Piecewise-Defined Functions, precalculus lesson

An electricity bill often charges one rate for the first 100 kilowatt hours of a month and a higher rate for every kilowatt hour after that. No single formula gives the bill, but two simple formulas do, each in charge of its own range of usage. Functions built like this show up in tax brackets, shipping prices and phone plans, and calculus uses them as its standard examples for studying what happens where one formula hands over to the next.

Key terms

Piecewise definition

Definition A piecewise-defined function $f$ is given by formulas $f_1, f_2, \ldots, f_n$ , called branches, together with sets of real numbers $D_1, D_2, \ldots, D_n$ , no two of which share a point, such that each formula $f_k$ is defined at every number of $D_k$ . The sets are usually described by inequalities, called the conditions. The domain of $f$ is the union $D_1 \cup D_2 \cup \cdots \cup D_n$ , and for each $x$ in it

\begin{gathered} f(x) = f_k(x) \\[0.4em] \text{for the one } k \text{ with } x \in D_k. \end{gathered}

It is written with a brace, one line per branch,

f(x) = \begin{cases} f_1(x), & x \in D_1, \\ \quad \vdots & \\ f_n(x), & x \in D_n. \end{cases}

A piecewise-defined function is still a single function, with exactly one output for each input in its domain.

In plain words One function, several formulas. Each formula is in charge of its own part of the domain, and the condition next to it tells you which part that is.

Example Suppose electricity costs 0.20 dollars per kilowatt hour for the first 100 kilowatt hours of a month and 0.30 dollars for each kilowatt hour after that. For a usage of $x$ kilowatt hours with $0 \le x \le 100$ , the bill is $0.2x$ dollars. For $x > 100$ , the first 100 kilowatt hours cost $20$ dollars and the remaining $x - 100$ cost $0.3$ dollars each. So the bill in dollars is

B(x) = \begin{cases} 0.2x, \\ \quad 0 \le x \le 100, \\[0.4em] 20 + 0.3(x - 100), \\ \quad x > 100. \end{cases}

A usage of 250 kilowatt hours costs $B(250) = 20 + 0.3 \cdot 150 = 65$ dollars. It does not cost $0.3 \cdot 250 = 75$ dollars, because only the usage above 100 is charged at the higher rate.

Branch domain

Definition The branch domain of a branch is the set $D_k$ of inputs for which that branch is used. It consists of the numbers that satisfy the branch’s condition and at which its formula is defined. The formula of a branch gives the value of $f$ only on its branch domain, even if the formula itself makes sense for other numbers too. The domain of $f$ is the union of the branch domains, so a number that satisfies a condition but makes that branch’s formula undefined, such as $0$ for the formula $1/x$ with the condition $x \le 0$ , is not in the domain of $f$ .

In plain words Each formula works only inside its own territory. Outside it the formula is switched off, whatever it would give.

Example Take

g(x) = \begin{cases} \dfrac{1}{x}, & x \le -1, \\ \sqrt{x}, & 0 \le x \le 4. \end{cases}

The branch domains are $(-\infty, -1]$ and $[0, 4]$ , so the domain of $g$ is $(-\infty, -1] \cup [0, 4]$ . The number $-\frac{1}{2}$ is not in the domain, even though $1/(-\frac{1}{2}) = -2$ makes sense, and $9$ is not in the domain, even though $\sqrt{9} = 3$ does. Inside the domain, $g(-2) = -\frac{1}{2}$ and $g(4) = 2$ .

Evaluation

Definition To evaluate a piecewise-defined function $f$ at a number $a$ , find the branch whose condition $a$ satisfies, so that $a \in D_k$ , and compute $f(a) = f_k(a)$ with that formula only. If $a$ satisfies none of the conditions, then $a$ is not in the domain and $f(a)$ is undefined.

In plain words First decide which branch you are in by checking the conditions, then compute. Never mix formulas.

Example Let

h(x) = \begin{cases} x + 3, & x < -1, \\ x^2, & -1 \le x \le 2, \\ 6 - x, & x > 2. \end{cases}

Each input below satisfies exactly one condition.

Input	Condition it satisfies	Value
$-3$	$x < -1$	$h(-3) = -3 + 3 = 0$
$-1$	$-1 \le x \le 2$	$h(-1) = (-1)^2 = 1$
$0.5$	$-1 \le x \le 2$	$h(0.5) = 0.5^2 = 0.25$
$2$	$-1 \le x \le 2$	$h(2) = 2^2 = 4$
$5$	$x > 2$	$h(5) = 6 - 5 = 1$

Endpoints

Definition The endpoints of a branch domain that is an interval are the finite numbers at its ends, such as $a$ and $b$ for the condition $a \le x < b$ , whether or not the condition includes them. An endpoint where one branch domain ends and the next one begins is called a break point, because the formula in use changes there. At an endpoint $c$ , a condition either includes $c$ , written with $\le$ or $\ge$ or a square bracket, or excludes it, written with $<$ or $>$ or a parenthesis. Since the conditions do not overlap, a break point that lies in the domain of $f$ is included by exactly one condition, and $f(c)$ comes from that branch.

In plain words The boundary point between two branches belongs to one side at most. The inequality signs tell you which side, if any, so read them carefully right at the boundary.

Example The function $g$ defined earlier has endpoints $-1$ , $0$ and $4$ but no break points, since no formula hands over to another at those numbers. The function $h$ above has break points $-1$ and $2$ . At $-1$ the condition $x < -1$ fails and $-1 \le x \le 2$ holds, so $h(-1) = (-1)^2 = 1$ . Using the first formula would give $-1 + 3 = 2$ , which is wrong for this $h$ . Had the conditions been written $x \le -1$ and $-1 < x \le 2$ , the value would be $2$ , so moving the equal sign changes the function. At $2$ the middle condition includes $2$ and $x > 2$ does not, so $h(2) = 4$ .

Open and closed graph points

Definition To graph a piecewise-defined function, graph each branch formula over its own branch domain only. At an endpoint $c$ of a branch domain where the formula $f_k$ is defined, the point $(c, f_k(c))$ is drawn as a closed (filled) dot if $c$ belongs to that branch domain, and as an open (hollow) dot if it does not. A closed dot is a point of the graph. An open dot marks a point that the piece comes arbitrarily close to but that this branch does not include. If another branch has a closed dot at the same spot, the point is on the graph after all and is drawn filled.

In plain words A filled dot means the point is on the graph. A hollow dot means the piece runs right up to that point but stops just short of it.

Example For $h$ from above, here are the ends of each piece.

Branch	Branch domain	Dots
$y = x + 3$	$(-\infty, -1)$	open dot at $(-1, 2)$
$y = x^2$	$[-1, 2]$	closed dots at $(-1, 1)$ and $(2, 4)$
$y = 6 - x$	$(2, \infty)$	open dot at $(2, 4)$

At $x = -1$ the graph has a hollow dot at height $2$ and a filled dot at height $1$ . At $x = 2$ the hollow dot of the last piece sits exactly on the filled dot of the parabola, so the graph runs through $(2, 4)$ without a break.

Jumps

Definition Let $c$ be a break point of $f$ such that one formula $f_L$ is used on an interval $(b, c)$ just to the left of $c$ and another formula $f_R$ is used on an interval $(c, d)$ just to the right. Suppose both formulas are defined at $c$ and their graphs have no breaks near $c$ , as with polynomials. The graph of $f$ has a jump at $c$ if $f_L(c) \ne f_R(c)$ , and the size of the jump is

f_R(c) - f_L(c),

positive for a step up and negative for a step down. If $f_L(c) = f_R(c)$ , the two pieces meet and there is no jump. The value $f(c)$ itself, which comes from whichever condition includes $c$ , does not decide whether there is a jump.

In plain words Follow the graph from the left up to $c$ , then from the right up to $c$ . If you arrive at two different heights, the graph jumps there, and you have to lift your pencil to draw it.

Example At the break point $-1$ of $h$ , the left formula gives $-1 + 3 = 2$ and the right formula gives $(-1)^2 = 1$ . These differ, so $h$ jumps at $-1$ , with size $1 - 2 = -1$ , a step down by $1$ . At $2$ the left formula gives $2^2 = 4$ and the right formula gives $6 - 2 = 4$ , so there is no jump. The electricity bill $B$ has no jump at $100$ either, since both of its formulas give $20$ there. Calculus describes the two heights at a break point with one-sided limits.

Overlapping and missing branch conditions

Definition The conditions of a piecewise definition should cover every intended input exactly once. If two conditions overlap at an input $a$ and their formulas give different values there, the rule assigns two outputs to $a$ and does not define a function. If the formulas agree at every shared input, the rule still defines a function, and removing the shared inputs from one of the conditions brings it to the standard form above, with branch domains that do not overlap. An input that satisfies no condition is not in the domain, even when it lies between two branch domains.

In plain words Two conditions that both claim a point must agree there, or the rule is broken. A point that no condition claims is simply missing from the domain.

Example The rule

k(x) = \begin{cases} x + 1, & x \le 2, \\ x^2, & x \ge 2 \end{cases}

claims $x = 2$ twice, with outputs $3$ and $4$ , so it is not a function. Changing the second condition to $x > 2$ repairs it, and then $k(2) = 3$ . The opposite slip is a gap. A rule with the conditions $x < 0$ and $x > 0$ says nothing about $0$ , so $0$ is not in its domain. If $0$ should be included, one of the two conditions must become $x \le 0$ or $x \ge 0$ .

Check yourself

A courier charges 6 dollars for a parcel of up to 2 kilograms, 10 dollars for more than 2 and up to 5 kilograms, and 15 dollars for more than 5 and up to 20 kilograms. Write the price as a piecewise function $S(w)$ of the weight $w$ , find $S(2)$ , $S(2.5)$ and $S(5)$ , and describe the jumps.

Show the answer

The price in dollars is

S(w) = \begin{cases} 6, & 0 < w \le 2, \\ 10, & 2 < w \le 5, \\ 15, & 5 < w \le 20. \end{cases}

So $S(2) = 6$ , because $2$ satisfies $0 < w \le 2$ , while $S(2.5) = 10$ and $S(5) = 10$ . The graph jumps up by $4$ at $w = 2$ , from a closed dot at $(2, 6)$ to an open dot at $(2, 10)$ , and up by $5$ at $w = 5$ , from a closed dot at $(5, 10)$ to an open dot at $(5, 15)$ .

Let $f(x) = x^2$ for $x < 1$ and $f(x) = 5 - x$ for $1 < x \le 4$ . Find the domain of $f$ and evaluate $f(1)$ , $f(4)$ and $f(5)$ where they are defined.

Show the answer

The branch domains are $(-\infty, 1)$ and $(1, 4]$ , so the domain is $(-\infty, 1) \cup (1, 4]$ . The input $1$ satisfies neither condition, so $f(1)$ is undefined, and the graph has open dots at $(1, 1)$ and $(1, 4)$ . The input $4$ satisfies $1 < x \le 4$ , so $f(4) = 5 - 4 = 1$ . The input $5$ satisfies no condition, so $f(5)$ is undefined.

One rule sets $p(x) = x^2$ for $x \le 1$ and $p(x) = 2x$ for $x \ge 1$ . Another sets $q(x) = x^2$ for $x \le 1$ and $q(x) = 2x - 1$ for $x \ge 1$ . Which of them defines a function?

Show the answer

Both rules claim $x = 1$ twice. For $p$ the two formulas give $1^2 = 1$ and $2 \cdot 1 = 2$ , two different outputs, so $p$ is not a function. For $q$ they give $1$ and $2 \cdot 1 - 1 = 1$ , the same output, so $q$ is a function with $q(1) = 1$ , and its graph has no jump at $1$ .

Piecewise-Defined Functions

Key terms

Piecewise definition

Branch domain

Evaluation

Endpoints

Open and closed graph points

Jumps

Overlapping and missing branch conditions

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus