Identify the bijective function from Z Into z a B c f(x)=2x+1 d f(x x 2 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 \[f :A \to B\] 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 \[\PageIndex{1}\] 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:

1. Interchange the role of \[x\] and \[y\] in the equation \[y=f[x]\]. That is, write \[x=f[y]\].
2. 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 \[x = 2y+1. \nonumber\] Solving for \[y\], we find \[y=\frac{1}{2}\,[x-1]\]. Therefore, the inverse function is \[{f^{-1}}:{\mathbb{R}}\to{\mathbb{R}}, \qquad f^{-1}[x]=\frac{1}{2}\,[x-1]. \nonumber\] 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 \[s :{\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]}\to{[-1,1]}\] defined by \[s[x]=\sin x\] is a bijection. Its inverse function is

\[s^{-1}:[-1,1] \to {\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]}, \qquad s^{-1}[x]=\arcsin x. \nonumber\]

The function \[\arcsin x\] is also written as \[\sin^{-1}x\], which follows the same notation we use for inverse functions.

hands-on 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

hands-on 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 \[f[x] = \cases{ 3x & if $x\leq 1$, \cr 2x+1 & if $x > 1$. \cr} \nonumber\] Find its inverse function.

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 \[f^{-1}[x] = \cases{ \mbox{???} & if $x\leq 3$, \cr \mbox{???} & if $x > 3$. \cr} \nonumber\] Next, we determine the formulas in the two ranges. We find

\[f^{-1}[x] = \cases{ \textstyle\frac{1}{3}\,x & if $x\leq 3$, \cr \textstyle\frac{1}{2} [x-1] & if $x > 3$. \cr} \nonumber\] The details are left to you as an exercise.

hands-on Exercise \[\PageIndex{3}\label{he:invfcn-03}\]

Find the inverse function of \[g :{\mathbb{R}}\to{\mathbb{R}}\] defined by \[g[x] = \cases{ 3x+5 & if $x\leq 6$, \cr 5x-7 & if $x > 6$. \cr} \nonumber\] 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.

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

hands-on 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?

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 \[\text{im}h = \{0,2,4,6,8\}\]. 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 \[f[n] = \cases{ 2n & if $n\geq0$, \cr -2n-1 & if $n < 0$. \cr} \nonumber\]

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 \[n = \cases{ 2m & if $m\geq0$, \cr -2m-1 & if $m < 0$. \cr} \nonumber\] We need to consider two cases.

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

Therefore, the inverse function is defined by \[f^{-1}:\mathbb{N} \cup \{0\} \to \mathbb{Z}\] by:

\[f^{-1}[n] = \cases{ \frac{2}{n} & if $n$ is even, \cr -\frac{n+1}{2} & if $n$ is odd. \cr} \nonumber\]

Verify this with some numeric examples.

hands-on Exercise \[\PageIndex{5}\label{he:invfcn-05}\]

The function \[f :{\mathbb{Z}}\to{\mathbb{N}}\] is defined as \[f[n] = \cases{ -2n & if $n < 0$, \cr 2n+1 & if $n\geq0$. \cr} \nonumber\] 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 \[\{0,1\}^n = \{[b_1,b_2,\ldots,b_n] \mid b_i\in\{0,1\} \mbox{ for each $i$, where $1\leq i\leq n$} \}. \nonumber\] Define \[F:{\wp[A]}\to{\{0,1\}^n}\] according to \[F[S] = [x_1,x_2,\ldots,x_n]\], where \[x_i = \cases{ 1 & if $a_i\in S$, \cr 0 & if $a_i\notin S$. \cr} \nonumber\] 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, \[F[\{a_2,a_5,a_8\}] = [0,1,0,0,1,0,0,1], \nonumber\] and \[F^{-1}\big[[1,1,0,0,0,1,1,0]\big] = \{a_1,a_2,a_6,a_7\}. \nonumber\] 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.

hands-on 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 \[f[x]=y \Leftrightarrow x=f^{-1}[y]. \nonumber\]
• 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!

1. \[f :{\mathbb{R}}\to{\mathbb{R}}\], \[f[x]=x^3-2x^2+1\].
2. \[g :{[\,2,\infty]}\to{\mathbb{R}}\], \[g[x]=x^3-2x^2+1\].
3. \[h:{\mathbb{R}}\to{\mathbb{R}}\], \[h[x]=e^{1-2x}\].
4. \[p :{\mathbb{R}}\to{\mathbb{R}}\], \[p[x]=|1-3x|\].
5. \[q:{[\,2,\infty]}\to{[\,0,\infty]}\], \[q[x]=\sqrt{x-2}\].

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 \[f[x] = \frac{3}{2}\,x+\frac{5}{2}, \qquad\mbox{and}\qquad g[x] = -\frac{3}{2}\,x+\frac{17}{2}. \nonumber\] 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 \[f[x] = \cases{ 3x+5 & if $x\leq 6$, \cr 5x-7 & if $x > 6$. \cr} \nonumber\]

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 \[g[x] = \cases{ x+3 & if $1\leq x< 2$, \cr 11-2x & if $2\leq x\leq 3$. \cr} \nonumber\] 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.

1. \[r :{\mathbb{Z}_{12}}\to{\mathbb{Z}_{12}}\], \[r[n]\equiv 7n\] [mod 12].
2. \[s :{\mathbb{Z}_{33}}\to{\mathbb{Z}_{33}}\], \[s[n]\equiv 7n+5\] [mod 33].
3. \[t :{\mathbb{Z}}\to{\mathbb{N}\cup\{0\}}\], \[t[n] = \cases{ 2n-1 & if $n > 0$, \cr -2n & if $n\leq0$,\cr}\]

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. \[\begin{array}{|c||*{8}{c|}} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \alpha[x]& g & a & d & h & b & e & f & c \\ \hline \end{array} \nonumber\] Find its inverse function.

Exercise \[\PageIndex{12}\label{ex:invfcn-12}\]

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

This page titled 6.6: Inverse Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Harris Kwong [OpenSUNY] .

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