Piecewise Functions

Prices set in bands

Many prices are set in bands. A postal service might charge $1 to send a letter that weighs up to 100 g, and $2 for a letter that weighs over 100 g and up to 250 g. The price is a function of the weight, but no single formula gives it: which rule applies depends on the band the weight falls in.

Write the weight in grams as w and the price in dollars as P(w). Then P(w) = 1 for $0<w\le 100$0 < w ≤ 100, and P(w) = 2 for $100<w\le 250$100 < w ≤ 250. A function defined by different rules on different parts of its domain is called a piecewise function. Each rule comes with an inequality that says which inputs it governs.

Each input belongs to exactly one piece

A letter that weighs exactly 100 g is in the first band, because "up to 100 g" includes 100. The second band says "over 100 g", which leaves 100 out. So P(100) = 1.

Suppose the second band were written "100 g and over" instead. Then a letter of exactly 100 g would be in both bands, and its price would be $1 and $2 at the same time. P(100) would have two outputs, and a function gives each input exactly one output. That is why the inequalities of a piecewise function are written so that every input in the domain lies in exactly one piece: at each boundary, one inequality includes the boundary value and the other leaves it out.

Find the piece, then use its rule

Here is a piecewise function with two rules: f(x) = 2x + 1 for x < 3, and f(x) = 10 − x for $x\ge 3$x ≥ 3. The number 3, where the rule changes, is the boundary.

To evaluate f, first decide which piece the input is in, then use that piece's rule and no other. For f(0): 0 < 3, so use 2x + 1, and f(0) = 2 × 0 + 1 = 1. The other rule would give 10 − 0 = 10, but 10 − x governs only the inputs from 3 upward, so 10 is not f(0).

For f(8): $8\ge 3$8 ≥ 3, so use 10 − x, and f(8) = 10 − 8 = 2. A negative input works the same way: −2 < 3, so f(−2) = 2 × (−2) + 1 = −3.

The graph of f. Left of x = 3 it is the line y = 2x + 1, which passes through (0, 1). From x = 3 on it is the line y = 10 − x, which passes through (8, 2). The two pieces meet at (3, 7).

The boundary value

For f(3), check each inequality. 3 < 3 is false, because 3 is not less than itself, so the first rule does not apply. $3\ge 3$3 ≥ 3 is true, because $\ge$≥ means "greater than or equal to". So the second rule answers: f(3) = 10 − 3 = 7. The boundary belongs to the rule whose inequality carries the "or equal to".

For this f, the first rule would also have given 2 × 3 + 1 = 7 at x = 3, so the two pieces happen to meet there. Usually they do not, and then the rule that owns the boundary decides the answer.

Here is a second function: f(x) = 2x + 1 for x < 1 and f(x) = 5 − x for $x\ge 1$x ≥ 1. Drag x along the axis, and the rule in charge lights up. At exactly x = 1 the $\ge$≥ gives the input to 5 − x, so f(1) = 4 and the filled dot is at (1, 4). The button moves the "or equal to" to the first rule, which then owns x = 1 and gives f(1) = 3. Here the two rules disagree at the boundary, so exactly one of them must own it.

Working backward from an output

Go back to f(x) = 2x + 1 for x < 3 and f(x) = 10 − x for $x\ge 3$x ≥ 3, and ask which inputs give the output 5. Try each rule, and keep a solution only if it lies in that rule's piece. The first rule: 2x + 1 = 5, so 2x = 4 and x = 2. It lies in its piece, because 2 < 3. The second rule: 10 − x = 5, so x = 5, and $5\ge 3$5 ≥ 3, so it counts too. Both inputs give 5: f(2) = 5 and f(5) = 5.

Now try f(x) = 9. The first rule gives 2x + 1 = 9, so x = 4, but 4 is not less than 3, so that rule does not apply at 4, and the solution is rejected. The second rule gives 10 − x = 9, so x = 1, but 1 is not 3 or more, so it is rejected too. No input gives 9. The graph shows why: its highest point is (3, 7).

The usual mistakes

Using the other rule. For f(0), 10 − 0 = 10 comes from the rule for $x\ge 3$x ≥ 3, and 0 is not in that piece. f(0) = 1.

Stopping partway through a rule. For f(2), 2 × 2 = 4 is only the multiply; the rule 2x + 1 then adds 1, so f(2) = 5.

Giving the boundary to the strict inequality. At x = 3, the rule for x < 3 does not apply, because 3 is not less than 3. The rule for $x\ge 3$x ≥ 3 owns the boundary.

Saying the boundary has no value. The $\ge$≥ includes 3, so f(3) has exactly one value, 7.

Bands on a graph

On a graph, each piece is drawn only over its own interval of inputs. At an end the piece includes, marked by $\le$≤ or $\ge$≥, the graph ends in a filled dot. At an end it does not include, marked by < or >, it ends in a hollow dot, which means that point is not on the graph.

For the postage price, the $1 band covers $0<w\le 100$0 < w ≤ 100, so it runs from a hollow dot at (0, 1), since a letter must weigh something, to a filled dot at (100, 1). The $2 band covers $100<w\le 250$100 < w ≤ 250, so it starts with a hollow dot at (100, 2) and ends with a filled dot at (250, 2). At w = 100 the only filled dot is at a height of 1, so P(100) = 1.

The postage price P(w) in dollars for a letter of w grams. The $1 band ends in a filled dot at (100, 1), and the $2 band starts with a hollow dot at (100, 2), because a letter of exactly 100 g is in the first band.