# Inverse Functions

Suppose $$f : A \to B$$ is a function whose domain is the set $$A$$ and whose codomain is the set $$B.$$ The function $$f$$ is called invertible if there exists a function $$f^{-1} : B \to A$$ with the domain $$B$$ and the codomain $$A$$ such that

${{f^{ – 1}}\left( y \right) = x\; \text{ if and only if }\;}\kern0pt{ f\left( x \right) = y,}$

where $$x \in A,$$ $$y \in B.$$

The function $$f^{-1}$$ is then called the inverse of $$f.$$

Not all functions have an inverse. If a function $$f$$ is not injective, different elements in its domain may have the same image:

$f\left( {{x_1}} \right) = f\left( {{x_2}} \right) = y_1.$

In this case, the converse relation $${f^{-1}}$$ is not a function because there are two preimages $${x_1}$$ and $${x_2}$$ for the element $${y_1}$$ in the codomain $$B.$$ So, to have an inverse, the function must be injective.

If a function $$f$$ is not surjective, not all elements in the codomain have a preimage in the domain. In this case, the converse relation $${f^{-1}}$$ is also not a function.

Thus, to have an inverse, the function must be surjective.

Recall that a function which is both injective and surjective is called bijective. Hence, to have an inverse, a function $$f$$ must be bijective. The converse is also true. If $$f : A \to B$$ is bijective, then it has an inverse function $${f^{-1}}.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Show that the function $$f:\mathbb{Z} \to \mathbb{Z}$$ defined by $$f\left( x \right) = x + 5$$ is bijective and find its inverse.

### Example 2

Show that the function $$g:\mathbb{R^{+}} \to \mathbb{R^{+}},$$ $$f\left( x \right) = x^2$$ is bijective and find its inverse.

### Example 3

The function $$f: \mathbb{R}\backslash\left\{ 3 \right\} \to \mathbb{R}\backslash\left\{ 1 \right\}$$ is defined as $$f\left( x \right) = \large{\frac{{x – 2}}{{x – 3}}}\normalsize.$$ Find the inverse function $$f^{-1}.$$

### Example 4

The function $$g: \mathbb{R} \to \mathbb{R}^{+}$$ is defined as $$g\left( x \right) = {e^{2x + 1}}.$$ Find the inverse function $$g^{-1}.$$

### Example 5

Consider the function $$f:\mathbb{Z}^2 \to \mathbb{Z}^2$$ defined as $$f\left( {x,y} \right) = \left( {2x – y,x + 2y} \right).$$ Find the inverse function $${f^{-1}}.$$

### Example 1.

Show that the function $$f:\mathbb{Z} \to \mathbb{Z}$$ defined by $$f\left( x \right) = x + 5$$ is bijective and find its inverse.

Solution.

It is easy to show that the function $$f$$ is injective. Using the contrapositive approach, suppose that $${x_1} \ne {x_2}$$ but $$f\left( {{x_1}} \right) = f\left( {{x_2}} \right).$$ Then we have:

${{x_1} + 5 = {x_2} + 5,}\;\; \Rightarrow {{x_1} = {x_2}.}$

This is a contradiction. Hence, the function $$f$$ is injective.

For any $$y \in \mathbb{Z}$$ in the codomain of $$f,$$ there exists a preimage $$x:$$

${y = f\left( x \right) = x + 5,}\;\; \Rightarrow {x = y – 5.}$

We see that the function $$f$$ is surjective, and consequently, it is bijective. The inverse function is given by

$x = {f^{ – 1}}\left( y \right) = y – 5.$

### Example 2.

Show that the function $$g:\mathbb{R^{+}} \to \mathbb{R^{+}},$$ $$f\left( x \right) = x^2$$ is bijective and find its inverse.

Solution.

By contradiction, let $${x_1} \ne {x_2}$$ but $$g\left( {{x_1}} \right) = g\left( {{x_2}} \right).$$ Then

${x_1^2 = x_2^2,}\;\; \Rightarrow {\left| {{x_1}} \right| = \left| {{x_2}} \right|.}$

Since the domain is restricted to the set of positive real numbers, we get $${x_1} = {x_2}.$$ This proves that the function $$g$$ is injective.

Take an arbitrary positive number $$y \in \mathbb{R^{+}}$$ in the codomain of $$g.$$ Find the preimage of the number:

${y = g\left( x \right) = {x^2},}\;\; \Rightarrow {x = \sqrt y .}$

It is clear that the preimage $$x$$ exists for any positive $$y,$$ so the function $$g$$ is surjective.

Since the function $$g$$ is injective and surjective, it is bijective and has an inverse $$g^{-1}$$ that is given by

$x = {g^{ – 1}}\left( y \right) = \sqrt y .$

### Example 3.

The function $$f: \mathbb{R}\backslash\left\{ 3 \right\} \to \mathbb{R}\backslash\left\{ 1 \right\}$$ is defined as $$f\left( x \right) = \large{\frac{{x – 2}}{{x – 3}}}\normalsize.$$ Find the inverse function $$f^{-1}.$$

Solution.

First we check that the function $$f$$ is bijective.

Let $${x_1} \ne {x_2},$$ where $${x_1},{x_2} \ne 1,$$ and suppose $$f\left( {{x_1}} \right) = f\left( {{x_2}} \right).$$ Then

$\require{cancel}{\frac{{{x_1} – 2}}{{{x_1} – 3}} = \frac{{{x_2} – 2}}{{{x_2} – 3}},}\;\; \Rightarrow {\left( {{x_1} – 2} \right)\left( {{x_2} – 3} \right) }={ \left( {{x_1} – 3} \right)\left( {{x_2} – 2} \right),}\;\; \Rightarrow {\cancel{{x_1}{x_2}} – 2{x_2} – 3{x_1} + \cancel{6} }={ \cancel{{x_1}{x_2}} – 3{x_2} – 2{x_1} + \cancel{6},}\;\; \Rightarrow {- 2{x_2} – 3{x_1} = – 3{x_2} – 2{x_1},}\;\; \Rightarrow {3{x_2} – 2{x_2} = 3{x_1} – 2{x_1},}\;\; \Rightarrow {{x_2} = {x_1}.}$

This is a contradiction. Hence, the function $$f$$ is injective.

Consider an arbitrary real number $$y$$ in the codomain of $$f.$$ Determine the preimage of the number $$y$$ by solving the equation for $$x:$$

${y = f\left( x \right) = \frac{{x – 2}}{{x – 3}},}\;\; \Rightarrow {x – 2 = y\left( {x – 3} \right),}\;\; \Rightarrow {x – 2 = xy – 3y,}\;\; \Rightarrow {xy – x = 3y – 2,}\;\; \Rightarrow {x\left( {y – 1} \right) = 3y – 2,}\;\; \Rightarrow {x = \frac{{3y – 2}}{{y – 1}}.}$

As you can see, the preimage $$x$$ exists for any $$y \ne 1.$$ Consequently, the function $$f$$ is surjective and, hence, it is bijective. The inverse function $$f^{-1}$$ is expressed as

$x = {f^{ – 1}}\left( y \right) = \frac{{3y – 2}}{{y – 1}}.$

### Example 4.

The function $$g: \mathbb{R} \to \mathbb{R}^{+}$$ is defined as $$g\left( x \right) = {e^{2x + 1}}.$$ Find the inverse function $$g^{-1}.$$

Solution.

We need to make sure that the function $$g$$ is bijective.

By contradiction, suppose $${x_1} \ne {x_2}$$ but $$g\left( {{x_1}} \right) = g\left( {{x_2}} \right).$$ It then follows that

${{e^{2{x_1} + 1}} = {e^{2{x_2} + 1}},}\;\; \Rightarrow {\ln {e^{2{x_1} + 1}} = \ln {e^{2{x_2} + 1}},}\Rightarrow {\left( {2{x_1} + 1} \right)\ln e = \left( {2{x_2} + 1} \right)\ln e,}\;\; \Rightarrow {2{x_1} + 1 = 2{x_2} + 1,}\;\; \Rightarrow {2{x_1} = 2{x_2},}\;\; \Rightarrow {{x_1} = {x_2}.}$

This proves that $$g$$ is injective.

Choose a positive real number $$y.$$ Solve the equation $$y = g\left( x \right)$$ for $$x:$$

${g\left( x \right) = y,}\;\; \Rightarrow {{e^{2x + 1}} = y,}\;\; \Rightarrow {2x + 1 = \ln y,}\;\; \Rightarrow {2x = \ln y – 1,}\;\; \Rightarrow {x = \frac{1}{2}\left( {\ln y – 1} \right).}$

The preimage $$x$$ exists for any $$y$$ in the codomain of $$g.$$ So, the function is surjective.

Since the function $$g$$ is injective and surjective, it is bijective and has an inverse $${g^{-1}},$$ which is given by

$x = {g^{ – 1}}\left( y \right) = \frac{1}{2}\left( {\ln y – 1} \right).$

### Example 5.

Consider the function $$f:\mathbb{Z}^2 \to \mathbb{Z}^2$$ defined as $$f\left( {x,y} \right) = \left( {2x – y,x + 2y} \right).$$ Find the inverse function $${f^{-1}}.$$

Solution.

Check the function $$f$$ for injectivity. Suppose that $$\left( {{x_1},{y_1}} \right) \ne \left( {{x_2},{y_2}} \right)$$ but $$f\left( {{x_1},{y_1}} \right) = f\left( {{x_2},{y_2}} \right).$$ Then

${\left( {2{x_1} – {y_1},{x_1} + 2{y_1}} \right) }={ \left( {2{x_2} – {y_2},{x_2} + 2{y_2}} \right),}\;\;\Rightarrow {\left\{ {\begin{array}{*{20}{l}} {2{x_1} – {y_1} = 2{x_2} – {y_2}}\\ {{x_1} + 2{y_1} = {x_2} + 2{y_2}} \end{array}} \right..}$

Solve the system of equation for $$\left( {{x_2},{y_2}} \right).$$ To eliminate $${y_2},$$ we multiply the first equation by $$2$$ and add both equations:

${\left\{ {\begin{array}{*{20}{l}} {2{x_1} – {y_1} = 2{x_2} – {y_2}}\\ {{x_1} + 2{y_1} = {x_2} + 2{y_2}} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {4{x_1} – 2{y_1} = 4{x_2} – 2{y_2}}\\ {{x_1} + 2{y_1} = {x_2} + 2{y_2}} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {5{x_1} = 5{x_2}}\\ {{x_1} + 2{y_1} = {x_2} + 2{y_2}} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {{x_1} = {x_2}}\\ {{x_1} + 2{y_1} = {x_2} + 2{y_2}} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {{x_1} = {x_2}}\\ {2{y_1} = 2{y_2}} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {{x_1} = {x_2}}\\ {{y_1} = {y_2}} \end{array}} \right..}$

Since $$\left( {{x_1},{y_1}} \right) = \left( {{x_2},{y_2}} \right),$$ we get a contradiction. So, the function $$f$$ is injective.

Check the surjectivity of the function $$f.$$ Let $$\left( {a,b} \right)$$ be an arbitrary pair of real numbers in the codomain of $$f.$$ Solve the equation $$f\left( {x,y} \right) = \left( {a,b} \right)$$ to express $$x,y$$ in terms of $$a,b.$$

${\left( {2x – y,x + 2y} \right) = \left( {a,b} \right),}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{c}} {2x – y = a}\\ {x + 2y = b} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{c}} {y = 2x – a}\\ {x + 2\left( {2x – a} \right) = b} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{c}} {y = 2x – a}\\ {x + 4x – 2a = b} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{c}} {y = 2x – a}\\ {5x = 2a + b} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{c}} {y = 2x – a}\\ {x = \frac{{2a + b}}{5}} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{c}} {x = \frac{{2a + b}}{5}}\\ {y = \frac{{2b – a}}{5}} \end{array}} \right..}$

Thus, we can always determine the preimage $$\left( {x,y} \right)$$ for any image $$\left( {a,b} \right).$$ Hence, the function is surjective and bijective.

The inverse of the function $${f^{-1}}$$ has already been found above. It is given by

${\left( {x,y} \right) = {f^{ – 1}}\left( {a,b} \right) }={ \left( {\frac{{2a + b}}{5},\frac{{2b – a}}{5}} \right).}$

We can check the result given that $$f\left( {x,y} \right) = \left( {a,b} \right):$$

${f\left( {x,y} \right) = \left( {2x – y,x + 2y} \right) }={ \left( {2 \cdot \frac{{2a + b}}{5} – \frac{{2b – a}}{5},}\right.}\kern0pt{\left.{\frac{{2a + b}}{5} + 2 \cdot \frac{{2b – a}}{5}} \right) }={ \left( {\frac{{4a + \cancel{2b} – \cancel{2b} + a}}{5},}\right.}\kern0pt{\left.{\frac{{\cancel{2a} + b + 4b – \cancel{2a}}}{5}} \right) }={ \left( {\frac{{5a}}{5},\frac{{5b}}{5}} \right) }={ \left( {a,b} \right).}$