Unit 3 · Topic 3.1

Periodic Phenomena

A periodic function repeats its outputs every P units of input, f(x + P) = f(x) , and the smallest such P is the period.

Practice 2: Multiple RepresentationsPractice 3: Communication and Reasoning

Part 1 · Hook

Why this matters

Your heart pushes out the same pressure wave with every beat. A lighthouse beam sweeps past the same window every few seconds, and the sea rises and falls about twice a day. Each of these repeats a pattern, so once you know one full cycle you can predict the output at any time at all, next minute or next year.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A table gives $g(2) = 7$ and $g(6) = 19$ . What is the average rate of change of $g$ on $[2, 6]$ ?

$3$
$12$
$\frac{1}{3}$
$6.5$

Show the answer

Change in output over change in input: $\frac{19 - 7}{6 - 2}$ $= \frac{12}{4}$ $= 3$ .

Correct: $3$ :
$12$ :
$\frac{1}{3}$ :
$6.5$ :

2. Over the equal intervals $[0, 1]$ , $[1, 2]$ and $[2, 3]$ , the average rates of change of $h$ are $5$ , $3$ and $1$ . What does this suggest about the graph of $h$ on $[0, 3]$ ?

It is concave down
It is concave up
It is decreasing
It is a straight line

Show the answer

The rates are positive, so $h$ is increasing, and they get smaller over equal intervals, so the graph is concave down.

Correct: It is concave down:
It is concave up:
It is decreasing:
It is a straight line:

Part 4 · See it

See it first

The depth of water in a tank, in feet, over 18 minutes. In each 6-minute cycle the depth rises in a straight line from 1 foot to 5 feet in 2 minutes, stays at 5 feet for 1 minute, then falls in a straight line back to 1 foot over 3 minutes. The first cycle, from t = 0 to t = 6, is shaded and drawn in green; the next two cycles are exact copies. A bracket marks the period, 6 minutes. — The tank's depth repeats every 6 minutes. The green cycle on $[0, 6]$ is all you need: every later cycle is a copy of it. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The outputs of $f$ repeat the same pattern every $P$ units of input $f$ is periodic, and the smallest such positive $P$ is the period: $f(x + P) = f(x)$ for every $x$
You know $f$ on one full cycle, an interval of length $P$ copy that piece of graph left and right, $P$ units at a time, to get the whole graph
You need the output at an input far from the known cycleadd or subtract whole periods until the input lands in the cycle: with $P = 6$ , $f(47)$ $= f(47 - 7 \cdot 6)$ $= f(5)$
Every feature of one cycle repeatsintervals where $f$ increases or decreases, concavity and average rates of change all repeat every $P$ units

Part 6 · Key ideas

Key ideas

A periodic function repeats its outputs: there is a positive number $P$ with $f(x + P) = f(x)$ for every input $x$ . The smallest such $P$ is the period.
One full cycle, any interval of length $P$ , holds all the information. Shift inputs by whole periods to reach any other input.
Over any interval one full period long, the average rate of change is $0$ , because the output ends where it started.
Two inputs can share an output without being a period apart: a cycle usually passes through each in-between value twice.

Part 7 · Misconception

A common mistake

The wrong idea: The period is the distance between two inputs that have the same output.

What actually happens: Inside one cycle the graph usually reaches the same output twice, once going up and once going down. The period is the length of the smallest interval after which the whole pattern repeats, so check that the graph matches from that point on, not only one value.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

A periodic function $f$

The graph shows a periodic function

f

[0, 12]

. On each piece,

f

is linear.

1. What is the period of $f$ ?

$4$
$2$
$6$
$12$

Show the answer

The highest points are at $x = 1$ , $5$ and $9$ , four units apart, and the whole pattern on $[1, 5]$ repeats on $[5, 9]$ . The period is $4$ .

Correct: $4$ : From one highest point to the next is $5 - 1 = 4$ , and the graph matches from there on.
$2$ : That is the distance from a highest point ( $x = 1$ ) to the next lowest point ( $x = 3$ ), only part of a cycle.
$6$ : Shifting by 6 does not keep the outputs: $f(1) = 5$ but $f(7) = -1$ .
$12$ : $f(x + 12) = f(x)$ does hold, but 12 is three periods. The period is the smallest positive value that works.

2. The pattern in the graph continues for all $x$ . What is $f(39)$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The period is $4$ and $39 = 9 \cdot 4 + 3$ , so $f(39) = f(3) = -1$ .

Answer: -1

3. What is the average rate of change of $f$ on $[5, 9]$ ?

$0$
$-\frac{3}{2}$
$\frac{5}{4}$
$-3$

Show the answer

$[5, 9]$ is exactly one period long, so $f(9) = f(5) = 5$ and the average rate of change is $\frac{5 - 5}{9 - 5} = 0$ : the secant line through $(5, 5)$ and $(9, 5)$ is horizontal.

Correct: $0$ : The output ends where it started, so the change in output is $0$ .
$-\frac{3}{2}$ : This uses $f(7) = -1$ as an endpoint, $\frac{-1 - 5}{9 - 5}$ . The interval ends at $x = 9$ , where $f(9) = 5$ .
$\frac{5}{4}$ : This is $f(9)$ divided by the length of the interval. A rate needs the change in output, $f(9) - f(5)$ .
$-3$ : That is the slope of the falling piece on $(5, 7)$ , not the slope of the secant line from $x = 5$ to $x = 9$ .

4. The pattern continues for all $x$ . On which interval is $f$ decreasing?

$(41, 43)$
$(43, 45)$
$(40, 42)$
$(42, 44)$

Show the answer

In the first cycle $f$ decreases on $(1, 3)$ . Adding ten periods ( $40$ ) gives $(41, 43)$ .

Correct: $(41, 43)$ : $(1, 3)$ plus $10 \cdot 4 = 40$ is $(41, 43)$ .
$(43, 45)$ : This is $(3, 5)$ plus $40$ , where $f$ rises from $-1$ to $5$ .
$(40, 42)$ : This is $(0, 2)$ plus $40$ : $f$ rises on $(40, 41)$ before it starts to fall.
$(42, 44)$ : This is $(2, 4)$ plus $40$ : $f$ falls only until $x = 43$ and then rises.

5. A function $g$ satisfies $g(x + 10) = g(x)$ and $g(x + 5) = g(x)$ for every $x$ , and no smaller positive number works. What is the period of $g$ ?

$5$
$10$
$15$
$50$

Show the answer

The period is the smallest positive $P$ with $g(x + P) = g(x)$ for all $x$ . Both 5 and 10 work, and 5 is smaller.

Correct: $5$ : The smallest shift that leaves every output unchanged is 5.
$10$ : 10 works because it is two periods; the period is the smallest shift that works.
$15$ : 15 also works (three periods), but it is not the smallest.
$50$ : This multiplies the two shifts; the period is the smallest shift that works, 5.

6. A periodic function $h$ has period 8. What is the average rate of change of $h$ on $[3, 11]$ ?

$0$
$\frac{h(11)}{8}$
$\frac{1}{8}$
It depends on the formula for $h$

Show the answer

$11 = 3 + 8$ , one period later, so $h(11) = h(3)$ and the average rate of change is $\frac{h(11) - h(3)}{8} = 0$ .

Correct: $0$ : One period apart, the outputs are equal, so the change in output is $0$ .
$\frac{h(11)}{8}$ : This divides one output by the length of the interval. A rate needs the change in output.
$\frac{1}{8}$ : This is one over the period. It is not a rate of change of $h$ .
It depends on the formula for $h$ : Whatever the formula, $h(11) = h(3)$ because the inputs are one period apart.

Part 9 · Summary

Summary

A periodic function repeats its outputs every

P

units of input,

f(x + P) = f(x)

, and the smallest such

P

is the period. Learn one cycle and you know the function everywhere: shift any input by whole periods into the cycle. Increasing and decreasing intervals, concavity and rates of change repeat every period, and the average rate of change over a full period is

0

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

A periodic function ff

Summary

What comes next

Connections

A periodic function $f$