Functions and their graphs (precalculus)

MotivationH

Knowing one graph means knowing many

A student in Lagos and her cousin in Toronto compare the weather by video call. She thinks in degrees Celsius, he in Fahrenheit. The conversion

F = \tfrac95\,C + 32

is a function: one input, exactly one output. Its graph is a straight line. To convert back, he needs the inverse function $C = \tfrac59(F - 32)$ , whose graph is the first line reflected in the diagonal. If the forecast says " $10$ degrees warmer tomorrow", that is another function, $C \mapsto C + 10$ . Doing one function after another is composition.

This lesson is about seeing functions whole. If you know the graph of $y = x^2$ , you know the graph of $y = 2(x - 3)^2 + 1$ without plotting a single new point. If you know the graph of $f$ , you know the graph of $f^{-1}$ for free. And from a formula like $\dfrac{2x + 1}{x - 3}$ you will be able to predict where its graph shoots off to infinity and what it looks like far away.

IntuitionH

A graph is a portrait of a function

In the lesson on functions we met a function as a rule that gives every input in its domain exactly one output. For real functions, we can draw all the pairs (input, output) at once as points $(x, f(x))$ in the plane. That picture is the graph.

The graph lets you read off properties at a glance:

the domain is the shadow of the graph on the $x$ -axis, and the range is its shadow on the $y$ -axis;
the zeros are where it meets the $x$ -axis;
increasing means the graph climbs as you move right;
a vertical line meets the graph of a function at most once (one input, one output) — the vertical line test;
a horizontal line meets it at most once exactly when the function is injective — the horizontal line test — and that, we will prove, is what an inverse needs.

DefinitionsH

Definitions

Definition— real function, natural domain, graph

A real function $f$ assigns to each number $x$ in its domain $D \subseteq \R$ exactly one real number $f(x)$ . When a function is given only by a formula, its natural domain is the set of all real $x$ for which the formula makes sense. The graph of $f$ is the set of points

\operatorname{graph} f = \{\, (x, f(x)) : x \in D \,\} \subseteq \R \times \R.

The range is the set of outputs $\{ f(x) : x \in D \}$ .

For example, $\sqrt{x - 1}$ has natural domain $[1, \infty)$ , $\dfrac{1}{x^2 - 4}$ has natural domain $\R \setminus \{-2, 2\}$ , and every polynomial has natural domain $\R$ .

Definition— increasing, even and odd

$f$ is strictly increasing on a set $S$ if $x < y$ implies $f(x) < f(y)$ for all $x, y \in S$ (and strictly decreasing if it implies $f(x) > f(y)$ ). On a domain symmetric about $0$ , $f$ is even if $f(-x) = f(x)$ for all $x$ , and odd if $f(-x) = -f(x)$ for all $x$ .

The graph of an even function is symmetric in the $y$ -axis ( $x^2$ , $|x|$ , $x^4 - 3x^2$ ). The graph of an odd function is symmetric under a half-turn about the origin ( $x^3$ , $1/x$ , $x^3 - 5x$ ). Most functions, such as $x^2 + x$ , are neither.

Definition— composition and inverse

The composite $g \circ f$ is the function $(g \circ f)(x) = g(f(x))$ . Its natural domain is the set of $x$ in the domain of $f$ with $f(x)$ in the domain of $g$ . A function $g : B \to A$ is an inverse of $f : A \to B$ if

g(f(x)) = x \text{ for all } x \in A \quad\text{and}\quad f(g(y)) = y \text{ for all } y \in B.

We then write $g = f^{-1}$ .

Definition— asymptotes and end behaviour (informal)

The line $y = L$ is a horizontal asymptote of $f$ if $f(x)$ gets as close to $L$ as we like once $x$ is large enough (or once $x$ is negative enough); we write $f(x) \to L$ as $x \to \infty$ (or $-\infty$ ). The line $x = a$ is a vertical asymptote if $|f(x)|$ grows beyond every bound as $x$ approaches $a$ from at least one side. The end behaviour of $f$ describes what $f(x)$ does as $x \to \pm\infty$ .

These last definitions use the idea of a limit informally. It is made precise in calculus, and the Lean proof below uses the precise version.

Worked examplesH

A gallery of function families

Example— The basic shapes

Linear $y = mx + c$ : a straight line with slope $m$ (see linear relationships).
Quadratic $y = ax^2 + bx + c$ : a parabola (see quadratics).
Powers $y = x^n$ : for even $n$ U-shaped and even; for odd $n$ S-shaped, odd and strictly increasing.
Absolute value $y = |x|$ : a V with its corner at the origin; $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$ .
Square root $y = \sqrt{x}$ on $[0, \infty)$ : half of a sideways parabola.
Reciprocal $y = 1/x$ on $x \ne 0$ : a hyperbola with asymptotes $x = 0$ and $y = 0$ .

Example— A piecewise function: a block tariff

Many electricity and water utilities charge in blocks. Here is an illustrative tariff in the style used by several municipalities in southern Africa: the first $50$ kWh in a month cost $1.20$ per kWh and each further kWh costs $1.80$ . The monthly bill for $x$ kWh is

B(x) = \begin{cases} 1.2\,x, & 0 \le x \le 50, \\ 60 + 1.8\,(x - 50), & x > 50. \end{cases}

The two pieces agree at $x = 50$ (both give $60$ ), so the graph has no jump, only a bend where the slope increases from $1.2$ to $1.8$ .

Example— A rational function

$r(x) = \dfrac{2x + 1}{x - 3}$ has natural domain $x \ne 3$ . Dividing, $r(x) = 2 + \dfrac{7}{x - 3}$ . So:

as $x \to \pm\infty$ , $\tfrac{7}{x - 3} \to 0$ and $r(x) \to 2$ : horizontal asymptote $y = 2$ ;
as $x \to 3$ , $\tfrac{7}{x - 3}$ becomes huge, positive just to the right of $3$ and negative just to the left: vertical asymptote $x = 3$ ;
the zero is at $2x + 1 = 0$ , so $x = -\tfrac12$ ; the $y$ -intercept is $r(0) = -\tfrac13$ .

Its graph is the hyperbola $y = 7/x$ stretched and moved: right by $3$ and up by $2$ .

Example— Holes are not asymptotes

$\dfrac{x^2 - 1}{x - 1} = \dfrac{(x - 1)(x + 1)}{x - 1} = x + 1$ for every $x \ne 1$ . Its graph is the line $y = x + 1$ with the single point $(1, 2)$ missing — a hole, not an asymptote. A zero of the denominator gives a vertical asymptote only if it does not cancel against the numerator.

Example— End behaviour of polynomials

For large $|x|$ the leading term wins. For $p(x) = -2x^3 + 50x^2 + 7$ ,

p(x) = -2x^3\left(1 - \frac{25}{x} - \frac{7}{2x^3}\right),

and the bracket tends to $1$ . So $p(x) \to -\infty$ as $x \to \infty$ and $p(x) \to +\infty$ as $x \to -\infty$ , like $-2x^3$ , even though $p(10) = 3007$ is positive.

Worked examplesH

Transforming graphs

Starting from the graph of $y = f(x)$ :

New function	What happens to the graph
$y = f(x) + k$	moves up by $k$
$y = f(x - h)$	moves right by $h$
$y = a\,f(x)$	stretches vertically by the factor $\lvert a \rvert$ (and reflects in the $x$ -axis if $a < 0$ )
$y = f(bx)$	squeezes horizontally by the factor $\lvert b \rvert$ (and reflects in the $y$ -axis if $b < 0$ )

Example— Building y = −2|x − 1| + 3 from y = |x|

Start with the V of $y = |x|$ , corner at $(0, 0)$ . Replace $x$ by $x - 1$ : the corner moves right to $(1, 0)$ . Multiply by $-2$ : the V flips upside down and becomes twice as steep. Add $3$ : the corner rises to $(1, 3)$ . The result is an upside-down V with its peak at $(1, 3)$ , crossing the $x$ -axis where $|x - 1| = \tfrac32$ , at $x = -\tfrac12$ and $x = \tfrac52$ .

Example— The order matters

$y = f(2x - 6)$ is $y = f\big(2(x - 3)\big)$ : squeeze by $2$ and shift right by $3$ , not by $6$ . Always factor out $b$ first to read off $h$ .

ProofH

Proofs

Theorem— Transformation rule

Let $b \ne 0$ . A point $(x, y)$ lies on the graph of $y = f(x)$ if and only if the point $\left(\tfrac{x}{b} + h,\ a y + k\right)$ lies on the graph of $y = a\,f\big(b(X - h)\big) + k$ . In particular ( $a = b = 1$ , $k = 0$ ), the graph of $y = f(x - h)$ is the graph of $f$ moved $h$ units to the right.

Proof

Write $g(X) = a f(b(X - h)) + k$ . If $y = f(x)$ and $X = \tfrac{x}{b} + h$ , then $b(X - h) = x$ , so $g(X) = a f(x) + k = ay + k$ : the moved point is on the graph of $g$ . Conversely, every point $(X, g(X))$ of the graph of $g$ arises this way, from the point $(x, f(x))$ with $x = b(X - h)$ . So the graph of $g$ is exactly the image of the graph of $f$ under the move.

For the shift: the output that $f$ produces at input $x_0$ , the function $x \mapsto f(x - h)$ produces when $x - h = x_0$ , that is, at $x = x_0 + h$ — later by $h$ . That is why subtracting $h$ moves the graph to the right.

∎

Theorem— Inverses and bijections

A function $f : A \to B$ has an inverse if and only if it is a bijection (injective and surjective). The inverse is then unique.

Proof

If $g$ is an inverse and $f(x_1) = f(x_2)$ , then $x_1 = g(f(x_1)) = g(f(x_2)) = x_2$ , so $f$ is injective. Every $y \in B$ equals $f(g(y))$ , so $f$ is surjective. Conversely, if $f$ is a bijection, each $y \in B$ is $f(x)$ for exactly one $x \in A$ . Define $g(y)$ to be that $x$ ; then $g(f(x)) = x$ and $f(g(y)) = y$ . If $g'$ is another inverse, then $g'(y) = g'(f(g(y))) = g(y)$ for every $y$ .

∎

Theorem— Strictly monotone functions are injective

If $f$ is strictly increasing (or strictly decreasing) on a set $S$ , then $f$ is injective on $S$ . So $f$ has an inverse from $S$ onto its range $f(S)$ , and if $f$ is strictly increasing so is $f^{-1}$ .

Proof

If $x \ne y$ in $S$ , say $x < y$ , then $f(x) < f(y)$ , so $f(x) \ne f(y)$ . For the inverse: let $y_1 < y_2$ be in $f(S)$ and suppose $f^{-1}(y_1) \ge f^{-1}(y_2)$ . Since $f$ is increasing, $f(f^{-1}(y_1)) \ge f(f^{-1}(y_2))$ , that is, $y_1 \ge y_2$ , a contradiction.

∎

This is how we invert functions that are not injective: restrict the domain to an interval where they are strictly monotone. $f(x) = x^2$ is not injective on $\R$ ( $f(-3) = f(3)$ ), but it is strictly increasing on $[0, \infty)$ , and there its inverse is $\sqrt{x}$ .

Theorem— The graph of the inverse is a reflection

If $g = f^{-1}$ , then $(x, y)$ lies on the graph of $f$ if and only if $(y, x)$ lies on the graph of $g$ . So the graph of $f^{-1}$ is the mirror image of the graph of $f$ in the line $y = x$ .

Proof

If $y = f(x)$ then $g(y) = g(f(x)) = x$ , so $(y, x)$ is on the graph of $g$ . If $x = g(y)$ then $f(x) = f(g(y)) = y$ . Finally, the reflection in the line $y = x$ is exactly the map $(x, y) \mapsto (y, x)$ : the midpoint $\left(\tfrac{x + y}{2}, \tfrac{x + y}{2}\right)$ of the two points lies on the line, and the segment joining them has direction $(1, -1)$ , perpendicular to the line.

∎

Example— Finding an inverse by algebra

To invert $f(x) = \dfrac{2x + 1}{x - 3}$ ( $x \ne 3$ ), solve $y = \dfrac{2x + 1}{x - 3}$ for $x$ :

y(x - 3) = 2x + 1 \;\Longrightarrow\; xy - 2x = 3y + 1 \;\Longrightarrow\; x = \frac{3y + 1}{y - 2} \quad (y \ne 2).

So $f^{-1}(y) = \dfrac{3y + 1}{y - 2}$ , defined for $y \ne 2$ . The value $y = 2$ , the horizontal asymptote of $f$ , is not in the range of $f$ , and it becomes the vertical asymptote of $f^{-1}$ . Reflection swaps the roles of horizontal and vertical.

Theorem— Even and odd parts

Every function $f : \R \to \R$ can be written in exactly one way as $f = e + o$ with $e$ even and $o$ odd, namely

e(x) = \frac{f(x) + f(-x)}{2}, \qquad o(x) = \frac{f(x) - f(-x)}{2}.

Proof

Existence: $e(-x) = \tfrac{f(-x) + f(x)}{2} = e(x)$ , $o(-x) = \tfrac{f(-x) - f(x)}{2} = -o(x)$ , and $e(x) + o(x) = f(x)$ .

Uniqueness: if $e_1 + o_1 = e_2 + o_2$ , then $d = e_1 - e_2 = o_2 - o_1$ is both even and odd, so $d(x) = d(-x) = -d(x)$ and $d(x) = 0$ for every $x$ .

∎

For a polynomial, the even part collects the even powers and the odd part the odd powers: $x^3 + 2x^2 - x + 5$ splits as $(2x^2 + 5) + (x^3 - x)$ .

Formal mathematicsF