Functions and Their Graphs - Maths: an open, verified university mathematics reference

Why this matters¶

A farmer has 40 metres of fencing and wants a rectangular pen against a long wall, so only three sides need fencing. How large can the pen be? Before we can answer, we need a precise way to say how the area depends on the width. Each width gives exactly one area, some widths make no sense at all (a negative width, or one that uses more than 40 metres of fence), and only some areas can actually occur.

Those three ingredients — a rule, the inputs it accepts and the outputs it produces — are what a function is. Almost everything in calculus is about functions: limits describe how their values behave, derivatives measure how fast they change, and integrals add up their values. This page fixes the vocabulary that the rest of the course uses, and in particular the exact meaning of “increasing”, which later results depend on.

Functions and their graphs¶

Think of a function as a machine: you feed it an input, and it returns exactly one output. The same input always gives the same output. Different inputs may give the same output.

In words. Every allowed input gets an output, and never more than one. The function is $f$ ; the number $f(x)$ is its value at one input $x$ . The graph is the curve $y = f(x)$ : the point above (or below) each input $x$ at height $f(x)$ .

Example. $f\colon \R \to \R$ , $f(x) = x^2$ , is a function: each real $x$ has exactly one square. Different inputs can share a value ( $f(-2) = f(2) = 4$ ); that is allowed.

Non-example. Assign to each $x \in \R$ the real numbers $y$ with $y^2 = x$ . This is not a function from $\R$ to $\R$ , and it fails in both ways: $x = 4$ gets two numbers, 2 and -2, and $x = -1$ gets none.

Because each $x$ has only one value, a vertical line $x = c$ meets the graph of a function at most once: at the point $(c, f(c))$ if $c$ is an allowed input, and nowhere otherwise. This is the vertical line test: a curve in the plane is the graph of a function of $x$ exactly when no vertical line meets it twice (Figure 1).

Two coordinate planes side by side. On the left, the parabola y = x squared, with a dashed vertical line at x = 1.5 that meets it at exactly one point. On the right, the sideways parabola y squared = x, opening to the right, with a dashed vertical line at x = 2.5 that meets it at two points, one above and one below the x-axis. — Figure 1:The vertical line test. Left: the graph of $y = x^2$ ; every vertical line meets it exactly once. Right: the curve $y^2 = x$ of the non-example; the vertical line $x = 2.5$ meets it twice, so it is not the graph of a function of $x$ .

The set $\{f(x) : x \in \dom f\}$ is written in set-builder notation, and most domains and ranges below are intervals or unions of intervals; both are introduced in Real Numbers and Intervals.

In words. The domain is what you may put in; the range is what actually comes out. The codomain is only a set that is promised to contain every value, and it may be larger than the range. To find the range you must show two things: every value $f(x)$ lies in the set you claim, and every element of that set is a value $f(x)$ for some $x$ .

Example. For $f(x) = \sqrt{x - 1}$ , the square root needs $x - 1 \ge 0$ , so $\dom f = [1, \infty)$ . Every square root is $\ge 0$ , and each $y \ge 0$ is the value at $x = y^2 + 1$ , because $\sqrt{y^2} = y$ for $y \ge 0$ (see the remark on square roots). So $\ran f = [0, \infty)$ .

Non-example. For $g\colon \R \to \R$ , $g(x) = x^2$ , the codomain is $\R$ , but the range is not: -1 is not in $\ran g$ , because no real $x$ has $x^2 = -1$ . Here $\ran g = [0, \infty)$ .

Reading a graph. The domain is the shadow of the graph on the $x$ -axis: the set of $x$ that have a point of the graph directly above or below them. The range is the shadow on the $y$ -axis: the set of heights that the graph reaches (Figure 2). A sketch suggests the domain and range; the algebra above confirms them.

The graph of y = square root of (x minus 1): a curve that starts at the point (1, 0) and rises ever more slowly to the right. A shaded band on the x-axis runs from 1 to the right, and a shaded band on the y-axis runs from 0 upwards. Dashed lines join the point (5, 2) of the graph to 5 on the x-axis and to 2 on the y-axis. — Figure 2:Domain and range as shadows, for $f(x) = \sqrt{x - 1}$ . The shadow of the graph on the $x$ -axis is $\dom f = [1, \infty)$ , and its shadow on the $y$ -axis is $\ran f = [0, \infty)$ . The point $(5, 2)$ of the graph casts 5 into the domain and $2 = f(5)$ into the range.

Sketching a graph. For a first sketch, find the domain, compute a few values (including any points where the formula changes, and where the graph crosses an axis), plot them, and join them in order of increasing $x$ , leaving gaps where $x$ is not in the domain. Where the formula changes, compare the heights that the two pieces give there before you join across that point: if they differ, the graph jumps, and no segment is drawn across the jump. For example, the function that is 0 for $x < 0$ and 1 for $x \ge 0$ has domain $\R$ , but its graph is two horizontal pieces, with no segment from $(-1, 0)$ to $(0, 1)$ .

Symmetry and monotonicity¶

Two kinds of shape recur throughout calculus. Some graphs look the same after a reflection: the graph of $y = x^2$ is unchanged by reflection in the $y$ -axis. Some functions only ever go up (or only ever go down) as $x$ moves to the right. Both properties save work: a symmetric graph needs to be sketched only for $x \ge 0$ , and a function that always goes up (one that is strictly increasing, in the words of the definition below) takes each value at most once.

In words. An even function takes the same value at $x$ and at $-x$ , so its graph is symmetric under reflection in the $y$ -axis. An odd function takes opposite values at $x$ and $-x$ , so its graph is symmetric under a half-turn about the origin: if $(x, y)$ is on the graph, so is $(-x, -y)$ . Both need a domain that is symmetric about 0, so a function whose domain is not symmetric is neither even nor odd. The condition on the values must hold for every $x$ in the domain; one $x$ where it fails is enough to show that a function is not even (or not odd).

Example. $x \mapsto x^4 - 3x^2$ is even and $x \mapsto x^3$ is odd. The names come from these examples: a polynomial with only even powers of $x$ is even, and one with only odd powers is odd.

Non-example. $f(x) = x^2 + x$ is neither even nor odd: $f(1) = 2$ and $f(-1) = 0$ , so $f(-1) \ne f(1)$ and $f(-1) \ne -f(1)$ . Most functions are neither even nor odd.

Definition 4 (Increasing, decreasing and monotone functions)

Let $f$ be a function and let $S \subseteq \dom f$ (usually an interval). We say that $f$ is

increasing on $S$ if for all $x_1, x_2 \in S$ with $x_1 < x_2$ we have $f(x_1) \le f(x_2)$ ;
strictly increasing on $S$ if for all $x_1, x_2 \in S$ with $x_1 < x_2$ we have $f(x_1) < f(x_2)$ ;
decreasing on $S$ if for all $x_1, x_2 \in S$ with $x_1 < x_2$ we have $f(x_1) \ge f(x_2)$ ;
strictly decreasing on $S$ if for all $x_1, x_2 \in S$ with $x_1 < x_2$ we have $f(x_1) > f(x_2)$ .

$f$ is monotone on $S$ if it is increasing on $S$ or decreasing on $S$ , and strictly monotone on $S$ if it is strictly increasing on $S$ or strictly decreasing on $S$ . When $S = \dom f$ we leave out “on $S$ ”.

In words. Increasing means “never goes down” as you move right through $S$ ; strictly increasing means “always goes up”. A strictly increasing function is increasing, but not the other way round: a function that is increasing may stay level for a while. The definition compares any two points of $S$ , not just neighbouring ones.

Example. $f(x) = 2x + 1$ is strictly increasing on $\R$ : if $x_1 < x_2$ , then $2x_1 + 1 < 2x_2 + 1$ . A constant function on a set $S$ with at least two points is both increasing and decreasing on $S$ , but neither strictly increasing nor strictly decreasing on $S$ . (On a set with only one point there are no $x_1 < x_2$ to compare, so every function is strictly increasing and strictly decreasing there.) The function

g(x) = \begin{cases} 0, & x < 0, \\ x, & x \ge 0, \end{cases}

(3)

is increasing on $\R$ but not strictly increasing, because $g(-2) = g(-1) = 0$ (Figure 3).

Two graphs side by side. On the left, a graph that runs along the x-axis at height 0 from x = -3 to x = 0 and then rises in a straight line to the point (3, 3). On the right, the cubic curve y = x cubed from x = -1.5 to x = 1.5, which rises everywhere and flattens out briefly near the origin without becoming level. — Figure 3:Left: the function $g$ above is increasing on $\R$ but not strictly increasing: it stays level for $x < 0$ . Right: $x \mapsto x^3$ is strictly increasing on $\R$ (Exercise 13 proves it). Its graph is very flat near 0, but no two points of it are at the same height.

Non-example. $h(x) = x^2$ is not monotone on $\R$ : $-1 < 0$ with $h(-1) > h(0)$ , but $0 < 1$ with $h(0) < h(1)$ . It is strictly decreasing on $(-\infty, 0]$ and strictly increasing on $[0, \infty)$ , which is why we always say where a function is increasing.

Worked examples¶

Example 1 (The natural domain of

\frac{\sqrt{x+2}}{x-3}

)

Find the natural domain of $f(x) = \dfrac{\sqrt{x+2}}{x-3}$ .

The square root is a real number only if $x + 2 \ge 0$ , that is, $x \ge -2$ .
The quotient is defined only if the denominator is not zero: $x - 3 \ne 0$ , that is, $x \ne 3$ .
Both conditions must hold, so the natural domain is the set of $x \ge -2$ with $x \ne 3$ :

\boxed{\dom f = [-2, 3) \cup (3, \infty)}

(4)

Check. $f(-2) = \frac{0}{-5} = 0$ is defined, so -2 belongs to the domain. At $x = -3$ the formula needs $\sqrt{-1}$ , and at $x = 3$ it divides by zero; neither is in the domain. ✓

Example 2 (The range of

x^2 - 4x + 5

)

Find the range of $f(x) = x^2 - 4x + 5$ , whose domain is $\R$ .

Complete the square. $x^2 - 4x + 5 = (x - 2)^2 + 1$ .
Every value is at least 1. A square is never negative, so $f(x) = (x-2)^2 + 1 \ge 1$ for every $x$ . Hence $\ran f \subseteq [1, \infty)$ .
Every $y \ge 1$ is a value. Given $y \ge 1$ , the number $y - 1 \ge 0$ has a square root (the remark on square roots). Put $x = 2 + \sqrt{y - 1}$ ; then $f(x) = \bigl(\sqrt{y-1}\bigr)^2 + 1 = y$ . Hence $[1, \infty) \subseteq \ran f$ .
The two inclusions together give

\boxed{\ran f = [1, \infty)}

(5)

Check. For $y = 10$ , step 3 gives $x = 2 + 3 = 5$ , and indeed $f(5) = 25 - 20 + 5 = 10$ . The smallest value is $f(2) = 4 - 8 + 5 = 1$ . ✓

Example 3 (Even, odd or neither?)

Decide whether each function is even, odd or neither.

$f(x) = x^3 - x$ . The domain $\R$ is symmetric, and $f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -f(x)$ , so $f$ is odd.
$g(x) = \dfrac{1}{x^2 - 1}$ . The domain is all $x$ except 1 and -1, which is symmetric about 0, and $g(-x) = \dfrac{1}{(-x)^2 - 1} = \dfrac{1}{x^2 - 1} = g(x)$ , so $g$ is even.
$h(x) = x^2 + x$ . As in the non-example above, $h(1) = 2$ and $h(-1) = 0$ , so $h$ is neither.
$k\colon [-1, 2] \to \R$ , $k(x) = x^2$ . The formula is the same as for an even function, but $2 \in \dom k$ while $-2 \notin \dom k$ . The domain is not symmetric about 0, and Definition 3 asks for a symmetric domain both for even and for odd. So $k$ is neither even nor odd.

Check. $f(2) = 8 - 2 = 6$ and $f(-2) = -8 + 2 = -6 = -f(2)$ ; $g(3) = \frac{1}{8} = g(-3)$ . ✓

Example 4 (A function that is increasing on each piece of its domain)

Show that $f(x) = \dfrac{x}{x + 1}$ is strictly increasing on $(-1, \infty)$ , and decide whether it is increasing on its whole domain.

Compare two points. Let $-1 < x_1 < x_2$ . Over a common denominator,
$f(x_2) - f(x_1) = \frac{x_2 (x_1 + 1) - x_1 (x_2 + 1)}{(x_1 + 1)(x_2 + 1)} = \frac{x_2 - x_1}{(x_1 + 1)(x_2 + 1)}.$
(6)
Signs. The numerator is positive because $x_1 < x_2$ . Both factors of the denominator are positive because $x_1, x_2 > -1$ . So $f(x_2) - f(x_1) > 0$ , that is, $f(x_1) < f(x_2)$ . Hence $f$ is strictly increasing on $(-1, \infty)$ .
The whole domain. $\dom f = (-\infty, -1) \cup (-1, \infty)$ . The same computation shows that $f$ is strictly increasing on $(-\infty, -1)$ as well (there both factors of the denominator are negative, so their product is positive). But $f$ is not increasing on its domain: $-2 < 0$ , while $f(-2) = \frac{-2}{-1} = 2 > 0 = f(0)$ .

\boxed{\text{strictly increasing on } (-\infty, -1) \text{ and on } (-1, \infty), \text{ but not increasing on } \dom f}

(7)

Check. $f(0) = 0$ , $f(1) = \frac{1}{2}$ , $f(3) = \frac{3}{4}$ : the values grow, as they should on $(-1, \infty)$ . ✓

Example 5 (The largest pen against a wall)

A rectangular pen is built against a long wall, with 40 m of fencing for the other three sides. Express its area as a function of the width $x$ (in metres) of the two sides that meet the wall, and find the domain and the range of that function.

Model. The two sides that meet the wall use $2x$ metres of fence, so the side parallel to the wall is $40 - 2x$ metres long. The area, in square metres, is
$A(x) = x(40 - 2x).$
(8)
Domain from the context. The formula makes sense for every real $x$ , but a pen needs $x > 0$ and $40 - 2x > 0$ , that is, $x < 20$ . So $\dom A = (0, 20)$ .
Range. Completing the square, $A(x) = 40x - 2x^2 = 200 - 2(x - 10)^2$ . Let $0 < x < 20$ . A square is never negative, so $(x - 10)^2 \ge 0$ . Also $(x - 10)^2 - 100 = x^2 - 20x = x(x - 20) < 0$ , because $x > 0$ and $x - 20 < 0$ . So $0 \le (x - 10)^2 < 100$ , and $0 < A(x) \le 200$ . Conversely, let $0 < a \le 200$ . Then $0 \le \frac{200 - a}{2} < 100$ , so $s = \sqrt{(200 - a)/2}$ is a real number with $s \ge 0$ and $s^2 < 100 = 10^2$ . Squaring keeps the order of non-negative numbers (the remark on square roots), so $s < 10$ . Hence $x = 10 - s$ lies in $(0, 10]$ , and $A(x) = 200 - 2s^2 = 200 - (200 - a) = a$ . So $\ran A = (0, 200]$ .

\boxed{A(x) = x(40 - 2x) \text{ m}^2, \quad \dom A = (0, 20), \quad \ran A = (0, 200]}

(9)

So the largest possible pen has area 200 m², when $x = 10$ m. Notice that the domain came from the situation, not from the formula: the natural domain of $x(40 - 2x)$ is all of $\R$ .

Check. $A(10) = 10 \cdot 20 = 200$ , and $A(5) = 5 \cdot 30 = 150 = 200 - 2 \cdot 5^2$ . ✓

Common mistakes¶

Rigorous track¶

What a “rule” is

Definition 1 speaks of a rule, which is not a precise mathematical object. The standard way to make it precise is to identify a function with its graph. A function from $A$ to $B$ is a subset $G \subseteq A \times B$ of ordered pairs such that

\forall x \in A \ \ \exists!\, y \in B : (x, y) \in G,

(10)

where $\exists!$ means “there exists exactly one”. The value $f(x)$ is that unique $y$ . With this definition a function is its set of input–output pairs, and the vertical line test is just the condition “exactly one $y$ for each $x$ ” drawn in the plane.

When are two functions equal?

Two functions $f$ and $g$ are equal when $\dom f = \dom g$ and $f(x) = g(x)$ for every $x$ in that common domain. The domain is part of the function, not an afterthought:

$x \mapsto x + 1$ on $\R$ and $x \mapsto \frac{x^2 - 1}{x - 1}$ (natural domain) are different functions: one is defined at 1 and the other is not.
$x \mapsto x^2$ on $\R$ is not monotone, but the “same” formula on $[0, \infty)$ is strictly increasing. Restricting the domain in this way is how inverse functions such as $\sqrt{x}$ are obtained later in the course.

Some books also count the codomain as part of a function, so that $x \mapsto x^2$ as a map $\R \to \R$ and as a map $\R \to [0, \infty)$ are different. Nothing in this course depends on that distinction.

Summary¶

A function $f\colon A \to B$ assigns to each $x \in A$ exactly one value $f(x)$ ; its graph is the set of points $(x, f(x))$ , and no vertical line meets it twice.
$\dom f$ is the set of inputs; $\ran f = \{f(x) : x \in \dom f\}$ is the set of values. A formula with no stated domain has its natural domain. To find a range, show both inclusions.
In a model, the domain comes from the situation, not only from the formula.
Even means that $\dom f$ is symmetric about 0 and $f(-x) = f(x)$ (mirror symmetry in the $y$ -axis); odd means that $\dom f$ is symmetric about 0 and $f(-x) = -f(x)$ (half-turn symmetry about the origin). A function whose domain is not symmetric is neither.
Increasing on $S$ : $x_1 < x_2$ implies $f(x_1) \le f(x_2)$ . Strictly increasing: $x_1 < x_2$ implies $f(x_1) < f(x_2)$ . Always say on which set.

Exercises¶

Exercise 13 (The cube is strictly increasing)

Show from Definition 4 that $x \mapsto x^3$ is strictly increasing on $\R$ .

Hint 1

For $x_1 < x_2$ , factor $x_2^3 - x_1^3 = (x_2 - x_1)\bigl(x_2^2 + x_1 x_2 + x_1^2\bigr)$ .

Hint 2

Complete the square in the second factor: $x_2^2 + x_1 x_2 + x_1^2 = \bigl(x_1 + \tfrac{1}{2} x_2\bigr)^2 + \tfrac{3}{4} x_2^2$ . When can this be zero?

Answer

$x_2^3 - x_1^3 = (x_2 - x_1)\bigl(\bigl(x_1 + \tfrac{1}{2}x_2\bigr)^2 + \tfrac{3}{4}x_2^2\bigr)$ , and both factors are positive when $x_1 < x_2$ .

Exercise 14 (Every function is even plus odd)

Let $f$ be any function with domain $\R$ . Show that there are exactly one even function $g$ and exactly one odd function $h$ , both with domain $\R$ , such that $f(x) = g(x) + h(x)$ for every $x$ . Find $g$ and $h$ when $f(x) = x^2 + x + 1$ .

Hint 1

Suppose $f = g + h$ works. Write down $f(x)$ and $f(-x)$ in terms of $g(x)$ and $h(x)$ .

Hint 2

Add and subtract the two equations of Hint 1 to find $g(x)$ and $h(x)$ . Then check that the functions you found really work.

Answer

$g(x) = \frac{f(x) + f(-x)}{2}$ and $h(x) = \frac{f(x) - f(-x)}{2}$ ; for $f(x) = x^2 + x + 1$ these are $g(x) = x^2 + 1$ and $h(x) = x$ .

Where this leads¶

These topics build on this one:

Polynomial and Rational Functions
Trigonometric Functions
The Limit of a Function
Combining and Transforming Functions (coming soon)

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