Calculus I › Applications of derivatives › full formula sheet
Increasing / decreasing
Read a function’s uphill and downhill stretches straight from its derivative’s sign.
On an interval I: increasing means x1 < x2 ⇒ f(x1) < f(x2); decreasing means x1 < x2 ⇒ f(x1) > f(x2).
Before this lesson: Definition of the derivative · Mean Value Theorem
Where it comes from
The problem: where does f(x) = x³ − 3x climb and where does it fall? The naive guess — “f > 0 means increasing” — is a trap. Exhibit A, f(x) = −x:
The right intuition: the derivative is a slope-o-meter. Positive slope means the graph climbs as x grows; negative slope means it falls. So the sign chart of f′ is the increasing/decreasing chart of f. For f(x) = x³ − 3x, f′(x) = 3x² − 3 = 3(x−1)(x+1):
The zeros of f′ (−1 and 1) are the fences between uphill and downhill country.
Before reading on: if f′(x) > 0 at every point of an interval, could f still dip down somewhere inside it? Which theorem from this unit settles the question?
Derivation
Why does f′’s sign dictate the direction? The Mean Value Theorem turns “slope” into “change” in one line.
What about f′ = 0 at isolated points? f(x) = x³ has f′(0) = 0 yet is increasing on all of ℝ. The proof only needs f′(c) > 0 for the c that MVT produces — a single flat point can’t flip the sign of a whole interval. “Increasing” tolerates flat spots; it forbids flat stretches.
Before reading on: f(x) = 1/x has f′ < 0 everywhere it is defined. Is it decreasing on its whole domain? Sketch it before you decide.
How to use it
The procedure, every time:
- Write the domain of f. Do this first — monotonicity is always claimed per interval, and domain breaks (like x = 0 for 1/x) split the answer.
- Compute f′ and find critical points: where f′ = 0 or f′ is undefined (but f is defined).
- Split the domain at the critical points into open intervals.
- Test the sign of f′ on each interval — plug in one test point, or read signs off a factored form.
- Conclude: f′ > 0 ⇒ increasing; f′ < 0 ⇒ decreasing, interval by interval.
The 1/x trap
f(x) = 1/x has f′(x) = −1/x² < 0 wherever defined — so it is decreasing on (−∞, 0) and decreasing on (0, ∞). But it is not decreasing on (−∞, 0) ∪ (0, ∞): f(−1) = −1 < 1 = f(1) violates the definition across the break. Never glue intervals across a domain gap.
When to reach for it
Curve sketching, finding where a function rises/falls, and as the setup for the first derivative test (next page) — which classifies what happens at the critical points.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the basic move: f(x) = x³ − 3x
- Domain. All reals — no breaks.
- Critical points. f′(x) = 3x² − 3 = 3(x−1)(x+1) = 0 at x = −1, 1. (Why factor? The factored form shows each factor’s sign at a glance.)
- Intervals. (−∞, −1), (−1, 1), (1, ∞).
- Signs. At x = −2: 3(4)−3 = 9 > 0. At x = 0: −3 < 0. At x = 2: 9 > 0. (Or read the factors: for x > 1 both factors positive; between −1 and 1 exactly one is negative.)
- Conclude. Increasing on (−∞, −1), decreasing on (−1, 1), increasing on (1, ∞).
Your turn: f(x) = x³ − 12x — where is it increasing / decreasing?
Answer: Increasing on (−∞, −2) and (2, ∞); decreasing on (−2, 2).
f′(x) = 3x² − 12 = 3(x² − 4) = 0 at x = ±2. f′(−3) = 15 > 0, f′(0) = −12 < 0, f′(3) = 15 > 0 ✓.
Example 2 — one critical point: f(x) = x² − 4x + 3
- Domain. All reals.
- Critical points. f′(x) = 2x − 4 = 0 at x = 2.
- Intervals. (−∞, 2), (2, ∞).
- Signs. f′(0) = −4 < 0; f′(3) = 2 > 0.
- Conclude. Decreasing on (−∞, 2), increasing on (2, ∞) — the parabola’s valley at x = 2.
Your turn: f(x) = x² + 6x + 5 — where is it increasing / decreasing?
Answer: Decreasing on (−∞, −3); increasing on (−3, ∞).
f′(x) = 2x + 6 = 0 at x = −3. f′(−4) = −2 < 0, f′(0) = 6 > 0 ✓.
Example 3 — the flat spot: f(x) = x³
- Critical points. f′(x) = 3x² = 0 at x = 0 — the only critical point.
- Signs. 3x² ≥ 0 everywhere, and > 0 except at x = 0. So f′ > 0 on (−∞, 0) and on (0, ∞).
- Conclude. Increasing on (−∞, 0) and on (0, ∞) — and since f is continuous at 0, increasing on all of ℝ. The flat spot at 0 is a single point, not a flat stretch.
Your turn: f(x) = −x³ — increasing or decreasing?
Answer: Decreasing on all of ℝ.
f′(x) = −3x² ≤ 0, and < 0 except at the single point x = 0. f is continuous at 0, so it decreases everywhere.
Example 4 — the domain trap: f(x) = 1/x
- Domain. x ≠ 0 — the break at 0 will matter.
- Critical points. f′(x) = −1/x², never 0; undefined at x = 0 (but f is undefined there too — not a critical point, a domain break).
- Intervals. (−∞, 0), (0, ∞).
- Signs. −1/x² < 0 on both — x² is always positive, the minus sign wins.
- Conclude. Decreasing on (−∞, 0) and decreasing on (0, ∞) — separately. Not decreasing on the union: f(−1) = −1 < 1 = f(1).
Your turn: f(x) = ex — where is it increasing / decreasing?
Answer: Increasing everywhere.
f′(x) = ex > 0 for all x — no critical points, no sign flips.
Memorization tips
- Sign of f′, not f: the derivative is the slope-o-meter. Tape over f and read only f′.
- Factor f′: factored form turns sign-testing into sign-reading — each factor flips at its own zero.
- Domain first: breaks in the domain split the answer before you even differentiate.
- Flat spots are fine: x³ proves an isolated f′ = 0 can’t stop an increasing function.
- Per interval, always: “increasing on (2, ∞)”, never “increasing at x = 3”.
- Never glue across gaps: 1/x is the eternal warning — decreasing on each piece, not on the union.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and increasing/decreasing is yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
How do you tell if a function is increasing or decreasing?
Check the sign of the derivative: f′ > 0 on an interval means f is increasing there; f′ < 0 means decreasing. Find where f′ = 0 or is undefined, split the domain at those points, and test the sign on each piece.
Does f(x) > 0 mean f is increasing?
No — that’s the trap. f(x) = −x has f(−1) = 1 > 0 yet is decreasing everywhere. Height (f) and slope (f′) are different things; only the sign of f′ decides uphill vs. downhill.
Can a function be increasing if f′ = 0 somewhere?
Yes, if the zero is isolated. f(x) = x³ has f′(0) = 0 but is increasing on all of ℝ — the flat spot at 0 is a single point, not a flat stretch.
Is 1/x decreasing on its whole domain?
No — only on each interval separately: decreasing on (−∞, 0) and on (0, ∞). Across the break, f(−1) = −1 < 1 = f(1), so it is not decreasing on (−∞, 0) ∪ (0, ∞). Monotonicity is claimed per interval.
Why does the sign of f′ control increasing/decreasing?
By the Mean Value Theorem, f(x2)−f(x1) = f′(c)(x2−x1) with x2−x1 > 0. So the sign of the change matches the sign of f′(c): positive derivative forces uphill, negative forces downhill.
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