Periodicity
A function f is periodic if there is a number T > 0 so that f(x + T) = f(x) for every x. The smallest such T is the period. The graph is one piece repeated again and again.
- sin x and cos x: period 2π.
- tan x: period π.
- sin(wx): period 2π ÷ w. So sin(2x) has period π.
- |sin x|: period π (the flipped parts make it repeat twice as often).
A function like x² is not periodic: its values keep growing and never repeat.
If f has period T, you only need to study one period and then copy it.
Monotonicity: increasing and decreasing
A function is increasing on an interval if the graph goes up as x grows, and decreasing if it goes down. Together these are called monotonic (one direction only).
The slope tells us. The slope of the curve at a point is written f′(x) (the derivative).
- f′(x) > 0 on an interval: increasing there.
- f′(x) < 0 on an interval: decreasing there.
Example: f(x) = x³ − 3x, f′(x) = 3x² − 3 = 3(x − 1)(x + 1). It is positive for x < −1 and x > 1 (increasing) and negative for −1 < x < 1 (decreasing).
Extrema: turning points
A local maximum is the top of a hill: the function goes up, then down. A local minimum is the bottom of a valley: down, then up. Together they are extrema (extremum for one).
At such a point the tangent line is flat, so f′(x) = 0. Find them like this:
- Solve f′(x) = 0. These are the candidates (critical points).
- Check the sign of f′ just before and just after. Plus then minus: maximum. Minus then plus: minimum. No change: neither (like x³ at 0).
For x³ − 3x: f′ = 0 at x = ±1. At x = −1 the slope goes + to −: local maximum, value 2. At x = 1 it goes − to +: local minimum, value −2.
Max and min on an interval [a, b]
On a closed interval [a, b] a smooth function has a largest and a smallest value. They can sit at a turning point inside, or at an end. So:
- Find the turning points (f′ = 0) that lie inside [a, b]. Ignore the ones outside.
- Calculate f at those points and at a and b.
- The biggest number is the maximum, the smallest is the minimum.
For f = x³ − 3x on [−2, 1.5]: f(−2) = −2, f(−1) = 2, f(1) = −2, f(1.5) = −1.125. Max 2 at x = −1; min −2 at x = −2 and x = 1.
A local max is not always the biggest on the interval: an end can be bigger.
Try it: predict, then check
In the 3D board choose "Max and min on [a, b]". Guess: if a = −1.5 and b = 2, where is the maximum? Slide a and b to check. At home: throw a small ball up (softly!) and watch: it rises, stops for a moment at the top (slope 0) and falls. The top is the turning point. Or clap in a steady rhythm: the time between two claps is the period.
Key formulas and definitions
- Periodic: f(x + T) = f(x) for all x
- sin(wx), cos(wx): T = 2π / w; tan(wx): T = π / w
- f′(x) > 0 → increasing; f′(x) < 0 → decreasing
- Turning point: f′(x) = 0 and the sign of f′ changes
- Max/min on [a, b]: compare f(a), f(b) and f at turning points inside
Worked examples
1. Find the period of y = sin(3x).
T = 2π ÷ w = 2π ÷ 3 = 2π/3.
2. Find the period of y = cos(x/2).
Here w = 1/2, so T = 2π ÷ (1/2) = 4π. The wave is stretched, so it repeats more slowly.
3. Is f(x) = x² periodic?
No. If f(x + T) = f(x) for all x, then at x = 0: T² = 0, so T = 0, but a period must be above 0. So x² is not periodic. The graph is a bowl that never repeats.
4. Where is f(x) = x² − 4x increasing and decreasing?
f′(x) = 2x − 4. It is 0 at x = 2. For x < 2, f′ < 0: decreasing. For x > 2, f′ > 0: increasing. So x = 2 is the minimum, with f(2) = −4.
5. Find the local maximum and minimum of f(x) = x³ − 3x.
f′ = 3x² − 3 = 0 gives x = ±1. At x = −1 the slope goes + to −: local maximum, f(−1) = −1 + 3 = 2. At x = 1 it goes − to +: local minimum, f(1) = 1 − 3 = −2.
6. Find the max and min of f(x) = x³ − 3x on [−2, 1.5].
Turning points ±1 are both inside. f(−2) = −8 + 6 = −2. f(−1) = 2. f(1) = −2. f(1.5) = 3.375 − 4.5 = −1.125. Largest 2 at x = −1, smallest −2 at x = −2 and x = 1.
7. Find the max and min of f(x) = x² − 4x + 1 on [0, 5].
f′ = 2x − 4 = 0 gives x = 2 (inside). f(2) = 4 − 8 + 1 = −3. f(0) = 1. f(5) = 25 − 20 + 1 = 6. Maximum 6 at x = 5 (an end), minimum −3 at x = 2.
8. Find the max and min of f(x) = x + 1/x on [1, 4].
f′ = 1 − 1/x². For x > 1, 1/x² < 1, so f′ > 0: increasing on the whole interval. No turning point inside (x = 1 is an end). Minimum f(1) = 2, maximum f(4) = 4 + 0.25 = 4.25.
Common mistakes
- Calling a function periodic after seeing it repeat once or twice. The repeat must hold for every x.
- Using the wrong period for sin(wx). It is 2π ÷ w, not 2π × w.
- Thinking every point with f′ = 0 is a max or min. x³ at 0 is flat for a moment but keeps going up.
- Forgetting to check the two ends of the interval. The biggest or smallest value is often at an end.