Chapter 8: Functions III: Mappings and Inverses

Chapter 4 introduced functions, their representations, and several basic function classes, while Chapter 5 showed how functions can be combined and composed. We now study how functions map their domains to their codomains, when those mappings can be reversed, and how inverse relationships can be used to solve non-linear equations. We conclude by determining whether a relation defines a function.

Injective, Surjective, and Bijective Functions

Functions can also be classified based on how they relate elements of their domain to elements of their codomain. While algebraic form determines a function’s shape or formula, mapping properties determine whether the function is injective, surjective, or both.

Definition: Injective Function

A function $f : A \to B$ is injective (or one-to-one) if it never assigns the same output value to two different inputs. In other words, each output in $B$ comes from at most one input in $A$ .

More formally, we can write:

$for all x_{1}, x_{2} \in A, f (x_{1}) = f (x_{2}) implies x_{1} = x_{2} .$

Or in plain words: If two inputs of a function give the same output, then those inputs must be equal.

Example: Checking Injectivity

The figure contrasts a function where each output comes from at most one input with one where two inputs share an output.

Injective and non-injective functions — Figure 8.1. Left: An injective function. Right: A non-injective function.

Let $f : R \to R$ be defined by:

$f (x) = 2 x + 3$

This function is injective because different $x$ -values always produce different $y$ -values. However, $f (x) = x^{2}$ is not injective on $R$ since $f (2) = f (- 2) = 4$ .

Definition: Surjective Function

A function $f : A \to B$ is surjective (or onto) if every element of the codomain $B$ appears as an output of the function. Equivalently,

$Range (f) = B .$

More formally, we can write:

$for every y \in B, there exists x \in A such that f (x) = y .$

Or in plain words: For every possible output value in the codomain, there exists at least one input value in the domain that produces it.

Example: Surjective Function

Surjectivity is about whether the function reaches every output in the codomain; the figure compares a function that covers the codomain with one that leaves gaps.

Surjective and non-surjective functions — Figure 8.2. Left: A surjective function that covers all possible $y$ -values in the codomain. Right: A non-surjective function which leaves gaps.

Let $f : R \to R$ be defined by:

$f (x) = 2 x + 1$

For any $y \in R$ , there exists $x = \frac{y - 1}{2}$ , so $f$ is surjective. However, $f (x) = x^{2}$ from $R \to R$ is not surjective because negative $y$ -values are never reached.

Definition: Bijective Function

A function $f : A \to B$ is bijective if it is both injective and surjective. This means that distinct inputs have distinct outputs, and every element of $B$ is produced by an input from $A$ .

Equivalently, a bijective function establishes a one-to-one correspondence between the sets $A$ and $B$ , so that nothing is repeated and nothing is left out.

Example: Bijective Function

A bijective function combines both requirements: no output is repeated, and none are missing.

Let $f : R \to R$ be defined by:

$f (x) = x + 1$

The function is bijective because each input produces a unique output (injective) and every real number occurs exactly once as an output (surjective).

Inverse Functions

An inverse function reverses the input-output mapping of a function. It is useful when we know an output $y$ and want to recover the input that produced it.

Definition: Inverse of a Function

A mapping diagram showing a function from a domain to a codomain and its inverse mapping back. — Figure 8.4. A function and its inverse reverse the direction of the mapping between domain and codomain.

Let $f : A \to B$ be a bijective function. Its inverse $f^{- 1} : B \to A$ satisfies

$f^{- 1} (f (x)) = x and f (f^{- 1} (y)) = y$

for all $x \in A$ and $y \in B$ . Thus, applying $f$ followed by $f^{- 1}$ , or vice versa, returns the original value.

The inverse allows us to solve $f (x) = y$ by applying $f^{- 1}$ :

$f (x) = y \Leftrightarrow x = f^{- 1} (y) .$

Warning: Existence and Notation

A function has an inverse on its stated codomain only if it is bijective. If it is not injective, an output may correspond to several inputs; if it is not surjective, some elements of the codomain cannot be reversed.
The notation $f^{- 1}$ represents the inverse function, not the reciprocal: $f^{- 1} (x) \neq = \frac{1}{f ( x )} .$

Definition: Identity Property of Inverses

For the notation below, $(f \circ g) (x) = f (g (x))$ means that $g$ is applied first and then $f$ , while the identity function on a set $A$ is defined by $id_{A} (x) = x$ . Function composition was introduced in Chapter 5.

The composition of a function and its inverse returns the identity function on the respective domains:

$f \circ f^{- 1} = id_{B}, f^{- 1} \circ f = id_{A} .$

The inverse reverses the domain and codomain of $f$ . Graphically, its graph is the reflection of the graph of $f$ across the line $y = x$ .

Method: Finding an Inverse Function

To find an inverse function:

Write the function as $y = f (x)$ .
Solve the equation for $x$ in terms of $y$ .
Rewrite the result as $f^{- 1} (y)$ , or rename the input variable.
Check that composing the two functions in either order returns the original input.

Example: Finding and Checking an Inverse Function

Let $f : R \to R$ be defined by

$f (x) = 3 x + 5.$

To find its inverse, solve $y = 3 x + 5$ for $x$ :

$y = 3 x + 5 \Leftrightarrow x = \frac{y - 5}{3} .$

Therefore,

$f^{- 1} (y) = \frac{y - 5}{3} .$

For example, $f (x) = 14$ gives

$x = f^{- 1} (14) = \frac{14 - 5}{3} = 3.$

We can check the formula in both directions:

$f^{- 1} (f (x)) = \frac{( 3 x + 5 ) - 5}{3} = x$

and

$f (f^{- 1} (y)) = 3 (\frac{y - 5}{3}) + 5 = y .$

A graph showing f(x)=3x+5, its inverse, and the reflection line y equals x. — Figure 8.5. For $f (x) = 3 x + 5$ , the graph of $f^{- 1} (x) = \frac{x - 5}{3}$ is the reflection of the graph of $f$ across the line $y = x$ .

Common Inverses

The table below shows frequently used inverse-function pairs with the domain and range of the original function. The restrictions shown are part of the function definition and ensure that the mapping is bijective.

Table 8.1. Common inverse-function pairs with domains and ranges.

Function $f (x)$	Inverse $f^{- 1} (x)$	Domain of $f$	Range of $f$
$f (x) = x + a$	$f^{- 1} (x) = x - a$	$R$	$R$
$f (x) = x - a$	$f^{- 1} (x) = x + a$	$R$	$R$
$f (x) = k x$ , $k \neq = 0$	$f^{- 1} (x) = \frac{x}{k}$	$R$	$R$
$f (x) = \frac{x}{k}$ , $k \neq = 0$	$f^{- 1} (x) = k x$	$R$	$R$
$f (x) = x^{n}$ , $n$ a positive odd integer	$f^{- 1} (x) = n x$	$R$	$R$
$f (x) = x^{n}$ , $n$ a positive even integer	$f^{- 1} (x) = n x$ (principal root)	$[0, \infty)$	$[0, \infty)$
$f (x) = e^{x}$	$f^{- 1} (x) = ln (x)$	$R$	$(0, \infty)$
$f (x) = b^{x}$ , $b > 0$ , $b \neq = 1$	$f^{- 1} (x) = lo g_{b} (x)$	$R$	$(0, \infty)$
$f (x) = ln (x)$	$f^{- 1} (x) = e^{x}$	$(0, \infty)$	$R$
$f (x) = lo g_{b} (x)$ , $b > 0$ , $b \neq = 1$	$f^{- 1} (x) = b^{x}$	$(0, \infty)$	$R$
$f (x) = sin (x)$	$f^{- 1} (x) = arcsin (x)$	$[- \frac{π}{2}, \frac{π}{2}]$	$[- 1, 1]$
$f (x) = cos (x)$	$f^{- 1} (x) = arccos (x)$	$[0, π]$	$[- 1, 1]$
$f (x) = tan (x)$	$f^{- 1} (x) = arctan (x)$	$(- \frac{π}{2}, \frac{π}{2})$	$R$

Logarithm Rules

Let $x > 0$ , $y > 0$ , and $n \in R$ . For a logarithm base $b > 0$ with $b \neq = 1$ , the most important rules are:

Table 8.2. Common logarithm rules.

Rule	Formula	Description
Logarithm Product Rule	$lo g_{b} (x y) = lo g_{b} (x) + lo g_{b} (y)$	The logarithm of a product equals the sum of the logarithms.
Logarithm Quotient Rule	$lo g_{b} (\frac{x}{y}) = lo g_{b} (x) - lo g_{b} (y)$	The logarithm of a quotient equals the difference of the logarithms.
Logarithm Power Rule	$lo g_{b} (x^{n}) = n lo g_{b} (x)$	A power in the argument becomes a multiplier in front of the logarithm.
Logarithm of 1	$lo g_{b} (1) = 0$	Any valid base raised to the power $0$ equals $1$ .
Logarithm of the Base	$lo g_{b} (b) = 1$	Any valid base raised to the power $1$ equals itself.
Inverse Property	$b^{l o g_{b} (x)} = x$	Exponential and logarithmic functions cancel each other.
Natural Log of $e$	$ln (e) = 1$	The natural logarithm means logarithm to base $e$ .
Change of Base	$lo g_{b} (x) = \frac{l o g _{k} ( x )}{l o g _{k} ( b )}$	Here $k > 0$ and $k \neq = 1$ .

The natural logarithm $ln (x)$ is $lo g_{e} (x)$ , where $e \approx 2.71828$ . These rules mirror the exponent rules introduced in Chapter 3.

Solving Non-Linear Equations

Many equations in mathematics involve non-linear functions such as exponentials and logarithms. The solving principles remain the same: we transform the equation into an equivalent one where the variable of interest is isolated, checking that the solution satisfies any domain restrictions.

With inverse-function pairs and logarithm rules now established, we can use them to undo the operation affecting the variable. In particular, when the variable appears in an exponent, we apply a logarithm to both sides, and when it appears inside a logarithm, we apply an exponential.

Example: Solving an Exponential Equation

Let us solve the equation $e^{2 x + 1} = w$ , $w > 0$ for $x$ .

$\Leftrightarrow \Leftrightarrow \Leftrightarrow e^{2 x + 1} ln (e^{2 x + 1}) 2 x + 1 x = w = ln (w) = ln (w) = \frac{1}{2} (ln (w) - 1)$

Example: Solving a Logarithmic Equation

Solve the equation $ln (x^{2} - 10) = 6$ for $x$ . Assume that $x^{2} > 10$ , as the logarithm otherwise is not defined. We obtain:

$\Leftrightarrow \Leftrightarrow \Leftrightarrow \Leftrightarrow ln (x^{2} - 10) e^{l n (x^{2} - 10)} x^{2} - 10 x^{2} x = 6 = e^{6} = e^{6} = e^{6} + 10 = \pm e^{6} + 10$

Both candidates satisfy the domain restriction because $x^{2} - 10 = e^{6} > 0$ , so both are solutions of the original equation.

Testing Relations

Method: Checking Whether a Relation Is a Function

To check whether a relation describes $y$ as a function of $x$ :

For a list of ordered pairs, check that each input $x$ has only one associated output $y$ .
Graphically, use the vertical line test: each vertical line should intersect the graph at most once.
Algebraically, try to solve for $y$ in terms of $x$ . If one input can produce more than one output, the relation is not a function of $x$ .

Graphically

A relation in which each $x$ -coordinate is matched with exactly one $y$ -coordinate describes $y$ as a function of $x$ . If the same $x$ -coordinate is associated with two different $y$ -coordinates, then the relation is not a function of $x$ .

Example: Checking Functional Relations

Which of the following relations describe $y$ as a function of $x$ ?

$R_{1} = {(- 2, 1), (1, 3), (1, 4), (3, - 1)}$
$R_{2} = {(- 2, 1), (1, 3), (2, 3), (3, - 1)}$

In $R_{1}$ , the input $x = 1$ is matched with both $y = 3$ and $y = 4$ . Hence, $R_{1}$ does not describe $y$ as a function of $x$ . In $R_{2}$ , every input has exactly one corresponding output, so $R_{2}$ does describe $y$ as a function of $x$ .

Two coordinate plots comparing a relation that is not a function with a relation that is a function. — Figure 8.6. $R_{1}$ fails because one input has two outputs; $R_{2}$ passes because each input has one output.

The Vertical Line Test

Definition: Vertical Line Test

A vertical line test diagram comparing a polynomial function with a circle that is not a function. — Figure 8.7. The polynomial passes the test, while the circle fails because some vertical lines meet it twice.

A relation describes $y$ as a function of $x$ if and only if every vertical line intersects its graph at most once.

If a vertical line intersects the graph more than once, the relation assigns more than one output to the same input and therefore violates the definition of a function.

An equation can describe a valid relation, such as a circle, without defining $y$ as a function of $x$ . Such relations may require another representation, for example an implicit or parametric description.

Algebraically

We can also check whether an equation defines a function by solving for one variable in terms of the other. If solving produces more than one output value for the same input, then the relation does not describe a single function.

Example: Equation That Is Not a Function

Does the equation $x^{2} + y^{2} = 1$ represent $y$ as a function of $x$ ?

First, solve for $y$ :

$y^{2} = 1 - x^{2} \Leftrightarrow y = \pm 1 - x^{2} .$

For inputs with $- 1 < x < 1$ , this gives two outputs, one positive and one negative. Therefore, the circle does not describe $y$ as a single function of $x$ . The upper and lower semicircles can instead be described separately by $y = 1 - x^{2}$ and $y = - 1 - x^{2}$ , each with domain $[- 1, 1]$ .

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