(i). The diagram shows a relation, $f_1$, between the sets $A = \{a, b, c, d\}$ and $B = \{1, 2, 3, 4\}$.

$f_1 : A \to B$ is a function as each element in $A$ has exactly one image in $B$.

We have that

$f_1(a) = 1$, $f_1(b) = 3$, $f_1(c) = 2$ and $f_1(d) = 4$.

The domain is the set, $A = \{a, b, c, d\}$.

The codomain is the set, $B = \{1, 2, 3, 4\}$.

The range is the set, $f_1[A] = \{1, 2, 3, 4\}$.

Note that the range is a subset of the codomain, i.e. $f_1[A] \subseteq B$.

In this case the range and the codomain are equal.

(ii). $f_2 : A \to B$ is a function as each element in $A$ has exactly one image in $B$.

We have that

$f_2(a) = 1, f_2(b) = 4, f_2(c) = 2$ and $f_2(d) = 4$.

The domain is the set, $A = \{a, b, c, d\}$.

The codomain is the set, $B = \{1, 2, 3, 4\}$.

The range is the set, $f_2[A] = \{1, 2, 4\}.$.

Note that the range is a subset of the codomain, i.e. $f_2[A] \subseteq B$ Compare with the function in part (i)..

(iii). $f_3 : A \to B$ is not a function as the element $a$ in $A$ has more than one image in $B$, i.e.

$f_3(a) = 1$ and $f_3(a) = 4$.

(iv). $f_4 : A \to B$ is a function as each element in $A$ has exactly one image in $B$.

(v). $f_5 : A \to B$ is not a function as the element $c$ in $A$ does not have an image in $B$.

Write each of the relations in Example 2 as a set of ordered pairs and determine whether or not they represent functions.

Solution

Referring back to Example 2 and treating each part of the question in turn.

(i). $\{(a, 1), (b, 3), (c, 2), (d, 4)\}$

This is a function as no two pairs have the same first component.

(ii). $\{(a, 1), (b, 4), (c, 2), (d, 4)\}$

This is a function as no two pairs have the same first component.

(iii). $\{(a, 1), (a, 4), (b, 2), (c, 3), (d, 4)\}$

This is not a function as the elements $(a, 1)$ and $(a, 4)$ have the same first component.

(iv). $\{(a, 4), (b, 3), (c, 3), (d, 4)\}$

This is a function as no two pairs have the same first component.

(v). This is not a function as the element $c$ in the domain is not mapped to an element in the codomain.

Draw the coordinate diagrams for each of the relations in Examples 2 and 3 and determine whether or not they represent functions.

Solution

Referring back to Examples 2 and 3 and treating each part of the question in turn.

(i) $f_1 = \{(a, 1), (b, 3), (c, 2), (d, 4)\}$

This is a function as each vertical line contains exactly one point of the set.

(ii) $f_2 = \{(a, 1), (b, 4), (c, 2), (d, 4)\}$

This is a function as each vertical line contains exactly one point of the set.

(iii) $f_3 = \{(a, 1), (a, 4), (b, 2), (c, 3), (d, 4)\}$

This is not a function as the vertical line through $a$ contains two points of the set.

(iv) $f_4 = \{(a, 4), (b, 3), (c, 3), (d, 4)\}$

This is a function as each vertical line contains exactly one point of the set.

(v) $f_5 = \{(a, 1), (b, 2), (d, 4)\}$

This is not a function as the vertical line through $c$ contains no point of the set.

(i). $y = x^2$ (ii). $y = \pm \sqrt{x}$

In the diagrams above:

(i). the line cuts the graph of $y = x^2$ in one point and so this is the graph of a function.

(ii). the line cuts the graph of $y = \pm \sqrt{x}$ in two points and so this is not the graph of a function.

We now introduce three very important classes of functions: injective functions, surjective functions and bijective functions.

(i). $g_1 : A \to B$ is one-to-one as all the elements in $A$ have distinct images in $B$, i.e.

$g_1(a) = 1,\; g_1(b) = 2, \; g_1(c) = 3$ and $g_1(d) = 4$.

(ii). $g_2 : A \to B$ is not one-to-one as two elements in $A$ have the same image in $B$, i.e.

$g_2(a) = g_2(d) = 3$.

Note that if two, or more, elements of the domain $A$, are mapped to the same element of $B$ the function is said to be many-to-one.

(iii). $g_3 : A \to B$ is one-to-one as all the elements in $A$ have distinct images in $B$. It does not matter that the element $3$ in $B$ is not mapped to by an element of $A$.

Write each of the relations in Example 5 as a set of ordered pairs and determine whether or not they are one-to-one functions.

Solution

Referring back to Example 5 and treating each part of the question in turn.

(i). $\{(a, 1), (b, 2), (c, 3), (d, 4)\}$

(ii). $\{(a, 3), (b, 1), (c, 2), (d, 3)\}$

(iii). $\{(a, 4), (b, 2), (c, 1), (d, 5)\}$

Only (i) and (iii) represent one-to-one functions. Case (ii) includes ordered pairs with a repeated second component. Hence, different inputs produce the same output and this is not allowed for a one-to-one function.

(i). $f(x) = e^x$ is one-to-one as no horizontal line intersects the curve in more than one point.

(ii). $g(x) = x^2 + x^3$ is not one-to-one as some horizontal lines cut the graph of $g(x)$ in more than one point.

Determine if the following functions are one-to-one:

(i). $f(x) = \frac{1}{x}, x \neq 0$.

(ii). $g(x) = x - x^3$

Solution

(i). Let $a_1$ and $a_2$ be any two real numbers such that $f(a_1) = f(a_2)$.

We need to show that $a_1 = a_2$.

Now $f(a_1) = \frac{1}{a_1}$ and $f(a_2) = \frac{1}{a_2}$.

Then $\frac{1}{a_1} = \frac{1}{a_2}$ gives that $a_1 = a_2$ proving that $f(x)$ is one-to-one.

(ii). We have $g(x) = x^2 - x^3 = x^2(1 - x)$.

We shall prove $g$ is not one-to-one by giving a counterexample.

If we can find two different elements in the domain that map to the same element then we have shown that the function is not one-to-one.

Let $a_1 = 0$ and $a_2 = 1$. Then

$g(0) = 0(1 - 0) = 0$ and

$g(1) = 1(1 - 1) = 0$.

Hence, $g(a_1) = g(a_2) = 0$, i.e. both $0$ and $1$ in the domain map to 0 in the range.

However, $a1 \neq a2$ and so $g(x)$ is not one-to-one.

(i). $h_1 : A \to B$ is onto as every element in $B$ is mapped to some element of $A$.

The range of $h_1, \{1, 2, 3\}$, is the same as the codomain of $h_1$.

(ii). $h_2 : A \to B$ is not onto as the element $3$ in $B$ is not the image of an element of $A$.

The range of $h_2, \{1, 2, 3\}$, is not the same as the codomain of $h_2, \{1, 2, 3, 4\}$.

The function $f : \mathbb{R} \to \mathbb{R}$ defined by, $f(x) = e^x$ is not onto as there are horizontal lines that do not intersect the curve. For example, the dashed line in the diagram does not intersect $f(x)$.

Note here that the domain and codomain both include all the real numbers. However, the function $e^x$ will always be positive and this is a problem as there are values in the codomain of $f$ that are not in the range of $f$ . Hence, $f(x) = e^x$ cannot be an onto function.

If however the codomain was restricted to the interval $(0, \; \infty)$ then the function is onto (and one-to-one).

Determine whether the following functions are onto.

(i). $f(x) = 3x - 4 \; \text{where} \; f : \mathbb{R} \to \mathbb{R}$.

(ii). $g(x) = x + 2 \; \text{where} \; g: \mathbb{Z}^+ \to \mathbb{Z}^+$. (Note that $\mathbb{Z}^+$ is the set of positive integers)

Solution

(i). Let $y \in \mathbb{R}$ be any element in the codomain.

We need to show that there is an $x \in \mathbb{R}$ (the domain) such that $f(x) = y$.

We have $y = 3x - 4$ so

$x = {\Large\frac{y + 4}{3}}$ is a real number, since $y$ is real, in the domain.

Now try to show that $f(x) = y$, i.e.

$$f\Bigg(\frac{y + 4}{3}\Bigg) = 3\Bigg(\frac{y + 4}{3}\Bigg) - 4$$

$= y + 4 - 4$

$= y$.

As we have found an element that maps to $y$ we have proved that $f$ is onto.

This is a relatively straightforward example as in general it can be much harder to prove algebraically that a function is onto than it is to prove a function is one-to-one.

(ii). We show this function is not onto using a counterexample.

If $g(x) = x + 2 = 2$ then we must have that $x = 0$.

However, $0$ is not a positive integer and therefore not in the domain of $g$.

Hence, the element 2 in the codomain is not mapped to by any element of the domain and we conclude that $g$ is not onto.

(i). $f : A \to B$ is bijective as it is one-to-one, no element of $B$ is the image of more than one element of $A$, and it is onto as every element in $B$ is the image of some element of $A$.

(ii). $g : A \to B$ is one-to-one but not onto as the element $3$ in $B$ is not mapped to some element of $A$.

(iii). $h : A \to B$ is onto but not one-to-one as the element $3$ in $B$ is the image of two elements, $c$ and $d$, in $A$.

(iv). $k : A \to B$ is neither one-to-one nor not onto.

It is not one-to-one as the element $4$ in $B$ is the image of two elements in $A$.

It is not onto as the element $3$ in $B$ is not mapped to by some element of $A$.

Which of the following functions have an inverse?

(i).

(ii).

(iii).

Solution

For a function to have an inverse it must be both one-to-one and onto. Only the function $f_2$ , shown in (ii), is one-to-one and onto and so only $f_2$ has an inverse. The function in (i) is neither one-to-one nor onto, while (iii) is onto but not one-to-one.

(i). Let $f : \mathbb{R} \to \mathbb{R}$ be defined by, $f(x) = 4x - 7$.

Determine the inverse function, $f^{-1}(x)$

Solution

Rewrite as: $y = 4x - 7$

Solve for $x$: $x = {\Large\frac{y + 7}{4}} = f^{-1}(y)$

Switch $x$ and $y$ to obtain the inverse: $y = {\Large\frac{x + 7}{4}} = f^{-1}(x)$.

So $f^{-1} : \mathbb{R} \to \mathbb{R}$ is defined by, $f^{-1}(x) = {\Large\frac{x + 7}{4}}$.

Note: If a point $(x,\; y)$ lies on the graph of a function then the point $(y,\; x)$ must lie on the graph of its inverse. For example, it is easily shown here that the point $(2, 1)$ lies on the graph of $f(x)$ and that the point $(1, 2)$ lies on the graph of its inverse $f^{-1}(x)$.

(ii). Let $f(x) = {\Large\frac{x + 6}{x - 3}},\; x \neq 3$. Determine the inverse function, $f^{-1}(x)$.

(Note that the condition $x \neq 3$ is included to avoid division by zero).

Solution

Rewrite as: $y = {\Large\frac{x + 6}{x - 3}}$

Solve for $x$: $y(x - 3) = x + 6$

$xy - 3y = x + 6$

$xy - x = 3y + 6$

$x(y - 1) = 3(y + 2)$

$x = {\Large\frac{3(y + 2)}{(y - 1)}} = f^{-1}(y), \; (y \neq 1)$

Switch $x$ and $y$ to obtain the inverse $y = {\Large\frac{3(x + 2)}{(x - 1)}} = f^{-1}(x), \; x \neq 1$.

Hence, the inverse here is, $f^{-1}(x) = {\Large\frac{3(x + 2)}{x - 1}}, \; x \neq 1$.

Note:

The condition $x \neq 1$ avoids division by zero.

Check: the point $(6,\; 4)$, say, lies on the graph of $f(x)$ and we can easily show that the point $(4,\; 6)$ lies on the graph of $f^{-1}(x)$.

In Unit 4 we described how the logarithmic and exponential functions are inverses of each other.

We illustrate with examples.

(i). Find the inverse of the function $f(x) = log_2(x - 4)$.

Note here that we must have $x > 4$ to ensure that we are taking the log of a strictly positive number.

Solution

Rewrite as: $y = log_2(x - 4)$

Solve for $x$: $2^y = 2^{log_2(x-4)}$

$2^y = x - 4$

$x = 2^y + 4 = f^{-1}(y)$

Switch $x$ and $y$ to obtain the inverse: $y = 2^x + 4 = f^{-1}(x)$.

Hence, the inverse here is, $f^{-1}(x) = 2^x + 4$.

Check: It is easily shown that the point $(5,\; 0)$, say, lies on the graph of $f(x)$ and that the point $(0,\; 5)$ lies on the graph of $f^{-1}(x)$.

(ii). Find the inverse of the function $f(x) = e^{3x-2}$.

Solution

Rewrite as: $y = e^{3x - 2}$

Solve for $x$: $ln(y) = ln(e^{3x-2})$

$ln(y) = 3x - 2$

$x = {\Large\frac{1}{3}}[ln(y) + 2] = f^{-1}(y)$

Switch $x$ and $y$ to obtain the inverse: $y = {\Large\frac{1}{3}}[ln(x) + 2] = f^{-1}(x), \; (x > 0)$.

Hence, the inverse here is, $f^{-1}(x) = {\Large\frac{1}{3}}[ln(x) + 2], \; (x > 0)$.

Note:

The condition $x > 0$ ensures we are taking the natural log of a strictly positive value.

Check: it is easily shown that the point $(1, \;e)$ lies on the graph of $f(x)$ and that the point $(e,\; 1)$ lies on the graph of $f^{-1}(x)$.

Let $g : \mathbb{R} \to \mathbb{R}$ be defined by, $g(x) = x^2$. Does $g$ have an inverse?

Solution

The function, $g : \mathbb{R} \to \mathbb{R}$ given by $g(x) = x^2$ is neither one-to-one nor onto and therefore does not have an inverse.

It is not one-to-one as, for example, $g(-2) = g(2) = 4$, i.e. the number $4$, in the range, is the image of more than one number in the domain.

It is not onto as the range of $g(x)$ is $[0, \; \infty)$ . For example, there is no number $x \in \mathbb{R}$, such that $x^2 = -1$.

Note that if we restrict the domain to strictly non-negative values of $x$, i.e. $x \ge 0$, then the function $g(x) = x^2$ will be both one-to-one and onto and therefore have an inverse. In this case $g^{-1}(x) = + \sqrt{x}$.

Similarly we could define $g(x)$ for $x \le 0$ to obtain $g^{-1}(x) = - \sqrt{x}$.

As a simple example, let $f : \mathbb{R} \to \mathbb{R}$ be defined by, $f(x) = 2x + 2$.

The inverse can be shown to be given by $f^{-1}(x) = {\Large\frac{x - 2}{2}}$.

Both functions are shown in the diagram. The graph of $f(x)$ and $f^{-1}(x)$ are symmetric across the line $y = x$.

Let $A = \{a, b, c, d\}, B = \{1, 2, 3, 4\}$ and $C = \{u, v, w, x, y\}$ and define the functions $f : A \to B$ and $g : B \to C$ as shown below.

Determine the composition $g \;o\; f$ and state the domain and range of $g \;o\; f$.

Solution

To evaluate $g \;o\; f$ we apply $f$ first followed by $g$ giving the following paths from $A$ to $C$.

$a \to 1 \to u, \; b \to 3 \to v, \; c \to 2 \to w, \; d \to 4 \to y$.

Combining the above gives:

For example, we have a path from $a$ in $A$ to $u$ in $C$, a path from $b$ in $A$ to $v$ in $C$, etc.

The relationship generated by the composite function is shown below.

The domain of $g \;o\; f$ is $\{a, b, c, d\}$ and the range is $\{u, v, w, y\}$.

Note:

If $f$ and $g$ are both one-to-one functions then $g \;o\; f$ is one-to-one. See Example 18 above where both functions are one-to-one.

If $f$ and $g$ are both onto functions then $g \;o\; f$ is onto.

Let $f$ and $g$ be functions, $\mathbb{R} \to \mathbb{R}$, defined as $f(x) = x + 3$ and $g(x) = 2x - 7$.

(i). Determine $g \;o\; f$ and verify your answer.

(ii). Determine $f \;o\; g$ and verify your answer.

Solution

(i). $(g \; o \; f)(x) = g(f(x))$

$= g(x + 3)$

$= 2(x + 3) - 7$

$=$ $\;2x\;-\;1$.

Verification: We should be able to take an arbitrary value of $x$, substitute it in $g \;o\; f$ and obtain the same result as if we were to first evaluate $f(x)$ at the chosen $x$ value and use the result as input to $g(x)$.

For example, choose $x = 2$ so that

$(g \;o\; f)(2) = 2 \times 2 - 1 = 3$.

Also, $f(2) = 2 + 3 = 5$. Using this value as input to $g$ gives, $g(5) = 2 \times 5 - 7 = 3$.

This is the same as the result obtained above for $(g\; o\; f)(2)$.

(ii). $(f \;o\; g)(x) = f(g(x))$

$= f(2x - 7)$

$= 2x - 7 + 3$

$=$$\;2x\; -\; 4$.

Note that $g\; o\; f \neq f\; o\; g$.

Verification:

Choose $x = 2$ so that

$(f\; o\; g)(2) = 2 \times 2 - 4 = 0$

Also, $g(2) = 2 \times 2 - 7 = -3$.

Using this value as input to $f$ gives, $f(-3) = -3 + 3 = 0$.

This is the same as the result obtained above for $(f\; o\; g)(2)$.

Let $f$ and $g$ be the functions, $\mathbb{R} \to \mathbb{R}$ defined as $f(x) = 3x - 4$ and $g(x) = x^2 + 2$.

Determine $g\; o\; f$ and $f\; o\; g$.

Solution

$(g\;o\;f)(x) = g(f(x))$$(f\; o\; g)(x) = f(g(x))$

$= g(3x - 4)$ $= f(x^2 + 2)$

$= (3x - 4)^2 + 2$ $= 3(x^2 + 2) -4$

$=$$\;9x^2 - 24x + 18$. $=$$\;3x^2 + 2$.

Clearly $g\; o\; f \neq f\; o\; g$.

In the examples given above we were able to compose the functions for all values of $x$. However, there are some pairs of functions that cannot be composed while some pairs can only be composed for specific values of $x$.

(i). Let $f(x) = - x^2$ and $g(x) = ln(x)$.

Show that the composite function $g \;o\; f$ does not exist.

Solution

First we note that $f(x) \le 0$ for all values of $x$ and that $g(x)$ is only defined for $x > 0$. Proceeding in the usual manner we have ,

$(g\; o\; f)(x) \;\;= g(f(x))$

$= g(-x^2)$

$= ln(-x^2)$.

This term however cannot be evaluated as the natural logarithm of a non-positive expression is not defined $(-x^2 \le 0$ for all $x$). Hence, the composite function $g\; o\; f$ does not exist.

(ii). Let $f(x) = x -1$ and $g(x) = \sqrt{x}$.

Determine the composite function $g \;o\; f$ and state any values for which it is undefined.

Solution

Note here that $f(x)$ is defined for all values of $x$ while $g(x)$ is only defined for $x \ge 0$. Proceeding in the usual manner we have,

$(g\; o\; f) \;\;= g(f(x))$

$= g(x - 1)$

$= \sqrt{x - 1}$

which is only valid for $x \ge 1$.

Hence, the composite function $(g\; o\; f)$ only exists for $x \ge 1$.

In Example 14 we found that the function $f(x) = 4x - 7$ has inverse $f^{-1}(x) = {\Large\frac{x + 7}{4}}$

We now prove this result by composing the functions:

$(f^{-1}\; o\; f)(x) \;\; = f^{-1}(f(x))$

$= f^{-1}(4x - 7)$

$= {\Large\frac{4x - 7 + 7}{4}}$

$= x$.

Similarly,

$(f\; o\; f^{1})(x) \;\; = f(f^{-1}(x))$

$= f \Big({\Large\frac{x + 7}{4}}\Big)$

$= 4 \Big({\Large\frac{x + 7}{4}}\Big) - 7$

$= x + 7 - 7$

$= x$.

Hence, $f(x) = 4x - 7$ and $f^{-1}(x) = {\Large\frac{x + 7}{4}}$ are inverses of each other.