Increasing, decreasing, constant, positive and negative intervals
A graph climbs in some places and falls in others, and it sits above the axis in some places and below it in others. How do you say exactly where?
Where you are 10 in Features of a graph
Why this exists
A graph tells a story (Graphs that tell a story): the river rose, held steady, then fell. To use that story you need to say exactly when. “It rose for a while” cannot be checked; “it rose from day 0 to day 3” can. This lesson names four kinds of stretch, where the outputs go up, go down, stay put, and where they sit above or below zero, and states each one as an interval of inputs.
Definition
Let f be a function and I an interval of inputs inside its domain.
- f is increasing on I if for any two inputs a < b in I, f(a) < f(b). Read from left to right, the graph rises.
- f is decreasing on I if for any two inputs a < b in I, f(a) > f(b). The graph falls.
- f is constant on I if every input in I gives the same output. The graph is a flat, horizontal piece.
- f is positive on I if f(x) > 0 for every x in I: the graph is above the x-axis. It is negative on I if f(x) < 0 for every x in I: below the axis.
Each interval is a set of inputs, x-values, written in Interval and set-builder notation. The outputs say how high the graph is; the interval says where along the x-axis the behavior happens.
The bracket convention. This lesson writes rising, falling and level stretches with round brackets at every end: at a turning point, at a corner where a flat piece starts, and at an end of the domain. That is a choice, not something the definition forces: a parabola opening down is also increasing up to and including its top, since putting the top in breaks no comparison. Leaving the ends out keeps each stretch clear of the single points where the graph changes what it is doing. Positive and negative are different: a sign belongs to one input at a time, so the brackets follow the actual outputs. A zero is never included, since 0 is neither positive nor negative, but an end of the domain with a nonzero output is included with a square bracket.
Intuition
Run a pencil along the graph from left to right. Where the pencil climbs, the function is increasing; where it drops, decreasing; where it runs level, constant. Then ask a separate question: is the pencil above the axis or below it?
The two questions are independent. On the left of the figure the graph is rising and negative at the same time, and near the top it is falling and still positive.
The zeros, the inputs where the graph meets the axis (Intercepts and zeros of a function), are where the sign can change. A graph without breaks cannot get from above the axis to below it without touching it, so the zeros cut the x-axis into stretches, each wholly positive or wholly negative.
Results
Lines. The function f(x) = mx + c is increasing on (-\infty, \infty) if m > 0, decreasing on (-\infty, \infty) if m < 0, and constant if m = 0.
Parabolas. Let f(x) = ax^2 + bx + c have its vertex at x = h (The vertex and the axis of symmetry). If a > 0, f is decreasing on (-\infty, h) and increasing on (h, \infty). If a < 0 it is the other way round.
Signs between zeros. If a graph has no breaks, its sign is the same all the way between two neighboring zeros.
Why it is true
Lines. Take inputs p < q. Then f(q) - f(p) = (mq + c) - (mp + c) = m(q - p). Since q - p > 0, the change in output has the sign of m: positive means a later input gives a larger output, negative a smaller one, zero the same one.
Parabolas. Written in vertex form, f(x) = a(x - h)^2 + k. Take p < q, both left of h. Then q is closer to h than p is, so (q - h)^2 < (p - h)^2, since a smaller distance has a smaller square. If a > 0, multiplying keeps the order, so f(q) < f(p): decreasing. Right of h the later input is farther from h, so its square and its output are larger: increasing. If a < 0, multiplying by a reverses both orders.
Signs between zeros. If f(p) > 0 and f(q) < 0, a graph drawn without lifting the pen must cross the axis somewhere between p and q, which is a zero between them. So between two neighboring zeros there is no sign change. (That a graph without breaks really must cross is a theorem with a name, met in a later course; here it is taken from the picture.)
Worked examples
A graph made of pieces
The graph of f is made of straight pieces with corners at (-2, -2), (1, 4) and (3, 4), and the arrowheads show that the end pieces carry on. Give the intervals where f is increasing, decreasing, constant, positive and negative.
Read from left to right. From the far left to x = -2 the graph falls; from -2 to 1 it rises; from 1 to 3 it is level at height 4; after 3 it falls and keeps falling. So f is
- decreasing on (-\infty, -2) and on (3, \infty),
- increasing on (-2, 1),
- constant on (1, 3).
The turning point at x = -2 and the corners at 1 and 3 belong to none of these.
The graph meets the axis at x = -3, -1 and 5, and has no breaks, so these zeros split the axis. The graph is above it far left, below between -3 and -1, above from -1 all the way to 5 (including the flat piece), and below after 5:
On (-2, -1) the function is increasing and negative at once.
An absolute value graph
Find the intervals for h(x) = 3 - |x + 1|, whose graph is drawn below.
The graph is an upside-down V with its corner at (-1, 3): increasing on (-\infty, -1) and decreasing on (-1, \infty). A turning point can be a sharp corner as well as a smooth top.
The rule agrees. For x \leq -1, x + 1 \leq 0, so |x + 1| = -(x + 1) and h(x) = 4 + x, a line of slope 1. For x \geq -1, |x + 1| = x + 1 and h(x) = 2 - x, slope -1.
The zeros come from each piece: 4 + x = 0 gives x = -4, and 2 - x = 0 gives x = 2. The top is above the axis, so h is positive on (-4, 2) and negative on (-\infty, -4) \cup (2, \infty).
What a table can promise
After a storm, Elodie reads the water level on a marker post beside a river at noon each day. The table gives L, the level in centimeters above the marker, t days after the storm; a negative level is below the marker. (The data are made up.)
| t | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| L | -35 | 20 | 65 | 80 |
| t | 4 | 5 | 6 | 7 |
|---|---|---|---|---|
| L | 50 | 10 | -15 | -25 |
What do the readings say about when the level rose, fell, and was above or below the marker? What can they not promise?
SolutionCompare neighboring readings. Each reading from day 0 to day 3 is higher than the one before, and each from day 3 to day 7 is lower. The level is above the marker on days 1 to 5 and below it on days 0, 6 and 7. A water level cannot jump, so it crossed the marker somewhere between day 0 and day 1, and again between day 5 and day 6.
The table promises nothing between noons. The water could have dipped overnight and recovered, so “rising on (0, 3)” is a reasonable reading, not a fact. The highest reading is on day 3, but the real peak could fall between readings, say at t = 3.5, above 80 cm. And the crossings lie somewhere in (0, 1) and in (5, 6); the table cannot say where. The honest summary: the level was probably rising for about the first 3 days and falling for the next 4, above the marker from some time in the first day until some time in the sixth. More frequent readings narrow this down; only a rule or a continuous record settles it.
A line and a parabola from their rules
Find the intervals for (a) \displaystyle L(x) = -\frac{3}{4}x + 6 and (b) f(x) = x^2 - 6x - 16.
Solution(a) The slope \displaystyle -\frac{3}{4} is negative, so L is decreasing on (-\infty, \infty). Its zero: \displaystyle -\frac{3}{4}x + 6 = 0 gives x = 8. Since the outputs shrink from left to right, they are positive before the zero and negative after: positive on (-\infty, 8), negative on (8, \infty). Check: L(0) = 6 > 0.
(b) The coefficient of x^2 is 1 > 0, so the parabola opens up. Its vertex is at \displaystyle x = -\frac{b}{2a} = \frac{6}{2} = 3, with f(3) = 9 - 18 - 16 = -25. So f is decreasing on (-\infty, 3) and increasing on (3, \infty).
Factorizing, x^2 - 6x - 16 = (x + 2)(x - 8), so the zeros are -2 and 8, one on each side of the axis of symmetry x = 3. The lowest point (3, -25) is below the axis, so the graph is below it between the zeros:
Check: f(0) = -16 < 0, and 0 is in (-2, 8).
A paraglider’s flight
The graph shows h, Eleazar’s altitude in meters above his launch site, t minutes after launch. Its straight pieces join (0, 0), (6, 180), (10, 120), (14, 240), (22, -240) and (24, -300); he lands at t = 24 on a field 300 m below the launch site. Give the intervals where h is increasing, decreasing, positive and negative, and say what each means.
The domain is [0, 24]. The graph rises on (0, 6) and (10, 14): Eleazar was climbing, in rising air, for the first 6 minutes and again from minute 10 to minute 14. It falls on (6, 10) and on (14, 24), when he was losing height. The bend at (22, -240) is not a turning point: the graph falls on both sides of it, only more slowly after it, so (14, 24) is one falling stretch.
For the signs, find the zero on the piece from (14, 240) to (22, -240). It drops 480 m in 8 minutes, 60 m per minute, so it reaches 0 after 240 \div 60 = 4 minutes, at t = 18. So h is positive on (0, 18) and negative on (18, 24]: Eleazar was above his launch site from just after launch until minute 18, and below it from then until he landed. The start is left out because h(0) = 0. Landing is not a zero, because h measures height against the launch site, not the ground: h(24) = -300 is negative, so 24 is in.
Boundaries
Only inside the domain. An interval of increase or of positive outputs is a set of inputs the function accepts (Domain and range from a graph). Outside the domain there is nothing to describe.
Breaks. A graph with a jump can change sign without a zero. Take p(x) = -x - 2 for x < 0 and p(x) = 3 - x for x \geq 0. Just left of 0 the outputs are near -2, negative; at 0 the graph jumps up to 3, positive. No input near 0 gives 0. Zeros mark every sign change only for a graph without breaks.
Two intervals are not one. The same p is decreasing on (-\infty, 0) and on (0, \infty), since each piece has slope -1. Yet the input -1 gives -1 and the later input 1 gives the larger output 2, so p is not decreasing across both. That is why rises and falls are listed with “and”, while signs are joined with \cup: a sign belongs to each input on its own, so a union of positive stretches is positive.
A table promises only its rows. Between two rows the function may do anything that a continuous record has not ruled out.
A stretch, not a point. Increasing, decreasing and constant describe intervals. A single input is none of the three.
Where this goes next
The inputs where increasing meets decreasing are the turning points of Turning points, maximums and minimums. On an interval where a function is increasing, the Average rate of change between any two inputs is positive, and Concavity: when the rate of change itself changes asks whether the rising speeds up or slows down. The positive and negative intervals of a quadratic are exactly the solutions of Quadratic inequalities, and Maximum, minimum and range of a quadratic function uses the vertex to find a parabola’s range. In calculus, the steepness of the graph at each single point is measured, and its sign finds these intervals without a picture.
Exercises
Reading graphs
For each graph, give the intervals where the function is increasing, decreasing and constant, and where it is positive and negative.
- Graph A, a parabola with its top at (-1, 4).
- Graph B, which touches the axis at x = -2, has a low point at (0, -4) and crosses the axis at x = 1.
- Graph C, with an open dot at (0, 2) and a filled dot at (0, -1).
Tables
A function g has these values.
| x | -3 | -2 | -1 | 0 |
|---|---|---|---|---|
| g(x) | 7 | 2 | -1 | -2 |
| x | 1 | 2 | 3 |
|---|---|---|---|
| g(x) | -1 | 2 | 7 |
- What do the rows of the table for g say about where g rises, falls and changes sign? What can they not promise?
- Ermengarde learns that g(x) = x^2 - 2. Give the exact intervals where g is increasing, decreasing, positive and negative, and compare them with your answer to exercise 4.
- A table gives k(0) = 1, k(1) = 3, k(2) = 4, k(3) = 6 and k(4) = 9, each row higher than the one before. Describe a graph that fits every row but is not increasing on (0, 4).
From a rule
Give the intervals where each function is increasing, decreasing or constant, and where it is positive and negative.
- y = 4x - 10
- \displaystyle y = -\frac{2}{5}x - 2
- f(x) = x^2 + 6x - 7
- g(x) = -x^2 + 8x
- \displaystyle k(x) = \frac{1}{2}(x + 1)^2
- y = -5
In context
- After a second storm, the river level in centimeters above the marker, t days later, is modeled by L(t) = -5t^2 + 40t - 35 for 0 \leq t \leq 8. When is the level rising and when is it falling? Answer with intervals and in words.
- For the river in exercise 13, when is the water above the marker and when is it below?
Explaining
- The greatest integer function y = \lfloor x \rfloor (Step functions and the greatest integer function) rounds every input down to an integer. Explain why it is constant on (0, 1) and on (1, 2) but not on (0, 2), and why it is not increasing on (0, 2) either.
- Give a rule for a function f that is increasing on (-3, 0) and on (0, 2) but not increasing on (-3, 2).
- Explain why a function that is increasing on an interval has at most one zero in it.
Answers
- Increasing on (-\infty, -1), decreasing on (-1, \infty), never constant. Zeros -3 and 1, top above the axis: positive on (-3, 1), negative on (-\infty, -3) \cup (1, \infty).
- Increasing on (-\infty, -2) and on (0, \infty), decreasing on (-2, 0). Positive on (1, \infty), negative on (-\infty, -2) \cup (-2, 1). At x = -2 the output is 0, so it is left out although the graph is below the axis on both sides of it.
- Increasing on (-\infty, 0) and on (2, \infty), constant on (0, 2), never decreasing on an interval even though it drops at the jump. Zeros -2 and 3. Positive on (-2, 0) \cup (3, \infty), negative on (-\infty, -2) \cup [0, 3): the filled dot says f(0) = -1, so 0 is in. The sign changes at the jump, where there is no zero.
- The rows fall from x = -3 to x = 0 and rise from x = 0 to x = 3. The outputs are positive at \pm 3 and \pm 2 and negative at -1, 0 and 1, so if g has no breaks it has a zero in (-2, -1) and one in (1, 2). The table cannot say where those zeros are, nor rule out a dip or bump between rows, nor promise that the lowest output is at 0.
- A parabola opening up with vertex at x = 0: decreasing on (-\infty, 0), increasing on (0, \infty). Zeros \displaystyle \pm\sqrt{2} \approx \pm 1.41, inside (-2, -1) and (1, 2) as the table said. Positive on \displaystyle (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty), negative on \displaystyle (-\sqrt{2}, \sqrt{2}).
- Join (0, 1), (1, 3) and (2, 4) with straight pieces, go down to (2.5, 2), then up to (3, 6) and (4, 9). Every row fits, but the graph falls on (2, 2.5). Rising rows are consistent with increasing; they do not prove it.
- Slope 4 > 0: increasing on (-\infty, \infty). Zero at \displaystyle x = \frac{5}{2}: positive on \displaystyle (\frac{5}{2}, \infty), negative on \displaystyle (-\infty, \frac{5}{2}).
- Slope \displaystyle -\frac{2}{5} < 0: decreasing on (-\infty, \infty). Zero at x = -5: positive on (-\infty, -5), negative on (-5, \infty).
- Opens up, vertex at x = -3: decreasing on (-\infty, -3), increasing on (-3, \infty). (x + 7)(x - 1) gives zeros -7 and 1: positive on (-\infty, -7) \cup (1, \infty), negative on (-7, 1).
- Opens down, vertex at \displaystyle x = -\frac{8}{2(-1)} = 4: increasing on (-\infty, 4), decreasing on (4, \infty). -x(x - 8) gives zeros 0 and 8: positive on (0, 8), negative on (-\infty, 0) \cup (8, \infty).
- Opens up, vertex at x = -1: decreasing on (-\infty, -1), increasing on (-1, \infty). The only zero is the vertex itself: positive on (-\infty, -1) \cup (-1, \infty), never negative.
- Constant on (-\infty, \infty) and negative on (-\infty, \infty); never increasing, decreasing or positive.
- Opens down, vertex at \displaystyle t = -\frac{40}{2(-5)} = 4: rising on (0, 4), falling on (4, 8). The water rose for the first 4 days after the storm, reaching L(4) = 45 cm above the marker, and fell for the next 4.
- L(t) = -5(t - 1)(t - 7), zeros 1 and 7. Above the marker on (1, 7), from day 1 to day 7; below on [0, 1) \cup (7, 8]. The ends are in, since L(0) = -35 and L(8) = -35 are both negative.
- On (0, 1) every input rounds down to 0, and on (1, 2) to 1, so each is constant. Across (0, 2) both 0 (at 0.5) and 1 (at 1.5) appear, so it is not constant there. It is not increasing either: 0.2 < 0.5 but both give 0, and increasing needs a larger output for every later input. The graph climbs only by jumping, never along a stretch.
- Let f(x) = x for x < 0 and f(x) = x - 5 for x \geq 0. Each piece has slope 1, so f is increasing on (-3, 0) and on (0, 2). But f(-1) = -1 and f(1) = -4: a later input gives a smaller output, so f is not increasing on (-3, 2).
- Suppose f is increasing on I and has two zeros a < b in I. Increasing forces f(a) < f(b), but both are 0. So there is at most one zero, and the function is negative before it and positive after it.