Continuous graphs, jumps and holes
Some graphs can be traced from end to end without lifting the pencil, and some cannot. What exactly happens at the places where the pencil has to leave the paper?
Where you are 10 in Features of a graph
Why this exists
Fill a bath and draw the depth of the water against time. The depth rises smoothly: it never skips from 10 centimeters to 12 without passing through every depth in between, and the drawn graph is one unbroken stroke. Now drop coins into a savings jar and draw the money in it against time. The amount sits still, then leaps by the value of a coin at the instant the coin lands. The graph is a row of flat pieces with gaps between them, and no pencil can draw it without lifting.
Both are graphs of functions, yet they behave very differently. The unbroken one lets you say “at some moment the depth was exactly 11 centimeters” without ever measuring it. The broken one forbids that: the jar may never hold certain amounts at all. To know which kind of statement a graph allows, you need to know where it breaks and how. Formulas break too, and the breaks are not always visible: the graph of \displaystyle f(x) = \frac{x^2 - x - 20}{x - 5} looks exactly like a straight line, but one single point of that line is missing. This lesson names the ways a graph can break and shows how to find each one from a picture or a formula.
Definition
In plain words, a graph is continuous over an interval of inputs when you can trace the whole of it over that interval without lifting your pencil. Where the pencil has to leave the paper, the graph is broken, and the function is discontinuous at that input.
To say this without a pencil, watch the heights the graph heads toward. Write x \to a (“x approaches a”) for inputs taking values closer and closer to a without equaling it, x \to a^- when they come from the left (values below a) and x \to a^+ when they come from the right. Write f(x) \to h when the outputs get as close to h as you like.
A function f is continuous at an input a when both of these hold:
- f(a) is defined, so the graph has a point above a, and
- as x \to a from both sides, f(x) \to f(a), so the graph heads straight into that very point.
At an end of an interval, only the side inside the interval is watched. A function is continuous over an interval when it is continuous at every input of it.
When this fails at an input a, the graph breaks there in one of three ways, drawn below.
- A hole: from both sides the graph heads toward the same point (a, h), but that point is not on the graph. Either f(a) is not defined at all, or f(a) is some other height, drawn as a lone dot away from the hole. A hole is also called a removable discontinuity.
- A jump: from the left the graph heads toward one height and from the right toward a different height. A jump is also called a jump discontinuity.
- A break at a vertical asymptote: on at least one side the outputs grow without bound, and the graph runs up or down along a vertical line. This kind is the subject of Vertical asymptotes.
An open dot marks a point the graph heads toward but does not include. A closed dot marks a point that is on the graph. A hole is drawn as an open dot; a jump is drawn as an open dot and a closed dot one above the other, or two open dots.
Intuition
Picture the pencil moving left to right along the graph. Over most inputs a small step in x moves the pencil a small distance, and the stroke stays on the paper. A break is a place where that fails.
At a hole the pencil arrives at a spot and finds nothing there. It can hop over the missing point and carry on at the same height, so the gap is a single point wide and the graph on each side lines up exactly. That is why a hole can be invisible when a computer plots the graph: one missing point among infinitely many takes up no width on the screen. It is also why a hole is called removable. Put one point back at the right height and the graph is whole.
At a jump the pencil arrives at one height and has to restart at another. No single point can repair that, because the two sides do not line up. The graph of a piecewise function is where jumps usually appear, at a boundary between two rules.
At a vertical asymptote the pencil is carried off the top or bottom of the page, and there is no height to arrive at on that side at all.
A sharp corner, as at the bottom of a V shape, is not a break. The pencil changes direction there but never leaves the paper.
Results
Polynomials never break. The graph of a polynomial, such as y = x^3 - 4x + 1, is continuous over every interval: it has no holes, no jumps and no asymptotes.
Where a rational function breaks. Let \displaystyle f(x) = \frac{P(x)}{Q(x)} with P and Q polynomials. Its graph is unbroken except at the zeros of Q. Cancel every factor common to P and Q, and look at a zero a of the original Q:
- if the canceled denominator is no longer zero at a, the graph has a hole at a, at the height the canceled formula gives at a;
- if the canceled denominator is still zero at a, the graph has a vertical asymptote at x = a.
Where a piecewise function breaks. Suppose each rule of a piecewise function is unbroken on its own interval. At a boundary a, let L be the height the left rule heads toward and R the height the right rule heads toward. The graph is unbroken at a exactly when L = R = f(a). If L \ne R, the graph jumps at a by |L - R|.
Holes can be filled, jumps cannot. A hole at (a, h) is removed by defining (or redefining) f(a) = h. No choice of the single value f(a) removes a jump.
An unbroken graph takes every height in between. If f is continuous over [a, b], then every height between f(a) and f(b) is the output of f at some input between a and b. This is the intermediate value theorem. A broken graph can skip heights.
Why it is true
Holes and asymptotes in a rational function. Suppose Q(a) = 0 and (x - a) cancels from P and Q, leaving \displaystyle \frac{p(x)}{q(x)} with q(a) \ne 0. For every x \ne a the factor x - a is not zero, so dividing it out is allowed, and \displaystyle f(x) = \frac{p(x)}{q(x)}. The right side is a quotient whose denominator is not zero near a, so its graph passes smoothly through x = a at height \displaystyle \frac{p(a)}{q(a)}. The graph of f agrees with it at every input except a, where f itself is not defined, because putting a into the original formula gives \displaystyle \frac{0}{0}. So the graph of f is the smooth graph with exactly one point removed: a hole.
If instead the canceled denominator is still zero at a, then for x near a the canceled denominator is a tiny number while the numerator is close to a number that is not zero (if the numerator were zero at a too, (x - a) would be a common factor and would have been canceled). A number not near zero divided by a tiny number is huge in size, so the outputs grow without bound: an asymptote.
Jumps. When L \ne R, inputs just left of a give heights near L and inputs just right of a give heights near R. A pencil tracing the graph would have to cover the vertical distance |L - R| while moving no distance across, which is a lift of the pencil. A value f(a) can equal L or R or neither, but not both, so at least one side still fails to head into the point (a, f(a)).
Polynomials. A polynomial is built from x and constants by adding and multiplying. Change x by a tiny amount and each power of x changes by a tiny amount, so the sum changes by a tiny amount: there is never a sudden leap, and since every real number is an allowed input there is never a missing point. The full proof makes “tiny” precise with limits; see Continuity.
Every height in between. Draw a horizontal line at a height k between f(a) and f(b). The pencil starts on one side of the line and finishes on the other, and it never leaves the paper, so somewhere it must cross the line. The input below that crossing has output k. A rigorous proof needs the completeness of the real numbers and is met in calculus.
Worked examples
Reading the breaks off a graph
The graph of a function g is drawn below. Find every input where the graph is broken, say what kind of break each one is, and give the value of g at each of those inputs.
Trace the graph from left to right and note each place the pencil must lift.
The first piece lies on the line y = x + 3. Near x = -2 it heads toward the height -2 + 3 = 1 from both sides, and the open dot at (-2, 1) says that point is not on the graph. Instead there is a closed dot at (-2, 4), so g(-2) = 4. The two sides line up, but the point they head into is missing: this is a hole at (-2, 1), with the function’s value placed somewhere else. Redefining g(-2) as 1 would remove it.
The first piece ends at an open dot at (1, 4), since 1 + 3 = 4. The next piece starts at a closed dot at (1, 2), so g(1) = 2. From the left the graph heads toward 4 and from the right toward 2. That is a jump at x = 1, down by 4 - 2 = 2.
At (4, -1) the graph changes from falling to rising. The falling piece y = 3 - x reaches 3 - 4 = -1 there, and the rising piece y = 2x - 9 starts from 2(4) - 9 = -1, the same point. The pencil turns but stays on the paper, so this corner is not a break.
The graph starts at x = -5 and stops at x = 6. These are the ends of the domain, not breaks: there is no graph beyond them to join. So the graph is broken at exactly two inputs, x = -2 (a hole, g(-2) = 4) and x = 1 (a jump, g(1) = 2).
A hole hidden in a formula
Find where the graph of \displaystyle f(x) = \frac{x^2 - x - 20}{x - 5} is broken, and describe the graph.
SolutionA fraction is undefined where its denominator is zero, so the only candidate is x = 5. Putting 5 in gives \displaystyle \frac{25 - 5 - 20}{0} = \frac{0}{0}, which is not a number, so 5 is not in the domain (see The implied domain of a formula).
To see what the graph does near 5, factor the numerator. Two numbers with product -20 and sum -1 are -5 and 4, so x^2 - x - 20 = (x - 5)(x + 4). For every x \ne 5 the factor x - 5 is not zero, so it may be divided out:
So the graph is the line y = x + 4 at every input except 5. The line would pass through (5, 9), and that one point is missing. A table from the original formula agrees:
| x | f(x) |
|---|---|
| 4.9 | 8.9 |
| 4.99 | 8.99 |
| 5 | undefined |
| 5.01 | 9.01 |
| 5.1 | 9.1 |
As x \to 5 from either side, f(x) \to 9, but f(5) does not exist. The graph has a hole at (5, 9) and no other break.
One consequence is easy to miss. Since x takes every value except 5, the height x + 4 takes every value except 9. The range of f is every real number except 9, so the hole shows in the range as well as the domain.
A jump at a boundary
Describe the graph of
near x = 3, and say whether it is broken there.
SolutionEach rule is a line, so each piece is unbroken on its own. The only place a break can happen is the boundary x = 3, where the rules change (see Graphing a piecewise function).
From the left, p follows 2x - 1, which heads toward 2(3) - 1 = 5. The input 3 itself is not allowed in that rule, since its condition is x < 3, so the left piece ends at an open dot at (3, 5).
From the right, and at 3 itself, p follows 11 - 3x. So p(3) = 11 - 9 = 2, a closed dot at (3, 2).
The two sides head toward different heights, 5 and 2, so the graph has a jump at x = 3, down by 5 - 2 = 3. Inputs close to 3 confirm it: p(2.9) = 4.8 while p(3.1) = 1.7.
Choosing a value so the pieces join
For which value of k is the graph of
unbroken? What happens when k = -1?
SolutionThe left rule is a polynomial and the right rule is a line, so each piece is unbroken; only the boundary x = 2 matters.
The left rule includes 2, so q(2) = 4 - 2^2 = 0, a closed dot at (2, 0). The right rule heads toward 2k + 6 as x \to 2^+. The graph is unbroken at 2 exactly when the right side heads into the same point:
With k = -3 the right piece is 6 - 3x, which starts from 6 - 6 = 0 at the closed dot, and the two pieces meet.
With k = -1 the right piece is 6 - x, which heads toward 6 - 2 = 4. The left side arrives at 0 and the right side at 4, so the graph jumps up by 4 at x = 2. No other value of k joins the pieces, because 2k + 6 = 0 has the single solution k = -3.
A hole and an asymptote in the same graph
Find every break in the graph of \displaystyle r(x) = \frac{x^2 - 9}{x^2 - x - 6}, and say which kind each one is.
SolutionBreaks can only occur where the denominator is zero. Factor it: two numbers with product -6 and sum -1 are -3 and 2, so
which is zero at x = 3 and x = -2. Factor the numerator as a difference of squares, x^2 - 9 = (x - 3)(x + 3). The factor x - 3 is common, and for x \ne 3 it is not zero, so it may be canceled:
At x = 3 the canceled denominator is 3 + 2 = 5, not zero. So the graph has a hole there, at the height the canceled formula gives: \displaystyle \frac{3 + 3}{3 + 2} = \frac{6}{5}. The hole is at \displaystyle \left(3, \frac{6}{5}\right).
At x = -2 the canceled denominator is still zero, while the numerator is -2 + 3 = 1. Near -2 the outputs are 1 divided by a tiny number: \displaystyle r(-2.01) = \frac{0.99}{-0.01} = -99 and \displaystyle r(-1.99) = \frac{1.01}{0.01} = 101. The outputs grow without bound, so the graph breaks at a vertical asymptote, the line x = -2.
Both inputs made the original denominator zero, yet they break the graph in different ways. Canceling is what tells them apart.
Heights a broken graph skips
Thandeka rents a locker at a station. It costs 4 dollars for each hour or part of an hour, for up to 3 hours. Draw the charge C against the time t in hours, find where the graph breaks, and use it to answer: what does she pay for 100 minutes, for exactly 2 hours, and for 2 hours and 1 minute? Is there a stay that costs 6 dollars?
SolutionAny stay up to and including 1 hour counts as one hour, any stay over 1 hour up to 2 hours counts as two, and so on:
Each piece is flat. At t = 1 the charge is still 4 (exactly one hour is one hour), so the first piece ends at a closed dot at (1, 4), and the next piece starts at an open dot at (1, 8). The same happens at t = 2. The graph jumps up by 4 at t = 1 and at t = 2, and is unbroken everywhere else.
100 minutes is \displaystyle \frac{100}{60} = \frac{5}{3} hours, between 1 and 2, so it costs \displaystyle C\!\left(\frac{5}{3}\right) = 8 dollars. Exactly 2 hours is in the second piece, since its condition is 1 < t \le 2, so it also costs 8 dollars. Two hours and one minute is just past 2, so it costs 12 dollars.
The charge goes from 4 to 8, yet 6 is never charged: the only outputs are 4, 8 and 12. A broken graph can skip a height. An unbroken graph cannot. If the bath in the opening was 10 centimeters deep at one moment and 12 at a later one, then at some moment in between it was exactly 11, by the intermediate value theorem in the Results. Functions like C, made of flat steps, are studied as step functions.
Boundaries
- The pencil test is a picture, not the definition. For graphs built from polynomials, fractions and pieces, as here, the two agree. Some functions wiggle infinitely often near a point, and a drawing cannot settle whether they break there; the definition with “f(x) \to f(a)”, made exact with limits in Continuity, decides.
- Continuous at every input of the domain does not mean the graph is one stroke. The function \displaystyle y = \frac{1}{x} is continuous at every input it accepts, yet its graph comes in two pieces, because 0 is not in its domain. Whether “the function is continuous” means “at every allowed input” or “over every interval” must be said.
- The ends of a domain are not breaks. The graph of \displaystyle y = \sqrt{x} starts at (0, 0) and there is nothing to its left to join. Likewise a graph drawn only over -5 \le x \le 6 simply stops at its ends.
- A corner is not a break. The graph of y = |x| turns sharply at the origin and is still continuous there.
- The original formula cannot give a hole’s height. At the hole of \displaystyle f(x) = \frac{x^2 - x - 20}{x - 5}, putting 5 in gives \displaystyle \frac{0}{0}, which says nothing. The canceled formula x + 4 gives the height 9, and it equals f(x) only for x \ne 5.
- Cancel completely before deciding. In \displaystyle \frac{x - 2}{(x - 2)^2} one factor of x - 2 cancels but one remains in the denominator, so x = 2 is an asymptote, not a hole.
- Dots are not a graph to be joined. A function defined only at whole-number inputs, such as the number of coins in a jar after each coin, has a graph of separate dots. Joining them adds points the function does not have.
Where this goes next
The third kind of break, where the outputs run off without bound, is the subject of Vertical asymptotes, and the rule for telling it apart from a hole is the first thing done in graphing a rational function, after finding its domain and zeros. The canceling itself is simplifying a rational expression. In calculus, “the outputs head toward h” becomes the limit, and continuity becomes the exact statement that the limit equals the value; see Continuity. The intermediate value theorem then becomes a tool: if a continuous function is negative at one input and positive at another, it has a zero between them, which is how the The intermediate value theorem hunts for roots.
Exercises
Reading breaks off a graph
Use Graphs A, B and C. Each is drawn only between x = -4 and x = 4.
- Where, if anywhere, is Graph A broken? Name the kind of break and say whether the function has a value there.
- Where is Graph B broken, and what kind of break is it? By how much does the height change? Is there a value for the function at that input that would make the graph unbroken?
- Is Graph C broken anywhere? Explain what happens at (0, -1).
Holes from a formula
For each function, find the break and describe the graph.
- \displaystyle f(x) = \frac{x^2 - 4x - 12}{x - 6}
- \displaystyle g(x) = \frac{x^2 + 7x + 10}{x + 2}
- \displaystyle h(x) = \frac{3x - 12}{x - 4}
- \displaystyle k(x) = \frac{2x^2 - 5x - 3}{x - 3}
- \displaystyle m(x) = \frac{x^3 - x}{x}
Hole or asymptote
Find every break in each graph, and say whether it is a hole (give the point) or a vertical asymptote (give the line).
- \displaystyle y = \frac{x + 1}{x^2 - 1}
- \displaystyle y = \frac{x - 2}{x^2 - 4x + 4}
- \displaystyle y = \frac{x^2 - 16}{x^2 + 4}
- \displaystyle y = \frac{x^2 + x - 2}{x^2 + 5x + 6}
Pieces that join or jump
- Let h(x) = x + 4 for x < 1 and h(x) = 3x + 2 for x \ge 1. Is the graph of h broken at x = 1?
- Let m(x) = x^2 for x \le -1 and m(x) = 2x + 5 for x > -1. Describe the graph at x = -1: which dots are open and closed, and is there a jump? If so, by how much?
- Let s(x) = kx - 1 for x < 2 and s(x) = x^2 + 3 for x \ge 2. Find the value of k that makes the graph unbroken.
- Let \displaystyle w(x) = \frac{x^2 - 1}{x - 1} for x \ne 1, and w(1) = c. Find the value of c that makes the graph of w unbroken.
- Kwabena sends parcels. A parcel up to and including 2 kilograms costs 5 dollars, over 2 up to 5 kilograms costs 9 dollars, and over 5 up to 10 kilograms costs 14 dollars. Where does the graph of the cost against the weight jump, and by how much each time? What does a parcel of exactly 5 kilograms cost? Is there a weight that costs 7 dollars?
Reasoning
- Are \displaystyle f(x) = \frac{x^2 - x - 20}{x - 5} and g(x) = x + 4 the same function? Explain, using their domains, their graphs and their ranges.
- Explain why no choice of the single value f(a) can remove a jump at x = a, while one choice always removes a hole.
- Write a formula for a function whose graph has a hole at (2, 5) and no other break. Explain why it works.
- The function \displaystyle y = \frac{1}{x} is continuous at every input in its domain. Can its graph be traced from x = -1 to x = 1 without lifting the pencil? Explain why these two facts do not contradict each other.
- Explain why the graph of a polynomial can have neither a hole nor a jump.
- A drink is at 70 degrees when it is poured and at 25 degrees thirty minutes later, and its temperature is a continuous function of time. Must it have been at exactly 40 degrees at some moment? Must it have been at 80 degrees at some moment? Explain.
Answers
- A hole at (2, 2). The line heads toward height 2 from both sides, but the open dot shows the point is missing and there is no other dot above x = 2, so the function has no value at 2. Elsewhere Graph A is unbroken.
- A jump at x = 0. From the left the graph heads toward 3 (the open dot at (0, 3)); the closed dot shows the value at 0 is -2, and the right piece starts there. The height drops by 3 - (-2) = 5. No value at 0 can mend it: the left side heads toward 3 and the right side toward -2, and a single point cannot be at both heights.
- Graph C is unbroken over the whole of -4 \le x \le 4. At (0, -1) the graph changes from falling to rising, a corner, but both sides reach the same point and the pencil never lifts. Its two ends are the ends of the drawn interval, not breaks.
- A hole at (6, 8). The numerator is (x - 6)(x + 2), so f(x) = x + 2 for x \ne 6. The graph is the line y = x + 2 with the point at height 6 + 2 = 8 removed.
- A hole at (-2, 3). The numerator is (x + 2)(x + 5), so g(x) = x + 5 for x \ne -2; the missing height is -2 + 5 = 3.
- A hole at (4, 3). The numerator is 3(x - 4), so h(x) = 3 for x \ne 4. The graph is the horizontal line y = 3 with one point missing.
- A hole at (3, 7). The numerator is (2x + 1)(x - 3), so k(x) = 2x + 1 for x \ne 3; the missing height is 2(3) + 1 = 7.
- A hole at (0, -1). The numerator is x(x^2 - 1), so m(x) = x^2 - 1 for x \ne 0. The graph is the parabola y = x^2 - 1 with its lowest point, the vertex, missing.
- A hole at \displaystyle \left(-1, -\frac{1}{2}\right) and an asymptote x = 1. The denominator is (x - 1)(x + 1); canceling x + 1 leaves \displaystyle \frac{1}{x - 1}. At -1 that gives \displaystyle \frac{1}{-2} = -\frac{1}{2}; at 1 the canceled denominator is still zero.
- No hole; an asymptote x = 2. The denominator is (x - 2)^2, so canceling one factor leaves \displaystyle \frac{1}{x - 2}, whose denominator is still zero at 2.
- No break at all. The denominator x^2 + 4 is at least 4 for every x, so it is never zero and the graph is unbroken everywhere. The zeros of the numerator, x = 4 and x = -4, are where the graph crosses the horizontal axis, not breaks.
- A hole at (-2, -3) and an asymptote x = -3. Numerator (x + 2)(x - 1), denominator (x + 2)(x + 3); canceling leaves \displaystyle \frac{x - 1}{x + 3}. At -2 that gives \displaystyle \frac{-3}{1} = -3; at -3 the canceled denominator is still zero.
- Not broken. The left rule heads toward 1 + 4 = 5 and h(1) = 3(1) + 2 = 5, so both sides meet at the closed dot (1, 5).
- A closed dot at (-1, 1), since m(-1) = (-1)^2 = 1, and an open dot at (-1, 3), where 2(-1) + 5 = 3. The graph jumps up by 3 - 1 = 2.
- k = 4. The right rule gives s(2) = 4 + 3 = 7 and the left rule heads toward 2k - 1, so 2k - 1 = 7.
- c = 2. For x \ne 1, \displaystyle w(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1, which heads toward 2; setting w(1) = 2 fills the hole at (1, 2).
- Jumps at 2 kilograms (up by 9 - 5 = 4 dollars) and at 5 kilograms (up by 14 - 9 = 5 dollars). A parcel of exactly 5 kilograms costs 9 dollars, since 5 is in “over 2 up to 5”. No weight costs 7 dollars: the only costs are 5, 9 and 14.
- No. Their formulas agree for every x \ne 5, but 5 is in the domain of g (with g(5) = 9) and not in the domain of f. The graph of f is the line y = x + 4 with a hole at (5, 9); the graph of g is the whole line. The range of g is every real number, while the range of f is every real number except 9. Two functions are the same only when their domains and their outputs both agree.
- At a jump the left side heads toward a height L and the right side toward a different height R. For the graph to be unbroken at a, both sides must head into the point (a, f(a)), so f(a) would have to equal L and R at once, which is impossible when L \ne R. At a hole both sides head toward the same height h, so setting f(a) = h satisfies both sides at once.
- One answer: \displaystyle y = \frac{(x - 2)(x + 3)}{x - 2}. The only zero of the denominator is 2. For x \ne 2 the formula equals x + 3, a line with no break, and at x = 2 that line would be at height 5, which is the missing point. Any formula \displaystyle \frac{(x - 2)\,p(x)}{x - 2} with p a polynomial and p(2) = 5 works the same way.
- No. The graph has a branch for negative inputs and a branch for positive inputs, and near 0 the outputs grow without bound, so the pencil must lift. There is no contradiction because 0 is not in the domain: “continuous at every input in the domain” says nothing about 0, while tracing from -1 to 1 asks about every input in between, 0 included.
- A hole needs an input the formula cannot accept, and a polynomial accepts every real number, since it only adds and multiplies. A jump needs the outputs to leap while the input barely moves, but changing x by a tiny amount changes each power of x, and so the whole sum, by a tiny amount. With no gap in the domain and no sudden leap, the graph cannot break.
- Yes for 40: it is between 25 and 70, and an unbroken graph from height 70 to height 25 takes every height in between, by the intermediate value theorem. Not necessarily for 80: the theorem only promises heights between the two measured ones. The drink could have warmed above 70 in between, but nothing forces it to.