Period, Monotonicity and Symmetry · Little Blue Book, Visualized

Try this

For $y = ∣ sin x ∣ + ∣ cos x ∣$ , the copy fits the graph when $T = π$ . Keep making $T$ smaller: what is the least positive period?
Switch to $y = sin (2 x + \frac{π}{3})$ , turn on “Axis of symmetry” and drag $a_{1}$ until you find an axis. Then turn on “Second axis” and find another one. Look at the reading card that appears last.
Same function: turn on “Center of symmetry” and drag $c$ to find a center. How far is it from the axis you found? What fraction of the period is that?
Switch to $y = sin x + sin 2 x$ . Can you find a center of symmetry? An axis of symmetry?
Turn on “Monotonic pieces”: solid pieces increase, dashed pieces decrease. Must the increasing and decreasing intervals be equally long? (Try $y = sin x + sin 2 x$ .)

For a sinusoidal function $y = A sin (ω x + φ)$ ( $ω > 0$ ), all of these questions come down to treating $u = ω x + φ$ as one block and using the properties of $sin u$ :

axes: u = \frac{π}{2} + k π, centers: u = k π, increasing (A > 0): - \frac{π}{2} + 2 k π ⩽ u ⩽ \frac{π}{2} + 2 k π .

Predict first

A function $f (x)$ is defined on $R$ , and its graph is symmetric about both lines $x = 1$ and $x = 3$ . Which number is certainly a period of $f (x)$ ?

2

4

6

it need not be periodic

Choose an answer first; then you can see the result.

Why: two reflections make one shift

Two axes ⇒ period; an axis and a center ⇒ period

Step 1 / 5

One axis of symmetry
The graph of $y = sin x + cos x$ is symmetric about $x = \frac{π}{4}$ : the point $x$ and its mirror image $2 a - x$ have the same function value.

$f (x) = f (2 a - x)$
A second axis
Two mirrors in a row
An axis and a center
Off by a sign, so do it twice

This has nothing to do with $f$ being a trig function: any function defined on $R$ with two different axes of symmetry is periodic. Trig functions just happen to be the most typical examples.

If there is only one axis

Break it

Counterexample lab: remove one condition

Conclusion: $f (x)$ is periodic

$y = sin x + cos x$ has the two axes $x = \frac{π}{4}$ and $x = \frac{5 π}{4}$ , so $2 π$ is a period.

Practice

Exercise 1Basic

The least positive period of $y = ∣ sin 2 x ∣$ is $k π$ . Find $k$ .

Exercise 2Basic

Of all the axes of symmetry of the graph of $y = sin (2 x + \frac{π}{3})$ , the one closest to the $y$ -axis is $x = k π$ . Find $k$ .

Exercise 3Advanced

On $[0, π]$ , the function $y = sin (\frac{π}{4} - 2 x)$ is increasing on the interval $[a, b]$ . If $a = k π$ , find $k$ .

Exercise 4Challenge

A function $f (x)$ defined on $R$ has a graph symmetric about the line $x = 1$ and also about the point $(3, 0)$ , and $f (1) = 2$ . Find $f (2029)$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Thinking sin|x| also has period 2π

y = sin ∣ x ∣

is periodic with least positive period

2 π

What went wrong? Think first, then open

Diagnosis For $x ⩾ 0$ it is just $sin x$ , but for $x < 0$ the graph is the mirror image of the right half. Take $x = - \frac{π}{2}$ : $f (- \frac{π}{2}) = 1$ , while $f (- \frac{π}{2} + 2 π) = sin \frac{3 π}{2} = - 1$ . One counterexample is enough.

y = sin ∣ x ∣

has just one axis of symmetry,

x = 0

, and it is not periodic. To show that “

T

is not a period”, it is enough to find one

x

with

f (x + T) \neq = f (x)

“A period must work for every x; one x is enough to break it”

Exhibit

Copying the increasing-interval recipe when ω is negative

The increasing intervals of

y = sin (\frac{π}{4} - 2 x)

come from solving

- \frac{π}{2} + 2 k π ⩽ \frac{π}{4} - 2 x ⩽ \frac{π}{2} + 2 k π

What went wrong? Think first, then open

Diagnosis The inner function $u = \frac{π}{4} - 2 x$ is decreasing in $x$ . While $sin u$ increases, $y$ actually decreases as $x$ grows, so this gives exactly the decreasing intervals.

First rewrite it as

y = - sin (2 x - \frac{π}{4})

. The increasing intervals of

y

are the decreasing intervals of

sin (2 x - \frac{π}{4})

\frac{π}{2} + 2 k π ⩽ 2 x - \frac{π}{4} ⩽ \frac{3 π}{2} + 2 k π

“Make ω positive first, then substitute the whole block”