1The idea in one paragraph

Read a graph from left to right. Where it climbs, the function is increasing; where it falls, it is decreasing. The derivative measures the gradient, and the gradient's sign is the direction: a positive gradient climbs, a negative one falls, and a zero gradient is flat for an instant. So the whole subtopic is one sentence: f is increasing where f′(x) > 0 and decreasing where f′(x) < 0. To find those intervals, find where f′(x) = 0, because the gradient usually changes sign only by passing through zero, then test the sign of f′ in between.

2What increasing and decreasing mean

A function is increasing on an interval if, as x increases through that interval, f(x) increases too: for any two values a < b in the interval, f(a) < f(b). It is decreasing if, as x increases, f(x) decreases: a < b gives f(a) > f(b).

Two things to fix in your head now.

It is always read left to right. "Increasing" describes what happens to y as x goes up. A curve that rises as you move left is decreasing.

It describes an interval, not the whole function. Most functions increase in some places and decrease in others. The answer to "where is f increasing?" is a set of x-values, written as inequalities, such as x < −1 or x > 3.

A small number of functions go one way the whole time. The exponential eˣ is increasing for every x; the function 1/x is decreasing on x < 0 and decreasing on x > 0 (though not across the gap, since f(−1) = −1 is less than f(1) = 1). Those are the exceptions. Mostly you will be finding the pieces.

3The sign of f′ gives the direction

Figure 1 shows the cubic f(x) = x³ − 3x² − 9x + 5 with tangents drawn at five points.

Where the curve climbs, its tangent slopes upwards, so the gradient is positive. Where it falls, the tangent slopes downwards, so the gradient is negative. At the top of the hump and the bottom of the dip, the tangent is horizontal and the gradient is zero. The derivative is the gradient, so:

f′(x) > 0 ⇒ f is increasing. f′(x) < 0 ⇒ f is decreasing. f′(x) = 0 ⇒ the tangent is horizontal: a stationary point.

A point where f′(x) = 0 is called a stationary point, because for that instant the function is neither rising nor falling. The hump and the dip in Figure 1 are stationary points. They are also turning points, because the function changes direction there. Section 6 shows that these are not always the same thing.

The same rule works for any rate of change. If T(t) is the temperature of a cup of coffee after t minutes, dT/dt < 0 means the coffee is cooling, and dT/dt = 0 at some instant means the temperature has momentarily stopped changing.

4Finding the intervals algebraically

Here is the method in full on the function from Figure 1. This is the standard Paper 1 question.

Find the values of x for which f(x) = x³ − 3x² − 9x + 5 is decreasing.

The derivative is zero only at x = −1 and x = 3, so it can only change sign there. Those two values split the number line into three intervals, and f′ has one sign throughout each. To find the sign, either test a value in each interval or use a sign diagram, like Figure 3.

So f is decreasing for −1 < x < 3, and increasing for x < −1 and for x > 3.

Figure 2 lines the two graphs up. The derivative y = 3x² − 6x − 9 is a positive quadratic, so it is negative between its roots and positive outside them. The graph of f is falling exactly where the graph of f′ is below the axis. The quick reasoning, "a positive quadratic is negative between its roots", is often all you need, and it is quicker than a table. Figure 3 is the sign diagram, which is the safest method when f′ has more than two factors.

Writing the answer. Inequalities are the clearest: −1 < x < 3. The IB also uses interval notation with reversed brackets for open intervals, ]−1, 3[, which means the same thing. Two cases need care.

• A function increasing on two separate pieces is written as two inequalities joined by "or", or listed: x < −1, x > 3. Do not write −1 > x > 3, which no number satisfies.
• At a stationary point itself f′ is zero, not positive. Strictly, f is still increasing on x ≤ −1 including the end point, because f(−1) is bigger than every value to its left. Mark schemes normally accept either x < −1 or x ≤ −1. Strict inequalities match the rule f′(x) > 0 and are never wrong.

5Reading a graph of f′

This is the question students dread, and it is easier than it looks, because you only have to read one thing: is the graph of f′ above or below the x-axis? The shape of f′, whether it is going up or down, does not tell you the direction of f. Only its sign does.

Figure 4(a) is a graph you might be given: y = f′(x), crossing the x-axis at −2, 1 and 3.

Read it from left to right.

• x < −2: f′ is below the axis, so f is decreasing.
• At x = −2, f′ crosses from negative to positive, so f stops falling and starts rising: f has a minimum.
• −2 < x < 1: f′ is above the axis, so f is increasing. Notice that f′ itself rises and then falls on this interval. That does not matter: f′ is positive throughout, so f climbs throughout. The hump in the graph of f′ is where f climbs most steeply, not where f has a maximum.
• At x = 1, f′ goes from positive to negative: f has a maximum.
• 1 < x < 3: f is decreasing, and at x = 3, where f′ goes from negative to positive, f has another minimum.
• x > 3: f is increasing.

Figure 4(b) shows a curve with those properties. You could not have drawn its exact height from f′ alone, because many curves share the same derivative (they differ by a vertical shift, as 5.5 explains), but its shape is fixed.

The summary rule:

Where f′ crosses the axis from + to −, f has a maximum. Where it crosses from − to +, f has a minimum. Where f′ touches the axis without crossing, f has a stationary point but does not turn.

6Stationary points that are not turning points

Setting f′(x) = 0 finds every stationary point, but not every stationary point is a hump or a dip.

Take f(x) = x³. Then f′(x) = 3x², which is zero at x = 0. But 3x² is positive on both sides of 0, so f is increasing to the left of 0 and increasing to the right. The curve flattens out for an instant at the origin and then carries on upwards, as Figure 5(a) shows. A stationary point where the curve does not turn is called a stationary point of inflexion; 5.7 and 5.8 come back to it.

Figure 5(b) is the same idea inside a longer question. For f(x) = x⁴ − 4x³ + 10, f′(x) = 4x³ − 12x² = 4x²(x − 3). The factor 4x² is never negative, so the sign of f′ is the sign of (x − 3):

So f is decreasing for x < 3 (the stationary point at x = 0 does not interrupt it) and increasing for x > 3, with a minimum at (3, −17). A squared factor in f′ is the tell: it gives a zero where the sign does not change.

This is also why the method in section 4 says test the sign, not assume the signs alternate. Alternating signs is a habit that works for simple quadratics and fails on exactly the questions designed to catch it.

7Showing a function is always increasing

"Show that f is increasing for all x" means: show that f′(x) > 0 for every x. There are two usual routes.

Every term is clearly positive. For f(x) = x³ + 5x − 2, f′(x) = 3x² + 5. Since x² ≥ 0 for every x, 3x² + 5 ≥ 5 > 0. So f′(x) > 0 for all x, and f is increasing for all x.

Complete the square, or use the discriminant. For f(x) = x³ − 3x² + 4x + 1:

Or: f′ is a quadratic with a positive leading coefficient and discriminant (−6)² − 4(3)(4) = 36 − 48 = −12 < 0, so it has no roots and is never zero; it is positive for all x. Either argument earns the reasoning mark if you finish with a sentence: "f′(x) > 0 for all x, therefore f is increasing for all x."

One subtlety. If f′(x) ≥ 0 and it equals zero only at isolated points, as 3x² does at x = 0, the function is still increasing: x³ is increasing everywhere. So when a question asks for values of a constant k that make f increasing for all x, a discriminant equal to zero is included. Try it Q3 is one of these.

8Intervals with technology

On Paper 2 you may be given a function you cannot easily differentiate by hand, or one where f′(x) = 0 has no neat solutions. Then:

1. Graph y = f(x) on the GDC over the domain given.
2. Use the calculator's maximum and minimum tools to find the x-coordinates of the turning points.
3. Read the direction between them from the graph, and write the intervals.

For f(x) = x − 2 sin x on 0 ≤ x ≤ 2π, Figure 6 shows the GDC output: a minimum at x ≈ 1.05 and a maximum at x ≈ 5.24.

So f is decreasing for 0 < x < 1.05, increasing for 1.05 < x < 5.24, and decreasing again for 5.24 < x < 2π (all to 3 s.f.). Once you have 5.6, you can check this exactly: f′(x) = 1 − 2 cos x, which is zero when cos x = 1/2, at x = π/3 ≈ 1.047 and x = 5π/3 ≈ 5.236.

Two cautions. The GDC finds turning points, so it will miss a stationary point of inflexion like the one in Figure 5(b); that one does not change the intervals anyway. And set the window to the domain the question gives, or you may report a turning point outside it.

A second method on the GDC is to graph y = f′(x) (with the calculator's numerical derivative) and find where it crosses the axis. That is the Figure 4 method with the machine drawing f′ for you.