The sign of f′ tells you where a curve climbs, where it dives, and exactly where it turns around.
A derivative is more than a slope at one point — it is a spotlight on the whole shape of a curve. Wherever the derivative f′ is positive, the function is climbing; wherever it is negative, the function is sliding downhill; and right where it crosses zero, the curve pauses to turn around. Read the sign of f′ along the number line and you have already mapped every hill and valley — no plotting of dozens of points required. In this lesson we turn that single idea into a toolkit: find the intervals of increase and decrease, locate the peaks and valleys, decide which is which with the first-derivative test, hunt down the largest and smallest values on a closed interval (where the endpoints matter as much as the critical points), and finally sketch a curve from its derivative alone.
You already know that f′(x₀) is the slope of the tangent at x₀ — the instant rate of change. Now ask a simpler, blunter question of that slope: is it positive or negative? That single bit of information decides the direction of the curve.
So we never need the actual value of f′, only its sign. A clean way to display that sign across the whole domain is a sign chart: a number line that marks each place f′ = 0 and labels the sign of f′ in every interval between. Green plus signs (↑ increasing) and red minus signs (↓ decreasing) make the shape jump out.
The direction of a curve is governed by the sign of its derivative: f′ > 0 ⇒ rising, f′ < 0 ⇒ falling, f′ = 0 ⇒ a flat (critical) moment. This is a statement about f′, not about f itself — a function can be positive while falling, or negative while rising.
To find where a curve rises or falls, follow three steps. (1) Compute the derivative. (2) Solve f′(x) = 0 — the solutions are the critical points; they cut the number line into intervals. (3) Test the sign of f′ in each interval (just pick one sample point inside). Where f′ > 0 the function increases; where f′ < 0 it decreases.
The derivative is f′(x) = 3x² − 3 = 3(x² − 1). Set it to zero: 3(x² − 1) = 0 ⇒ x² = 1 ⇒ x = −1 or x = 1. These two critical points split the line into three pieces. Test a point in each:
• On (−∞, −1), try x = −2: f′(−2) = 3(4 − 1) = +9 > 0 → increasing.
• On (−1, 1), try x = 0: f′(0) = 3(0 − 1) = −3 < 0 → decreasing.
• On (1, ∞), try x = 2: f′(2) = 3(4 − 1) = +9 > 0 → increasing.
Conclusion: f increases on (−∞, −1), decreases on (−1, 1), increases on (1, ∞).
Watch the sign chart of x³ − 3x: at x = −1 the curve switches from rising to falling — that is a local maximum, a peak. At x = 1 it switches from falling to rising — a local minimum, a valley. At both, the tangent is flat: f′ = 0.
A point where f′ = 0 (or where f′ does not exist) is called a critical point. Every local peak or valley of a smooth curve occurs at a critical point — so to find the turning points, we only ever look among the critical points. Their heights are the extreme values:
peak height = f(−1) = (−1)³ − 3(−1) = 2 · valley height = f(1) = (1)³ − 3(1) = −2.
A flat tangent does not guarantee a peak or valley. Take g(x) = x³: its derivative g′(x) = 3x² is zero at x = 0, yet g keeps rising on both sides (g′ stays positive). The sign of g′ never changes, so x = 0 is a critical point but neither a maximum nor a minimum — just a momentary flattening. A turning point needs a genuine sign change, which is exactly the next test.
To classify a critical point, look at how f′ changes sign as you cross it, left to right:
| f′ just left | f′ just right | verdict |
|---|---|---|
| + (rising) | − (falling) | local maximum (a peak) |
| − (falling) | + (rising) | local minimum (a valley) |
| + | + | no change → neither |
| − | − | no change → neither |
In words: + to − is a max, − to + is a min, and no change is neither. For x³ − 3x, at x = −1 the sign goes + → − (a max); at x = 1 it goes − → + (a min). For x³, at x = 0 it goes + → + (no change — neither), exactly as the watch-out warned.
A local peak is the tallest point in its neighborhood. But on a closed interval [a, b] we often want the global (absolute) maximum and minimum — the single highest and lowest values over the whole interval. Here is the catch: those can sit at a critical point or at an endpoint. A curve can keep climbing right up to the edge x = b, so the highest point is the endpoint, not any interior turn.
On a closed interval [a, b], the global max and min of a continuous function are found among a short list of candidates: every critical point inside (a, b), plus the two endpoints a and b. Evaluate f at each candidate and simply compare — the biggest value is the global max, the smallest is the global min.
Critical points (from f′ = 3x² − 3 = 0) are x = −1 and x = 1, both inside the interval. The candidates are the critical points and the endpoints: x = −2, −1, 1, 1.5. Evaluate:
f(−2) = −2 · f(−1) = 2 · f(1) = −2 · f(1.5) = −1.125
Global maximum = 2 at x = −1 (a critical point). Global minimum = −2, reached at both x = −2 (an endpoint) and x = 1 (a critical point). Miss the endpoints and you might miss the true extreme.
Now everything comes together. To sketch a curve by hand, you no longer plot a fog of points — you assemble its shape from the sign chart:
For f(x) = x³ − 3x: rise to the peak (−1, 2), fall to the valley (1, −2), then rise again — a smooth S-shaped cubic. The derivative dictated the whole silhouette.
"Increasing" is decided by f′ > 0 — not by f being above the x-axis. On (1, 2) the cubic x³ − 3x is still negative in value, yet f′ = 3x² − 3 > 0 there, so the curve is climbing. Sign of the derivative sets direction; sign of the function only tells you above/below the axis.
| Question | What to do | The rule |
|---|---|---|
| Where does f rise / fall? | Sign of f′ | f′ > 0 rise · f′ < 0 fall |
| Where are the turns? | Solve f′(x) = 0 | critical points |
| Peak or valley? | First-derivative test | +→− max · −→+ min · no change neither |
| Global max/min on [a, b]? | Compare candidates | critical points and endpoints a, b |
| Sketch the curve | Stitch the sign chart | rise · turn · fall · turn · rise |
Next, in Derivatives at Work: Optimization, we put this to use — the biggest box, the cheapest can, the most profit all live where f′ = 0 (after we check the endpoints). It rests on the differentiation rules from The Differentiation Toolbox.
Six questions to lock it in. Tap the answer you think is right.
This lesson is the heart of curve analysis in AP Calculus AB/BC and aligns with HSF-IF.B.4 and HSF-IF.B.6: students read intervals of increase and decrease, relative maxima and minima, and end behavior directly from the function's rate of change. The recurring traps are worth naming at home: a flat tangent (f′ = 0) does not by itself create a peak or valley — only a sign change of f′ does (x³ is the classic counterexample) — and on a closed interval the absolute extreme is just as likely to sit at an endpoint as at an interior critical point. Every slope, critical point, and extreme value in the figures here is computed by the page's math engine, never estimated from the drawing.