📈Increasing, Decreasing & Extrema

Technician Fluxion keeps one instrument on the bench of the Tangent Laboratory: a needle that reads the slope of whatever curve is clamped under it. The needle does not say how high the curve is. It says which way the curve is going, right now, at this x.

That single reading is the whole of this craft.

f'(x) > 0 means the curve is climbing at x — f is increasing there f'(x) < 0 means the curve is falling at x — f is decreasing there f'(x) = 0 means the tangent is flat for that instant

Read the middle line carefully, because it is the one students misuse. A negative derivative does not mean a small function. f(x) = 1000 - x has f'(x) = -1 everywhere: the function is enormous and going down the whole time. Height and direction are different questions, and f' only ever answers the second one.

The consequence is the point of the topic. If f' keeps one sign across a whole stretch of the number line, then f keeps going one way across that whole stretch — no peaks, no valleys, no surprises. So to describe the shape of f you do not need to plot a hundred points. You need to know where f' changes sign, and nothing else.

A continuous f' can only change sign by passing through zero or by breaking off entirely. Those two events are the only candidates, and they have a name.

A critical number of f is an x in the domain of f where

f'(x) = 0, or f'(x) has no value.

All three parts of that sentence do work.

f'(x) = 0. Differentiate, set the derivative equal to zero, solve. For a polynomial this is the whole job, because a polynomial is defined and differentiable everywhere.

f(x) = x³ - 3x² - 9x + 2 f'(x) = 3x² - 6x - 9 = 3(x + 1)(x - 3)

so the critical numbers are x = -1 and x = 3. Notice the factoring: reading a root off a factor means asking what makes that factor zero, so (x + 1) gives -1, not +1.

f' has no value. A derivative can simply stop existing at a point while the function sails on through it. The standard example is a cusp:

f(x) = x^2/3 has f'(x) = 2 / (3 x^1/3)

That fraction has 2 on top, so it is never zero — and at x = 0 its denominator is zero, so it has no value at all. Yet f(0) = 0 is perfectly well defined. The graph comes to a sharp point at the origin, and x = 0 is a critical number. A sharp corner, like the one in f(x) = |x - 2| at x = 2, is the same story.

In the domain of f. This is the guard that stops the previous rule running wild. For

f(x) = x + 9/x, f'(x) = 1 - 9/x²

f' has no value at x = 0 — but neither does f, so x = 0 is not in the domain and is not a critical number. Setting 1 - 9/x² = 0 gives x² = 9, so the critical numbers are x = -3 and x = 3, and that is the complete list.

A critical number is a *candidate*, not a promise. Every peak and every valley of a differentiable function sits at a critical number, but a critical number does not have to be a peak or a valley. Which is why there is a test.

Once the critical numbers are known, they cut the number line into pieces, and f' cannot change sign inside a piece. So one test value per piece settles the sign of the whole piece. Write the results in a row — that is a sign chart.

For f'(x) = 3(x + 1)(x - 3), with critical numbers -1 and 3, test x = -2, x = 0 and x = 4:

f' is positive for x < -1, negative for -1 < x < 3, positive for x > 3

Now read the chart at each critical number, left to right:

• ◆positive then negative: the curve climbed, flattened, and fell. That is a local maximum.
• ◆negative then positive: the curve fell, flattened, and climbed. That is a local minimum.
• ◆no change of sign: the curve flattened for an instant and carried straight on. That is neither.

So this f has a local maximum at x = -1 and a local minimum at x = 3. And the intervals come free from the same chart: f is increasing for x < -1 and x > 3, and decreasing for -1 < x < 3.

The third case is not a curiosity. It happens whenever a factor of f' is repeated an even number of times:

f'(x) = 12(x - 3)² (x - 4)

The squared factor cannot change sign — it touches zero and comes straight back positive — so the sign of f' before and after x = 3 is the same. x = 3 is a genuine critical number with a genuinely horizontal tangent, and it is neither a maximum nor a minimum. At x = 4 the sign really does turn, from negative to positive, so there is a local minimum there.

Writing the answer down. An interval of increase or decrease is a set of x-values, written with plain inequalities and joined with "and" when there is more than one piece: increasing for x < -1 and x > 3. Two separate intervals are not one big interval — between them the curve is going the other way, and saying "increasing for x < 3" would be flatly untrue on the middle stretch.

A local maximum beats its immediate neighbours. An absolute maximum beats everything on the interval. They are different claims, and on a closed interval [a, b] the second one has a short, complete recipe.

On [a, b], with f continuous, the extreme values can only be reached at

1. a critical number inside the interval, or 2. an endpoint, x = a or x = b.

That is the whole list. So work out f at every candidate, put the heights side by side, and pick the largest and the smallest. This is the candidates test, and it needs no sign chart and no second derivative — just arithmetic and a comparison.

f(x) = 2x³ - 21x² + 72x - 2 on [1, 11] f'(x) = 6x² - 42x + 72 = 6(x - 3)(x - 4)

Candidates: the endpoints 1 and 11, and the critical numbers 3 and 4, which both lie inside.

f(1) = 51 f(3) = 79 f(4) = 78 f(11) = 911

The absolute maximum value is 911, at the endpoint x = 11. The absolute minimum value is 51, at the other endpoint. The interior critical numbers, which are the ones the sign chart makes a fuss about, produced neither.

That is the trap the whole idea exists to catch. x = 3 is a perfectly good local maximum — it beats everything nearby — and it is nowhere near the highest point of the interval, because the curve turns round again at x = 4 and climbs past it. A local high point is not automatically the absolute high point, and the only way to know is to compare the endpoints against it.

Two more habits worth building. Say whether the answer wanted is the value (a height, f of something) or the place (an x): they are different numbers and the question always specifies. And discard any critical number that falls outside the interval — it is not a candidate at all, however interesting it is elsewhere.

Why does the recipe work? Because a continuous function on a closed interval really does reach a highest and a lowest value — that is the Extreme Value Theorem — and if that value is not at an end, it is a local extreme, and so it sits at a critical number. Continuity and a closed interval are the entire price of entry; the theorem asks nothing about derivatives, and it says nothing about *where* the extremes are.

The last idea connects the average behaviour of f across an interval with the instantaneous behaviour somewhere inside it.

The Mean Value Theorem. If f is continuous on [a, b] and differentiable between a and b, then there is at least one c strictly between a and b with

f'(c) = (f(b) - f(a)) / (b - a)

In words: at some instant, the actual slope equals the average slope. Drive 120 km in 2 hours and, at some moment, your speedometer read exactly 60 km/h — the theorem is a promise about that moment, and it never says which moment.

Finding c is a two-sided calculation. Work out the number on the right first, then set f'(x) equal to it and solve.

f(x) = 3x² + 8x on [0, 5] average rate of change = (f(5) - f(0)) / (5 - 0) = (115 - 0) / 5 = 23 f'(x) = 6x + 8, so 6c + 8 = 23 and c = 5/2

For a parabola c always lands on the midpoint of the interval, because f' is a straight line and a straight line hits its average value halfway along. Do not carry that habit any further: for a cubic the derivative is curved, the equation to solve is a quadratic, and the answer is whatever it is — often not the midpoint, and often one of two roots, only one of which lies inside the interval. The other root is not an answer; the theorem promises a c strictly between a and b.

Rolle's theorem is the special case where the two endpoint heights are equal. If f is continuous on [a, b], differentiable between a and b, and f(a) = f(b), then somewhere inside f'(c) = 0. Geometrically: leave at one height, come back to the same height, and at some point you must have turned round.

All three conditions have to hold before either theorem promises anything, and each one can go missing:

• ◆f(a) and f(b) come out different, so Rolle does not apply (the Mean Value Theorem still does).
• ◆f has a sharp corner inside the interval, as f(x) = |x - 2| does at x = 2, so it is not differentiable there and neither theorem applies.
• ◆f has a break in it, so it is not continuous and neither theorem applies.

Checking the conditions is not a formality. A theorem whose conditions are not met makes no claim at all, true or false.