← Unit 5

Increasing, Decreasing & Extrema

5.3 · 5.4  —  where a curve climbs, where it falls, and how the single sign of f′ pinpoints every peak and valley — the First-Derivative Test.
A function’s whole shape comes down to one question, asked over and over: is it heading up or down right now? The first derivative answers it. Where f′ > 0 the graph climbs; where f′ < 0 it falls; and at the exact handoff — where f′ changes sign — sits a peak or a valley. Master that one sign and you can read a function’s rises, falls, and extrema without ever plotting a single point.

Watch the test find them

The curve is f; the graph stacked beneath it is f′. Drag the point — its tangent tilts, glowing green where the slope is positive and red where it’s negative — while a marker rides f′ below. Every peak and valley of f sits exactly where f′ crosses zero. Press Run the test and it sweeps across, reads the sign of f′ on each side of every critical point, and stamps the verdict: +→− a max, −→+ a min, no flip → a shelf.

f′>0 · rising f′<0 · falling ▲ max ▽ min ○ shelf
at x
f′(x)
slope

Why the sign of f′ is the whole story

It looks like a rule you just memorize — but it falls straight out of last lesson’s Mean Value Theorem. Four steps take you from “positive slope” to the full First-Derivative Test.

The First-Derivative Test, precisely

Everything you need to classify a peak or valley lives in the sign of f′ on either side of a critical point.

The test the procedure

Find the critical points — where f′=0 or f′ DNE. At each one, check the sign of f′ just to its left and just to its right:

No sign change ⇒ not an extremum — just a level shelf.

The sign chart how to organize it

Mark every critical point on a number line; they cut it into intervals. Pick one test value inside each, plug into f′, and record a + or .

The pattern of signs names every increasing/decreasing interval and flags every extremum at once — exactly what the f′ lane traces beneath the curve above.

When a critical point isn’t an extremum

A critical point is only a candidate. Two classic traps — one where f′=0 but nothing turns, one where f′ doesn’t even exist yet there’s a genuine minimum:

A shelf, not a peak

For f(x)=x³, f′(x)=3x² is zero at x=0 — but f′ is positive on both sides. The sign never changes (+ → +), so the graph just keeps climbing. Critical, but not an extremum.

A corner that does count

For f(x)=|x|, f′ is −1 then +1 — undefined right at x=0. Yet the sign flips − → +, so x=0 is a true minimum. Never skip the points where f′ fails to exist.

Four cards to keep

Increasing / decreasing the definition

On an interval where f′>0, f is increasing; where f′<0, it is decreasing. State intervals in terms of x.

Critical points candidates

Interior points where f′(x)=0 or f′(x) does not exist. Every local extremum is a critical point — but not every critical point is an extremum.

Max vs. min don’t swap them

+ → − means the graph rose then fell — a local maximum. − → + means it fell then rose — a local minimum. Match the arrows to the picture.

Local vs. absolute scope

The first-derivative test finds local (relative) extrema — biggest/smallest nearby. For the absolute extremum on a closed interval, still compare those against the endpoints.

Practice

Quick check

Where it goes wrong

Reading f instead of f′

Increase, decrease, and extrema are decided by the sign of the derivative — not by whether f itself is positive or negative.

Trusting every f′=0

You need a sign change. \(f(x)=x^3\) has \(f'(0)=0\) but keeps rising — a shelf, not a peak.

Skipping f′-undefined points

Corners and cusps — like \(|x|\) or \(x^{2/3}\) at 0 — are critical points too, and can be real extrema.

Swapping max and min

\(+\to-\) is a max (up then down); \(-\to+\) is a min (down then up). Keep the order straight.

Carry these forward