$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 6.6: Inverse Functions

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

A bijection is a function that is both one-to-one and onto. Naturally, if a function is a bijection, we say that it is bijective. If a function $$f :{A}{B}$$ is a bijection, we can define another function $$g$$ that essentially reverses the assignment rule associated with $$f$$. Then, applying the function $$g$$ to any element $$y$$ from the codomain $$B$$, we are able to obtain an element $$x$$ from the domain $$A$$ such that $$f(x)=y$$. Let us refine this idea into a more concrete definition.

Definition: inverse function

Let $$f :{A}\to{B}$$ be a bijective function. Its inverse function is the function $${f^{-1}}:{B}\to{A}$$ with the property that $f^{-1}(b)=a \Leftrightarrow b=f(a).$ The notation $$f^{-1}$$ is pronounced as “$$f$$ inverse.” See Figure [fig:invfcn] for a pictorial view of an inverse function.

Why is $${f^{-1}}:{B}\to{A}$$ a well-defined function? For it to be well-defined, every element $$b\in B$$ must have a unique image. This means given any element $$b\in B$$, we must be able to find one and only one element $$a\in A$$ such that $$f(a)=b$$. Such an $$a$$ exists, because $$f$$ is onto, and there is only one such element $$a$$ because $$f$$ is one-to-one. Therefore, $$f^{-1}$$ is a well-defined function.

If a function $$f$$ is defined by a computational rule, then the input value $$x$$ and the output value $$y$$ are related by the equation $$y=f(x)$$. In an inverse function, the role of the input and output are switched. Therefore, we can find the inverse function $$f^{-1}$$ by following these steps:

• Interchange the role of $$x$$ and $$y$$ in the equation $$y=f(x)$$. That is, write $$x=f(y)$$.
• Solve for $$y$$. That is, express $$y$$ in terms of $$x$$. The resulting expression is $$f^{-1}(x)$$.
• Be sure to write the final answer in the form $$f^{-1}(x) = \ldots\,$$. Do not forget to include the domain and the codomain, and describe them properly.

Example $$\PageIndex{1}\label{invfcn-01}$$

To find the inverse function of $$f :{\mathbb{R}}\to{\mathbb{R}}$$ defined by $$f(x)=2x+1$$, we start with the equation $$y=2x+1$$. Next, interchange $$x$$ with $$y$$ to obtain the new equation $x = 2y+1.$ Solving for $$y$$, we find $$y=\frac{1}{2}\,(x-1)$$. Therefore, the inverse function is ${f^{-1}}:{\mathbb{R}}\to{\mathbb{R}}, \qquad f^{-1}(x)=\frac{1}{2}\,(x-1).$ It is important to describe the domain and the codomain, because they may not be the same as the original function.

Example $$\PageIndex{2}\label{eg:invfcn-02}$$

The function $$s :{\big[-\frac{\pi}{2},\frac{\pi}{2}\big]}{[-1,1]}$$ defined by $$s(x)=\sin x$$ is a bijection. Its inverse function is $\textstyle {s^{-1}}:{[-1,1]}\to{\big[-\frac{\pi}{2},\frac{\pi}{2}\big]}, \qquad s^{-1}(x) = \arcsin x.$ The function $$\arcsin x$$ is also written as $$\sin^{-1}x$$, which follows the same notation we use for inverse functions.

Exercise $$\PageIndex{1}\label{he:invfcn-01}$$

The function $$f :{[-3,\infty)}\to{[\,0,\infty)}$$ is defined as $$f(x)=\sqrt{x+3}$$. Show that it is a bijection, and find its inverse function

Exercise $$\PageIndex{2}\label{he:invfcn-02}$$

Find the inverse function of $$g :{\mathbb{R}}\to{(0,\infty)}$$ defined by $$g(x) = e^x$$.

Remark

Exercise caution with the notation. Assume the function $$f :{\mathbb{Z}}\to{\mathbb{Z}}$$ is a bijection. The notation $$f^{-1}(3)$$ means the image of 3 under the inverse function $$f^{-1}$$. If $$f^{-1}(3)=5$$, we know that $$f(5)=3$$. The notation $$f^{-1}(\{3\})$$ means the preimage of the set $$\{3\}$$. In this case, we find $$f^{-1}(\{3\})=\{5\}$$. The results are essentially the same if the function is bijective.

If a function $$g :{\mathbb{Z}}\to{\mathbb{Z}}$$ is many-to-one, then it does not have an inverse function. This makes the notation $$g^{-1}(3)$$ meaningless. Nonetheless, $$g^{-1}(\{3\})$$ is well-defined, because it means the preimage of $$\{3\}$$. If $$g^{-1}(\{3\})=\{1,2,5\}$$, we know $$g(1)=g(2)=g(5)=3$$.

In general, $$f^{-1}(D)$$ means the preimage of the subset $$D$$ under the function $$f$$. Here, the function $$f$$ can be any function. If $$f$$ is a bijection, then $$f^{-1}(D)$$ can also mean the image of the subset $$D$$ under the inverse function $$f^{-1}$$. There is no confusion here, because the results are the same.

Example $$\PageIndex{3}\label{eg:invfcn-03}$$

The function $$f :{\mathbb{R}}\to{\mathbb{R}}$$ is defined as $f(x) = \cases{ 3x & if x\leq 1, \cr 2x+1 & if x > 1. \cr}$ Find its inverse function.

Solution

Since $$f$$ is a piecewise-defined function, we expect its inverse function to be piecewise-defined as well. First, we need to find the two ranges of input values in $$f^{-1}$$. The images for $$x\leq1$$ are $$y\leq3$$, and the images for $$x>1$$ are $$y>3$$. Hence, the codomain of $$f$$, which becomes the domain of $$f^{-1}$$, is split into two halves at 3. The inverse function should look like $f^{-1}(x) = \cases{ \mbox{???} & if x\leq 3, \cr \mbox{???} & if x > 3. \cr}$ Next, we determine the formulas in the two ranges. We find f^{-1}(x) = \cases{ \textstyle\frac{1}{3}\,x & if x\leq 3, \cr\noalign{\smallskip} \textstyle\frac{1}{2} (x-1) & if x > 3. \cr} The details are left to you as an exercise.

Exercise $$\PageIndex{3}\label{he:invfcn-03}$$

Find the inverse function of $$g :{\mathbb{R}}\to{\mathbb{R}}$$ defined by $g(x) = \cases{ 3x+5 & if x\leq 6, \cr 5x-7 & if x > 6. \cr}$ Be sure you describe $$g^{-1}$$ properly.

Example $$\PageIndex{4}\label{eg:mod10fcn}$$

The function $$g :{\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$ is defined by $$g(x)\equiv 7x+2$$ (mod 10). Find its inverse function.

Solution

From $$x=g(y)\equiv7y+2$$ (mod 10), we obtain $y \equiv 7^{-1}(x-2) \equiv 3(x-2) \pmod{10}.$ Hence, the inverse function $${g^{-1}}:{\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$ is defined by $$g^{-1}(x)\equiv3(x-2)$$ (mod 10).

Exercise $$\PageIndex{4}\label{he:invfcn-04}$$

The function $$h:{\mathbb{Z}_{57}}\to{\mathbb{Z}_{57}}$$ defined by $$h(x)\equiv 49x-3$$ (mod 57). Find its inverse function.

Example $$\PageIndex{5}\label{eg:invfcn-05}$$

Define $$h:{\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$ according to $$h(x)=2(x+3)\bmod10$$. Does $$h^{-1}$$ exist?

Solution

Since $$2^{-1}$$ does not exist, we suspect the answer is no. In fact, $$h(x)$$ is always even, and it is easy to verify that $$\image{h} = \{0,2,4,6,8\}$$. Since $$h$$ is not onto, $$h^{-1}$$ does not exist.

Example $$\PageIndex{6}\label{eg:invfcn-06}$$

Find the inverse function of $$f :{\mathbb{Z}}\to{\mathbb{N}\cup\{0\}}$$ defined by $f(n) = \cases{ 2n & if n\geq0, \cr -2n-1 & if n < 0. \cr}$

Solution

In an inverse function, the domain and the codomain are switched, so we have to start with $${f^{-1}}:{\mathbb{N}\cup\{0\}}\to{\mathbb{Z}}$$ before we describe the formula that defines $$f^{-1}$$. Writing $$n=f(m)$$, we find $n = \cases{ 2m & if m\geq0, \cr -2m-1 & if m < 0. \cr}$ We need to consider two cases.

If $$n=2m$$, then $$n$$ is even, and $$m=\frac{n}{2}$$.

If $$n=-2m-1$$, then $$n$$ is odd, and $$m=-\frac{n+1}{2}$$.

Therefore, the inverse function is defined by ${f^{-1}}:{\mathbb{N}\cup\{0\}}\to{\mathbb{Z}}, \qquad f^{-1}(n) = \cases{ \frac{n}{2} & if n is even, \cr -\frac{n+1}{2} & if n is odd. \cr}$ Verify this with some numeric examples.

Exercise $$\PageIndex{5}\label{he:invfcn-05}$$

The function $$f :{\mathbb{Z}}\to{\mathbb{N}}$$ is defined as $f(n) = \cases{ -2n & if n < 0, \cr 2n+1 & if n\geq0. \cr}$ Find its inverse.

Let $$A$$ and $$B$$ be finite sets. If there exists a bijection $$f :{A}{B}$$, then the elements of $$A$$ and $$B$$ are in one-to-one correspondence via $$f$$. Hence, $$|A|=|B|$$. This idea provides the basis for some interesting proofs.

Example $$\PageIndex{7}\label{eg:invfcn-07}$$

Let $$A=\{a_1,a_2,\ldots,a_n\}$$ be an $$n$$-element sets. Recall that the power set $$\wp(A)$$ contains all the subsets of $$A$$, and $\{0,1\}^n = \{(b_1,b_2,\ldots,b_n) \mid b_i\in\{0,1\} \mbox{ for each i, where 1\leq i\leq n} \}.$ Define $$F:{\wp(A)}\to{\{0,1\}^n}$$ according to $$F(S) = (x_1,x_2,\ldots,x_n)$$, where $x_i = \cases{ 1 & if a_i\in S, \cr 0 & if a_i\notin S. \cr}$ Simply put, $$F(S)$$ is an ordered $$n$$-tuple whose $$i$$th entry is either 1 or 0, indicating whether $$S$$ contains the $$i$$th element of $$A$$ (1 for yes, and 0 for no).

It is clear that $$F$$ is a bijection. For $$n=8$$, we have, for example, $F(\{a_2,a_5,a_8\}) = (0,1,0,0,1,0,0,1),$ and $F^{-1}\big((1,1,0,0,0,1,1,0)\big) = \{a_1,a_2,a_6,a_7\}.$ The function $$F$$ defines a one-to-one correspondence between the subsets of $$A$$ and the ordered $$n$$-tuples in $$\{0,1\}^n$$. Since there are two choices for each entry in these ordered $$n$$-tuples, we have $$2^n$$ such ordered $$n$$-tuples. This proves that $$|\wp(A)|=2^n$$, that is, $$A$$ has $$2^n$$ subsets.

Exercise $$\PageIndex{6}\label{he:invfcn-06}$$

Consider the function $$F$$ defined in Example 6.6.7. Assume $$n=8$$. Find $$F(\emptyset)$$ and $$F^{-1}\big( (1,0,1,1,1,0,0,0)\big)$$.

## Summary and Review

• A bijection is a function that is both one-to-one and onto.
• The inverse of a bijection $$f :{A}{B}$$ is the function $${f^{-1}}:{B}\to{A}$$ with the property that $f(x)=y \Leftrightarrow x=f^{-1}(y).$
• In brief, an inverse function reverses the assignment rule of $$f$$. It starts with an element $$y$$ in the codomain of $$f$$, and recovers the element $$x$$ in the domain of $$f$$ such that $$f(x)=y$$.

Exercise $$\PageIndex{1}\label{ex:invfcn-01}$$

Which of the following functions are bijections? Explain!

$$f :{\mathbb{R}}\to{\mathbb{R}}$$, $$f(x)=x^3-2x^2+1$$.

$$g :{[\,2,\infty)}\to{\mathbb{R}}$$, $$g(x)=x^3-2x^2+1$$.

$$h:{\mathbb{R}}\to{\mathbb{R}}$$, $$h(x)=e^{1-2x}$$.

$$p :{\mathbb{R}}\to{\mathbb{R}}$$, $$p(x)=|1-3x|$$.

$$q:{[\,2,\infty)}\to{[\,0,\infty)}$$, $$q(x)=\sqrt{x-2}$$.

Exercise $$\PageIndex{2}\label{ex:invfcn-02}$$

For those functions that are not bijections in the last problem, can we modify their codomains to change them into bijections?

Exercise $$\PageIndex{3}\label{ex:invfcn-03}$$

Let $$f$$ and $$g$$ be the functions from $$(1,3)$$ to $$(4,7)$$ defined by $f(x) = \frac{3}{2}\,x+\frac{5}{2}, \qquad\mbox{and}\qquad g(x) = -\frac{3}{2}\,x+\frac{17}{2}.$ Find their inverse functions. Be sure to describe their domains and codomains.

Exercise $$\PageIndex{4}\label{ex:invfcn-04}$$

Find the inverse function $$f :{\mathbb{R}}\to{\mathbb{R}}$$ defined by $f(x) = \cases{ 3x+5 & if x\leq 6, \cr 5x-7 & if x > 6. \cr}$

Be sure you describe $$f^{-1}$$ correctly and properly.

Exercise $$\PageIndex{5}\label{ex:invfcn-05}$$

The function $$g :{[\,1,3\,]}\to{[\,4,\,7]}$$ is defined according to $g(x) = \cases{ x+3 & if 1\leq x< 2, \cr 11-2x & if 2\leq x\leq 3. \cr}$ Find its inverse function. Be sure you describe it correctly and properly.

Exercise $$\PageIndex{6}\label{ex:invfcn-06}$$

Find the inverse of the function $$r :{(0,\infty)}\to{\mathbb{R}}$$ defined by $$r(x)=4+3\ln x$$.

Exercise $$\PageIndex{7\label{ex:invfcn-07}$$

Find the inverse of the function $$s :{\mathbb{R}}\to{(-\infty,-3)}$$ defined by $$s(x)=4-7e^{2x}$$.

Exercise $$\PageIndex{8}\label{ex:invfcn-08}$$

Find the inverse of each of the following bijections.

1. $$h:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}$$, $$h(1)=e$$, $$h(2)=c$$, $$h(3)=b$$, $$h(4)=a$$, $$h(5)=d$$.
2. $$k :{\{1,2,3,4,5\}}\to{\{1,2,3,4,5\}}$$, $$k(1)=3$$, $$k(2)=1$$, $$k(3)=5$$, $$k(4)=4$$, $$k(5)=2$$.

Exercise $$\PageIndex{9}\label{ex:invfcn-09}$$

Find the inverse of each of the following bijections.

1. $$u:{\mathbb{Q}}\to{\mathbb{Q}}$$, $$u(x)=3x-2$$.
2. $$v:{\mathbb{Q}-\{1\}}\to{\mathbb{Q}-\{2\}}$$, $$v(x)=\frac{2x}{x-1}$$.
3. $$w:{\mathbb{Z}}\to{\mathbb{Z}}$$, $$w(n)=n+3$$.

Exercise $$\PageIndex{10}\label{ex:invfcn-10}$$

Find the inverse of each of the following bijections.

$$r :{\mathbb{Z}_{12}}\to{\mathbb{Z}_{12}}$$, $$r(n)\equiv 7n$$ (mod 12).

$$s :{\mathbb{Z}_{33}}\to{\mathbb{Z}_{33}}$$, $$s(n)\equiv 7n+5$$ (mod 33).

$$t :{\mathbb{Z}}\to{\mathbb{N}\cup\{0\}}$$, $$t(n) = \cases{ 2n-1 & if$$n > 0$$\cr -2n & if$$n0$$\cr}$$

Exercise $$\PageIndex{11}\label{ex:invfcn-11}$$

The images of the bijection $${\alpha}:{\{1,2,3,4,5,6,7,8\}}\to{\{a,b,c,d,e,f,g,h\}}$$ are given below. $\begin{array}{|c||*{8}{c|}} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \alpha(x)& g & a & d & h & b & e & f & c \\ \hline \end{array}$ Find its inverse function.

Exercise $$\PageIndex{12}\label{ex:invfcn-12}$$

Below is the incidence matrix for the bijection $${\beta}: {\{a,b,c,d,e,f\}}\to{\{x,y,z,u,v,w\}}$$. $\begin{array}[t]{cc} & \begin{array}{cccccc} u & v & w & x & y & z \end{array} \\ \begin{array}{c} a \\ b \\ c \\ d \\ e \\ f \end{array} & \left(\begin{array}{cccccc} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \end{array}\right) \end{array}$ Find its inverse function.