}\). Because \((0,5)\) is a point that \(f\) passes through, the line \(f^{-1}\) will pass through the point \((5,0)\). Thus \(f^{-1}(x)=\frac{1}{4}(x-5).\)
No we do an example that is not a linear function.
Example 3
Find the inverse of \(f\) where \(f\) is given by \[f(x) = \dfrac{2x - 6}{x+1}.\] What are the domain and range of \(f\) and \(f^{-1}\)?
This is a rational function, so find the inverse by solving for \(y\) in the equation \[x=\frac{2y-6}{y+1}.\]
\[ \begin{align*} x&=\frac{2y-6}{y+1}\\ x(y+1)&=2y-6\\ xy+x&=2y-6\\ xy-2y&=-6-x\\ y(x-2)&=-6-x\\ y&=\frac{-6-x}{x-2}. \end{align*} \]
Thus \(f^{-1}(x)=\frac{-6-x}{x-2}\).
The domain of \(f\) is \((-\infty,-1)\cup(-1,\infty)\) and the domain of \(f^{-1}\) is \((-\infty,2)\cup(2,\infty)\). To find the range for \(f\), find all \(b\) for which \[b=\frac{2x-6}{x+1}.\] If you solve for \(x\), you’ll get \[x=\frac{-6-b}{b-2},\] which is actually the formula for \(f^{-1}\). So the range of \(f\) is the domain of \(f^{-1}\). Similarly, the range of \(f^{-1}\) will be the domain of \(f\).
Thus \(\mathcal{D}(f)=(-\infty,-1)\cup(-1,\infty)\), \(\mathcal{R}(f)=(-\infty,2)\cup(2,\infty)\), \(\mathcal{D}(f^{-1})=(-\infty,2)\cup(2,\infty)\), and \(\mathcal{R}(f^{-1})=(-\infty,-1)\cup(-1,\infty)\).
Here are some important facts about an invertible function that we saw were true in the pervious example.
Invertible Function Facts
Fact: If \(f\) is an invertible function, then \[\mathcal D(f) = \mathcal R(f^{-1})\quad \text{and} \quad \mathcal D(f^{-1}) = \mathcal R(f).\]
This should make sense because a point \((x,y)\) in \(f\) becomes the point \((y,x)\) in \(f^{-1}\).
Invertible Function Properties
Suppose \(f\) is an invertible function and \(f^{-1}\) is its inverse function.
For any \(x\) in \(\mathcal D(f)\), \[f^{-1}(f(x)) = x.\]
For any \(y\) in \(\mathcal R(f)\), \[f(f^{-1}(y)) = y.\]
Note however that \(f\circ f^{-1}\) and \(f^{-1}\circ f\) may have different domains and so may not be equal.
This should make sense as reflecting a point twice along the line \(y=x\) should bring it back to where it originally was.
However, it is important to pay attention to the domain. See this example.
Example 4
For each \(x\) in \((-\infty, 0]\), take \(f\) to be the function that is given by \[f(x) = -x.\] Compute \(f\circ f^{-1}\) and \(f^{-1}\circ f\).
Note that \(\mathcal{D}(f)=(-\infty, 0]\) and \(\mathcal{R}(f)=[0,\infty)\) (since \(f(x)=-x\)).
First find the inverse of \(f\). You should get \(f^{-1}(x)=-x\). So therefore \(\mathcal{D}(f^{-1})=[0,\infty)\) and \(\mathcal{R}(f^{-1})=(-\infty, 0]\)
Thus
\[\left(f\circ f^{-1}\right)(x)=f(-x)=-(-x)=x\]
for \(x \in [0,\infty)\) and
\[\left(f^{-1}\circ f\right)(x)=f^{-1}(-x)=-(-x)=x\]
for \(x \in (-\infty, 0].\)
In the pervious examples, we had functions given by formulas. What if we do not have a formula? In this case, the idea extends more generally like this:
Let \(f\) be a function from a nonempty set \(X\) to a nonempty set \(Y\).
For every \(x\) in the domain of \(f\), there is a unique \(y\) so that \((x,y)\) is in \(f\). This is to say that \(f\) is single valued.
The inverse relation \(f^{-1}\) is the set \[f^{-1} = \{(y,x)\colon (x,y) \in f\}.\]
For any point \(y\) in the range of \(f\), define \(f^{-1}(y)\) by \[f^{-1}(y) = \{x\in \mathcal D(f)\colon f(x) = y\}.\]
A function \(f\) is invertible if the relation \(f^{-1}\) defines a function from \(Y\) to \(X\).
In this case, the inverse relation \(f^{-1}\) gives the unique \(x\) value in the domain of \(f\) for a given \(y\) value in the range of \(f\).
If \(f\) is invertible, redefine \(f^{-1}\) to be the function that is given by \[f^{-1}(y) = x\] rather than the relation that is given by \[f^{-1}(y) = \{x\}.\]
Now let’s see it with this next example. Pay attention to how we redefine \(f\) to ensure it is invertible.
Example 5
Suppose that the function \(f\) is given by \[f = \{(a,3), (b, 2), (c, 3), (d, 3), (e, 5), (g, 7), (h, 2)\}.\] Determine \(f^{-1}(3)\), \(f^{-1}(2)\), and \(f^{-1}(e)\). Eliminate some points in \(f\) so that it becomes invertible.
\[f^{-1} = \{(3,a), (2, b), (3, c), (3, d), (5, e), (7, g), (2, h)\}.\]
\[f^{-1}(3)=\{a,c,d\}\]
\[f^{-1}(2)=\{b,h\}\]
\[f^{-1}(e)=\{\}\]
There are many ways to make \(f\) invertible. Just make sure \(y\) coordinate is repeated. Here are two options.
\(f = \{(a,3), (b, 2), (e, 5), (g, 7)\}.\)
\(f=\{ (c, 3), (e, 5), (g, 7), (h, 2)\}\)
Restricting Domain to Guarantee Invertibility
A function may not be invertible on one domain, but may be invertible if restricted to a smaller domain. In this case, there may be many choices for the domain to use as the domain of the restricted function. Different choices of domain lead to different inverse functions.
Example 6
Find some restrictions of the function \({\rm pow}_2\) that guarantee that \({\rm pow}_2\) is invertible and for each of these restrictions, find a formula for the inverse function.
Does not pass the horizontal line test. Here is one restriction.
\(f\big|_{[0,\infty)}\)
Reflect over \(y=x\) to get its inverse
The inverse is \(f^{-1}\big|_{[0,\infty)}(x)=\sqrt{x}.\)
Another restriction is \(f\big|_{(-\infty,0]}\)
Reflect over \(y=x\) to get its inverse
The inverse is \(f^{-1}\big|_{(-\infty,0]}(x)=-\sqrt{x}.\)
Another restriction is \(f\big|_{[-2,-1]\cup[0,1)\cup(2,\infty)}\)
Reflect over \(y=x\) to get its inverse
The inverse is \[ \begin{align*} f^{-1}=\begin{cases} \sqrt{x}&\text{ if } x\in[0,1]\\ -\sqrt{x}&\text{ if } x\in[1,4]\\ \sqrt{x}&\text{ if } x\in(4,\infty)\\ \end{cases} \end{align*} \]
This next theorem summarizes which kind of functions are invertible.
Invertible Functions
For any natural number \(n\), if \(n\) is odd, then \({\rm pow}_n\) is invertible.
For any natural number \(n\), if \(n\) is even, then \({\rm pow}_n\) is invertible when restricted to \([0, \infty)\).
Define the function \({\rm pow}_n^{-1}\) as the inverse function of \({\rm pow}_n\).
Note: Denote \({\rm pow}_n^{-1}\) by \({\rm pow}_{\frac{1}{n}}\) so that \(x^{\frac{1}{n}}\) is defined to be \({\rm pow}_{\frac{1}{n}}(x)\).
Let’s look at the square root function.
Example 7
Define, evaluate, and sketch the square root function.
\(\text{pow}_{1/2}=\sqrt{x}\)
The points \((0,0),(1,1),(2,4),(3,9)\) are in \(f=\rm{pow_2}\) so the points \((0,0),(1,1),(4,2),(9,3)\) are in \(f^{-1}=\rm{pow_{1/2}}\)
Let’s look at the inverse of \({\rm pow_3}\).
Example 8
Reflect the function \({\rm pow_3}\) across \({\rm pow}_1\). Is \({\rm pow}_3^{-1}\) a function? If so, identify the function and try to evaluate it at \(-27\), \(1\), \(8\)
The points \((-3,-27),(-2,-8),(-1,-1),(0,0),(1,1),(2,8),(3,27)\) are in \({\rm pow_3}\) so
\((-27,-3),(-8,-2),(-1,-1),(0,0),(1,1),(8,2),(27,3)\) are in \({\rm pow}_3^{-1}\).
The function \(f^{-1}\) is a function and \(f^{-1}={\rm pow}_{1/3}=x^{1/3}\).
Finally, let’s look at a more complicated example, that is a transformed \({\rm pow_3}\).
Example 9
Take \(f\) to be the function that is given by \[f(x) = 5(x + 2)^3 + 4.\] Reflect \(f\) across \({\rm pow}_1\) to obtain a new set of points is a function. What is the function?
Solve for \(y\) in the equation \(x=5(y+2)^3+4\).
\[ \begin{align*} x&=5(y+2)^3+4\\ x-4&=5(y+2)^3\\ \frac{x-4}{5}&=(y+2)^3\\ \left(\frac{x-4}{5}\right)^{1/3}&=y+2\\ \left(\frac{x-4}{5}\right)^{1/3}-2&=y. \end{align*} \]