Piecewise Functions

Intuition

The rules that run everyday life rarely fit into a single formula. A parking garage might charge a flat $5$ dollars for a stay of one hour or less, then $8$ dollars for anything longer: one function, but a different formula depending on how long you stay. A function built that way, from several formulas each in charge of its own stretch of the domain, is a piecewise function. You already met three of them earlier in this chapter without the name: the absolute value, the floor, and the ceiling all switch rules at certain points. The last lesson’s hyperbola also broke into two separate branches, but for a different reason: one formula governed every allowed input, and the break was only a forbidden input falling out of the domain. A piecewise function splits on purpose, handing different inputs to different formulas by design. This lesson gives that pattern its name and settles the two skills that matter most, picking the right piece when you evaluate and drawing the graph with the correct open and closed dots.

Learning goals

Evaluate a piecewise function by choosing the piece whose condition holds
Graph each piece over its own interval with a closed or open dot at the boundary
Compare the two boundary heights to tell a jump from a join
Find the domain and range from each piece's restricted interval
Recognize absolute value, floor, and ceiling as piecewise functions from earlier lessons

What a piecewise function is

Take the parking garage from the introduction: $5$ dollars for a stay of one hour or less, $8$ dollars for anything longer. Call the charge $C(h)$ for a stay of $h$ hours. There are two rules, and each one owns its own share of the possible stays:

C(h) = \begin{cases} 5 & \text{if } 0 < h \le 1, \\ 8 & \text{if } h > 1. \end{cases}

Every value of $C$ is in dollars. To use it, decide which condition your stay satisfies, then read off that piece’s charge.

Stay $h$	Condition true	Charge $C(h)$
$0.5$ hour	$0 < h \le 1$	$5$
$1$ hour	$0 < h \le 1$	$5$
$2.5$ hours	$h > 1$	$8$

A piecewise function is defined by two or more formulas, called pieces, each paired with a condition that says which inputs it governs. The pieces are collected under one large brace, exactly as $C$ is above. A larger piecewise function works the same way with more pieces. Here is one with three:

f(x) = \begin{cases} x + 3 & \text{if } x < -1, \\ x^2 & \text{if } -1 \le x \le 2, \\ 8 - x & \text{if } x > 2. \end{cases}

Read the brace as a short list of instructions. Given an input $x$ , run down the conditions on the right, find the single one that $x$ satisfies, and apply the matching formula on the left. Nothing about the notation is mysterious; it is just a compact way to say “the value is whichever of these lines applies to your input.”

To evaluate a piecewise function at a number, always decide the condition first and compute second. Take $x = 3$ . Scanning the conditions, $3$ is not less than $-1$ , and it does not lie between $-1$ and $2$ . But $3$ is greater than $2$ , so the third piece owns it and $f(3) = 8 - 3 = 5$ . Take $x = 0$ instead. It satisfies $-1 \le 0 \le 2$ , the middle condition, so $f(0) = 0^2 = 0$ . Choosing the formula before substituting is the entire skill. Plugging $0$ into the first or third piece would produce a number the function never actually returns at $0$ .

The care is needed at a boundary, an input where one condition ends and the next begins. Look at $x = -1$ . The first condition asks for $x < -1$ , which $-1$ fails, and the second asks for $-1 \le x \le 2$ , which $-1$ passes thanks to the “or equal to.” So the boundary $x = -1$ belongs to the middle piece alone, and $f(-1) = (-1)^2 = 1$ . The inequality signs are not decoration. The strict $<$ on one side and the $\le$ on the other are exactly what hand the shared point to a single piece. Those two signs also keep the other piece from claiming that point.

Worked example 1 Evaluate a three-piece function

Use the function

f(x) = \begin{cases} x + 3 & \text{if } x < -1, \\ x^2 & \text{if } -1 \le x \le 2, \\ 8 - x & \text{if } x > 2, \end{cases}

and find $f(-4)$ , $f(-1)$ , $f(2)$ , and $f(5)$ . For each input, name the condition first.

The input $-4$ satisfies $x < -1$ , so the first piece applies and $f(-4) = -4 + 3 = -1$ .

The input $-1$ is a boundary. It fails $x < -1$ but satisfies $-1 \le x \le 2$ , so the middle piece owns it and $f(-1) = (-1)^2 = 1$ .

The input $2$ is the other boundary. It satisfies $-1 \le x \le 2$ (again by the “or equal to”), so the middle piece applies and $f(2) = 2^2 = 4$ .

The input $5$ satisfies $x > 2$ , so the third piece applies and $f(5) = 8 - 5 = 3$ .

f(-4) = -1, \quad f(-1) = 1, \quad f(2) = 4, \quad f(5) = 3.

Both boundaries went to the middle piece, because both of its inequalities carry the “or equal to.”

Check your understanding

Find $g(3)$ for the function

$g(x) = \begin{cases} 2x + 1 & \text{if } x < 3, \\ 4 - x & \text{if } x \ge 3. \end{cases}$

1

7

4

-1

Graphing a piecewise function

To graph a piecewise function, graph each piece on its own interval and nowhere else. For the running $f$ , you draw the line $y = x + 3$ only to the left of $-1$ , and the parabola $y = x^2$ only from $-1$ to $2$ . The line $y = 8 - x$ is drawn only to the right of $2$ . The one new question a piecewise graph raises is what to draw at the ends of each interval, and the inequality signs answer it.

At a boundary the piece that includes the endpoint (its condition uses $\le$ or $\ge$ ) genuinely reaches that point. Because that piece reaches the point, you mark its end with a closed dot, a filled circle. The piece that excludes the endpoint (its condition uses $<$ or $>$ ) approaches the point but never arrives. Because that piece never arrives, you mark its end with an open dot, a hollow circle. A filled dot says “this exact point is on the graph.” A hollow dot says “this piece stops just short of here, so its own boundary input is decided by another condition instead.”

Watch how that plays out on a simple two-piece function:

g(x) = \begin{cases} x + 1 & \text{if } x \le 1, \\ 4 & \text{if } x > 1. \end{cases}

The first piece is the line $y = x + 1$ drawn up to $x = 1$ . Its condition includes the boundary ( $x \le 1$ ), so it truly reaches the point $(1, 2)$ , drawn as a closed dot. The second piece is the constant $y = 4$ drawn for every input past $1$ . Its condition excludes the boundary ( $x > 1$ ), so at $x = 1$ it stops just short with an open dot at $(1, 4)$ . Between those two dots the graph leaps straight from height $2$ up to height $4$ without passing through the values in between. That leap is a jump.

A piecewise function that jumps. The line piece y equals x plus 1 runs up to x equals 1, where its condition includes the boundary, so it reaches the point (1, 2) with a closed (filled) dot. The constant piece y equals 4 governs every input past 1, so at x equals 1 it stops just short with an open (hollow) dot at (1, 4). The two pieces reach different heights at x equals 1, height 2 and height 4, so the graph jumps and cannot be drawn without lifting the pen.

Whether a piecewise graph jumps or connects at a boundary comes down to one comparison, as long as both pieces actually settle on a definite height there: do the two pieces reach the same height? Plug the boundary into both formulas and see. For $g$ , the left piece gives $1 + 1 = 2$ while the right piece gives $4$ . The heights differ, so the graph cannot be drawn without lifting the pen, and a jump appears. When the two pieces land on the same height instead, the closed dot and the open dot fall on the very same point. That shared point fills the gap, and the graph connects into an unbroken curve. No limits or calculus are needed for this; you simply evaluate both pieces at the boundary and compare the two numbers, provided both pieces actually have a number to give there.

That last condition is not automatic. Every piece in this lesson is a line, a constant, or a polynomial, and each of those has a definite value at every input, so the plug-in test above always applies here. It would not work for every possible piece. The last lesson’s rational functions can fail to have a height at a boundary in two different ways: a hole, where the formula is simply undefined at that one input, or a vertical asymptote, where the height runs off without settling on any number as the input gets close. Either way there is no finite height to plug in, so the jump-or-connect comparison above does not apply, and deciding what happens there takes the separate tools the last lesson built for exactly that purpose.

Compare $g$ with a function whose pieces agree at the boundary:

h(x) = \begin{cases} 2x & \text{if } x < 1, \\ x + 1 & \text{if } x \ge 1. \end{cases}

At $x = 1$ the first piece heads toward $2 \cdot 1 = 2$ and the second piece sits at $1 + 1 = 2$ . Same height. The excluded point from the left piece and the included point from the right piece are both $(1, 2)$ . Because they are the same point, they merge into one filled dot and the graph has no gap. It still turns a sharp corner, because the slopes differ ( $2$ on the left, $1$ on the right), but a corner is not a jump. You can trace the whole graph without lifting the pen.

A piecewise function that connects. The piece y equals 2x runs up to the boundary x equals 1, and the piece y equals x plus 1 begins there. Both pieces reach height 2 at x equals 1, so the hollow endpoint of the left piece lands on the filled endpoint of the right piece and they merge into a single closed dot at (1, 2). The graph has a corner because the slopes differ, but no jump, so it can be drawn without lifting the pen.

Worked example 2 Describe the graph at each boundary

Sketch the shape of

p(x) = \begin{cases} -2 & \text{if } x < 0, \\ x - 2 & \text{if } 0 \le x \le 3, \\ 3 & \text{if } x > 3, \end{cases}

reading off the dots and deciding jump or connect at each boundary.

At $x = 0$ , the left piece is the constant $-2$ , which excludes $0$ (its condition is $x < 0$ ), so it gives an open dot at $(0, -2)$ . The middle piece includes $0$ and gives $0 - 2 = -2$ , a closed dot at $(0, -2)$ . Both pieces reach height $-2$ , so the dots land on the same point and the graph connects at $x = 0$ .

At $x = 3$ , the middle piece includes $3$ and gives $3 - 2 = 1$ , a closed dot at $(3, 1)$ . The right piece is the constant $3$ , which excludes $3$ (its condition is $x > 3$ ), so it gives an open dot at $(3, 3)$ . The heights $1$ and $3$ differ, so the graph jumps at $x = 3$ .

\text{connects at } x = 0, \qquad \text{jumps at } x = 3.

So $p$ runs flat at height $-2$ , then rises along a line of slope $1$ from $(0, -2)$ up to the closed dot $(3, 1)$ . From there the graph leaps up to the open dot $(3, 3)$ and stays flat at height $3$ .

Check your understanding

At the boundary $x = 2$ , describe the graph of

$k(x) = \begin{cases} x^2 & \text{if } x \le 2, \\ 6 & \text{if } x > 2. \end{cases}$

A Closed dot at

(2, 4)

, open dot at

(2, 6)

; the graph jumps B Open dot at

(2, 4)

, closed dot at

(2, 6)

; the graph jumps C Closed dot at

(2, 4)

, open dot at

(2, 6)

; the graph connects D A single closed dot at

(2, 6)

; the graph connects

Domain and range

The domain of a piecewise function is the set of inputs it accepts, and it is the union of the intervals attached to the pieces. When the conditions partition the domain, as they should, those intervals fit together with no overlap and no gap. Because those intervals fit together that way, you can read the domain straight off the conditions. For $g$ , the conditions $x \le 1$ and $x > 1$ together account for every real number, so the domain of $g$ is all real numbers.

The range is the set of outputs, and you build it one piece at a time. Find the values each piece produces over its own interval, then take the union. A piece is only responsible for the part of its graph that actually gets drawn, so restrict each formula to its interval before reading off its outputs. For $g$ , the line $y = x + 1$ on $x \le 1$ produces every height up to and including $2$ (reached at the boundary $x = 1$ ). The constant piece contributes only the single height $4$ . The range of $g$ is therefore $y \le 2$ together with $y = 4$ , with everything strictly between $2$ and $4$ skipped over by the jump.

Worked example 3 Find the domain and range

Work out the domain and range of

r(x) = \begin{cases} 1 & \text{if } x < 0, \\ x + 2 & \text{if } 0 \le x \le 2, \\ 6 & \text{if } x > 2. \end{cases}

For the domain, list the intervals: $x < 0$ , then $0 \le x \le 2$ , then $x > 2$ . Together they cover every real number with no overlap, so the domain is all real numbers.

For the range, take the pieces one at a time. The first piece is the constant $1$ , contributing only the value $1$ . The middle piece $x + 2$ runs over the closed interval $0 \le x \le 2$ , so its outputs run from $0 + 2 = 2$ up to $2 + 2 = 4$ , giving every height in $2 \le y \le 4$ . The third piece is the constant $6$ , contributing only the value $6$ . Union the three:

\text{range: } y = 1, \quad \text{or} \quad 2 \le y \le 4, \quad \text{or} \quad y = 6.

The range is not a single interval. It is a stack of separate pieces. The heights between $1$ and $2$ , and between $4$ and $6$ , are never reached, because the two jumps skip straight over them.

Check your understanding

Find the range of the function

$q(x) = \begin{cases} x + 1 & \text{if } x \le 1, \\ 4 & \text{if } x > 1. \end{cases}$

y \le 2

y = 4

y \le 2

C all real numbers D

2 \le y \le 4

The special functions were piecewise all along

This lesson is a capstone for the chapter, because three functions you studied earlier in this chapter were piecewise the whole time. The clearest case is the absolute value. Its rule keeps a nonnegative input as is and flips the sign of a negative input, which is precisely a two-piece definition:

\lvert x \rvert = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0. \end{cases}

At the boundary $x = 0$ the top piece gives $0$ and the bottom piece heads toward $0$ as well, so the two pieces meet at the same height. That agreement is exactly why the V of the absolute value graph connects at its vertex instead of jumping. Same height at the boundary means no gap, just the sharp corner where the slopes switch from $-1$ to $+1$ .

Worked example 4 Write an absolute value as a piecewise function

Rewrite $\lvert x - 3 \rvert$ without the bars, as a piecewise function.

The bars keep their contents when the inside is zero or positive and negate the contents when the inside is negative, so split on the sign of $x - 3$ . The inside is nonnegative when $x - 3 \ge 0$ , that is $x \ge 3$ , and negative when $x < 3$ :

\lvert x - 3 \rvert = \begin{cases} x - 3 & \text{if } x \ge 3, \\ -(x - 3) & \text{if } x < 3. \end{cases}

Tidy the lower piece by distributing the sign, $-(x - 3) = 3 - x$ , so the function reads $x - 3$ for $x \ge 3$ and $3 - x$ for $x < 3$ . Check the boundary: at $x = 3$ the top piece gives $0$ and the bottom formula also heads toward $3 - 3 = 0$ . The two pieces therefore meet, so the graph is the usual V with its corner at $(3, 0)$ .

The floor and ceiling are piecewise too, only with infinitely many pieces. The floor holds the constant value $n$ on each interval $n \le x < n + 1$ . So writing one line per integer gives a piecewise definition with a separate constant piece for every whole-number step. Unlike the absolute value, neighboring floor pieces reach different heights at each integer boundary, one unit apart, so the floor jumps by one at every integer. That is why its graph is a staircase of separate steps rather than a connected curve. A piecewise function with constant pieces like this, whose graph is a row of flat segments, is called a step function. The floor and ceiling are the headline examples.

The ceiling has the same step shape, built from its own pieces. It holds the constant value $n$ on the interval $n - 1 < x \le n$ , the mirror image of the floor’s interval: instead of owning its left endpoint, each ceiling piece owns its right endpoint. That is why the ceiling’s steps land one row higher than the floor’s on most of each interval, and why the two functions agree only exactly at the integers.

Check your understanding

Why is the floor function a piecewise function?

A It holds one constant value on each interval between consecutive integers, with a different constant on every interval B It has exactly two pieces, one for zero or positive inputs and one for negative inputs C It is one smooth formula whose graph happens to look like steps D Its neighboring pieces always meet at the same height, so the graph never jumps

Check your understanding

The floor holds its constant value $n$ on $n \le x < n + 1$ , owning its left endpoint. On which interval does the ceiling hold the constant value $n$ ?

n - 1 < x \le n

, owning its right endpoint B

n \le x < n + 1

, the same interval as the floor C

n - 1 \le x < n

, owning its left endpoint D

n < x < n + 1

, owning neither endpoint

Piecewise models in the real world

Piecewise functions are the natural language for any rule that changes at thresholds, which is most real pricing. Postage climbs in jumps as a package crosses weight limits. A parking garage bills one amount for the first hour and another after that. A phone plan charges a flat monthly fee up to some data cap, then a per-gigabyte rate beyond it. Income tax applies one rate to the first band of income and steeper rates to higher bands. In every case a single function describes the cost, and its formula depends on which band the input falls into.

Worked example 5 A shipping cost that jumps

A store sets its shipping charge by the order weight $w$ , in pounds:

C(w) = \begin{cases} 5 & \text{if } 0 < w \le 2, \\ 8 & \text{if } 2 < w \le 5, \\ 12 & \text{if } w > 5. \end{cases}

Every value is in dollars. Find the charge on orders weighing exactly $2$ pounds, $3$ pounds, and $5.5$ pounds, and describe the graph.

An order of exactly $2$ pounds sits on a boundary. The first condition $0 < w \le 2$ includes $2$ because of the “or equal to,” so the first tier owns it and the charge is $5$ dollars. An order of $3$ pounds satisfies $2 < w \le 5$ , so it costs $8$ dollars. An order of $5.5$ pounds satisfies $w > 5$ , so it costs $12$ dollars.

C(2) = 5, \qquad C(3) = 8, \qquad C(5.5) = 12.

Because each tier is a constant, the graph is flat within a tier and jumps up at each boundary, exactly a step function like the floor. The boundary weights $2$ and $5$ each belong to the lower tier, which is why a package right at the limit still ships at the cheaper price. Reading the inequality the other way, so that $2$ pounds cost $8$ dollars, would overcharge every order that lands exactly on a limit.

Worked example 6 Build a piecewise fare from a description

A taxi charges a flat $3$ dollars for any trip up to and including one mile, then $2$ dollars for each mile beyond the first. Write the fare $F(d)$ as a function of the distance $d$ in miles (with $d > 0$ ), and find $F(1)$ and $F(4)$ .

Up to one mile the fare is the flat charge, $3$ dollars, so the first piece is the constant $3$ on $0 < d \le 1$ . Past one mile, you still pay the $3$ dollars plus $2$ dollars for each of the $d - 1$ extra miles, so the second piece is $3 + 2(d - 1)$ on $d > 1$ :

F(d) = \begin{cases} 3 & \text{if } 0 < d \le 1, \\ 3 + 2(d - 1) & \text{if } d > 1. \end{cases}

Evaluate. A one-mile trip is on the boundary, which the first piece includes, so $F(1) = 3$ dollars. A four-mile trip uses the second piece, $F(4) = 3 + 2(4 - 1) = 3 + 6 = 9$ dollars.

F(1) = 3, \qquad F(4) = 9.

Check the boundary for a jump. At $d = 1$ the first piece gives $3$ , and the second formula heads toward $3 + 2(1 - 1) = 3$ as well, the same height, so this fare connects with no sudden leap at one mile. The graph still turns a corner there, since the rate changes from flat to rising, but a corner is not a jump. That matches the design, since the extra-mile rate only starts adding once you pass the first mile.

Common mistakes

Substituting before checking the condition. Always find which condition the input satisfies first, then use that piece. Plugging $x = 0$ into a piece meant for $x > 2$ produces a value the function never returns at $0$ .
Getting boundary ownership backward. At a boundary, the piece that owns the point is the one whose condition carries the “or equal to.” For $x \le 1$ against $x > 1$ , the input $x = 1$ belongs to the first piece, not the second.
Overlapping or gapped conditions. Writing $x \le 1$ together with $x \ge 1$ claims $x = 1$ twice; if the two formulas disagree there, the rule offers two outputs, which is not a function. Writing $x < 1$ together with $x > 1$ leaves $x = 1$ with no piece at all, undefined there even though $1$ was meant to be an input. The safest habit, and the one this lesson always uses, is conditions that partition the domain, one piece per input.
Open and closed dots reversed. The closed (filled) dot goes on the piece that includes the boundary ( $\le$ or $\ge$ ). The open (hollow) dot goes on the piece that excludes it ( $<$ or $>$ ). The filled dot always sits where the function genuinely equals that value.
Drawing a vertical segment across a jump. A jump is a gap, shown by an open dot and a closed dot at different heights, not a vertical line connecting them. The function never takes the in-between values.
Assuming a piecewise function always jumps. It jumps only when the two pieces reach different heights at the boundary. When they reach the same height, the graph connects; compare the two values, do not guess (as long as both pieces actually have a height there, see the note above about rational pieces).
Reading the range from the whole formulas. Restrict each piece to its own interval before collecting outputs. The line $y = x + 1$ has range all real numbers on its own, but on $x \le 1$ it only reaches $y \le 2$ .

Key takeaways

A piecewise function is defined by several formulas (pieces), each with a condition on $x$ , collected under a brace with $\begin{cases}\end{cases}$ . A well-formed piecewise function’s conditions partition the domain, so exactly one piece owns each input.
To evaluate $f(a)$ , first decide which condition $a$ satisfies, then apply that piece. At a boundary, the inequality signs ( $\le$ or $\ge$ versus $<$ or $>$ ) decide which piece claims the shared point.
To graph, draw each piece over only its own interval. Put a closed dot where a piece includes the boundary and an open dot where it excludes it.
The graph jumps at a boundary when the two pieces reach different heights there, and connects when they reach the same height. Compare the two values; no limits are needed, just arithmetic, as long as both pieces actually reach a definite height there, true for every line, constant, and polynomial piece in this lesson. A rational piece can instead be undefined at the boundary (a hole) or run off without bound (an asymptote), and then this comparison does not apply.
The domain is the union of the pieces’ intervals. The range is the union of the outputs each piece produces once restricted to its interval, which can leave gaps.
The absolute value is piecewise ( $x$ for $x \ge 0$ , $-x$ for $x < 0$ ). The floor and ceiling are step functions, piecewise with a constant piece on every integer interval, jumping by one at each integer.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Absolute value and piecewise functions (Khan Academy)
Piecewise Functions (Math is Fun)
Piecewise Function (Wolfram MathWorld)
Piecewise-defined functions (free textbook) (OpenStax)
Piecewise function (Wikipedia)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Why each input must belong to exactly one piece

Why each input must belong to exactly one piece#

A function is a rule that assigns to every input in its domain exactly one output. A piecewise definition can only be a genuine function when its conditions do two things. Those conditions must never assign one input to two conflicting pieces, and they must never leave an intended input with no piece at all.

Suppose some input $a$ satisfied two conditions whose formulas disagree, giving one piece the value $p(a)$ and another the value $q(a)$ with $p(a) \ne q(a)$ . Then the rule would offer two different outputs at the same input, and a rule that returns two values is not a function. So the conditions must never overlap in a way that hands a single input two different results. The safe design, and the one this lesson always uses, is to make the conditions mutually exclusive, so no input satisfies more than one; overlap only becomes an error when it hands the same input two disagreeing outputs.

Now suppose instead that some input we intend to feed in satisfied no condition at all. Then the rule would give it no output, and the function would simply be undefined there. So the conditions must also cover every input in the intended domain. Building a piecewise definition that avoids both problems, the way every example in this lesson does, means its conditions partition the domain: each input lands in one piece and only one.

This is the whole reason for the careful inequality signs at a boundary. Where two pieces meet at a value $b$ , exactly one of them must claim $b$ . Writing one condition as $x \le b$ and the next as $x > b$ hands $b$ to the first piece and to no other. Writing them as $x \le b$ and $x \ge b$ would be an error, because then $b$ satisfies both and the function risks two values. Writing them as $x < b$ and $x > b$ would be an error too, because then $b$ satisfies neither and the function has a gap.

A bit of history (optional)

Pull a guitar string aside with one finger and hold it there. Look at the shape just before you let go. It is not a curve at all. It is two straight segments meeting at a sharp corner, one rule to the left of your finger and a different rule to the right.

In 1747 the French mathematician Jean d’Alembert wrote down an equation governing how such a string moves. That should have settled the matter. Instead it opened an argument that ran for fifty years, and the argument was about the shape your finger had just made.

Was a bent string even allowed as a starting position? For d’Alembert it was not. A function meant a single formula, and a plucked string is two formulas taped together at a crease. He wanted the starting shape produced by one rule, smoothly, with no corner in it anywhere. His opponents made the obvious objection. Anybody can pluck a string, so the mathematics had better cope.

The pluckers won, and the victory cost mathematics its old idea of what a function is. A rule was allowed to be assembled from pieces, each governing its own territory. The bent string is a piecewise function, drawn exactly as you have been drawing them. Its two pieces meet at the same height, so the graph connects rather than jumping. That is the test you applied at every boundary here.