Calculus I › Applications of derivatives › full formula sheet
The first derivative test
Critical points are suspects — the sign change is the verdict.
A critical point c has f′(c) = 0 or f′(c) undefined (with f defined at c). Local max at c means f(c) ≥ nearby values; local min means f(c) ≤ nearby values.
Before this lesson: Increasing / decreasing · Definition of the derivative
Where it comes from
The problem: f(x) = x³ − 3x has critical points at x = −1 and x = 1. Which is the peak, which the valley — or are they neither? The naive rule, “f′ = 0 means max or min,” is executed by a single counterexample — f(x) = x³ at 0:
The right intuition is a walk: hike uphill then downhill and you crossed a peak; hike downhill then uphill and you crossed a valley. The sign of f′ is your trail log — + means climbing, − means descending — so the change in the sign at a critical point is the verdict: +→− peak, −→+ valley, no change acquittal.
Before reading on: f climbs toward c, then falls away. What must the sign of f′ do at c — and what does that make the point?
Derivation
Let c be a critical point, f continuous near c, and suppose f′ changes from + to − at c. The increasing/decreasing theorem does the rest.
And with no sign change? Then f climbs through c (x³ at 0) or falls through c (−x³ at 0) — c is a terrace point, neither max nor min. The test’s silence is itself an answer.
Before reading on: f′(0) = 0 for f(x) = x³. Max, min, or neither? Commit to an answer before you read the verdict.
How to use it
The procedure, every time:
- Find all critical points — f′ = 0 and f′ undefined (with f defined). Missing the undefined ones is the classic omission.
- Build the sign chart of f′ around each critical point (same chart as the increasing/decreasing page).
- Read the verdict at each critical point: +→− local max; −→+ local min; no change neither.
- Compute f(c) for the max/min values — the test classifies the location; f gives the height.
Its edge over the second derivative test
This test never goes silent: it works when f′(c) is undefined (cusps) and when f″(c) = 0. The price is a full sign chart. Use the second derivative test for speed when it applies; fall back here when it doesn’t.
Repeated factors don’t change signs
In f′(x) = 12x²(x−1), the x² factor is ≥ 0 on both sides of 0 — an even power never flips a sign. Only odd-power factors change the verdict (see Example 4).
Worked examples
Four classifications, 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
- Critical points. f′(x) = 3x² − 3 = 3(x−1)(x+1) = 0 at x = −1, 1.
- Sign chart. + on (−∞, −1), − on (−1, 1), + on (1, ∞).
- Verdicts. At x = −1: +→− → local max, f(−1) = −1 + 3 = 2. At x = 1: −→+ → local min, f(1) = 1 − 3 = −2.
Your turn: f(x) = x³ − 12x — classify each critical point.
Answer: Local max at x = −2 (f = 16); local min at x = 2 (f = −16).
f′ = 3x² − 12 = 0 at x = ±2; signs +/−/+. At −2: +→− max, f(−2) = −8 + 24 = 16. At 2: −→+ min, f(2) = 8 − 24 = −16 ✓.
Example 2 — the acquittal: f(x) = x³
- Critical points. f′(x) = 3x² = 0 at x = 0.
- Sign chart. 3x² ≥ 0 everywhere: + on both sides of 0.
- Verdict. No sign change → neither — a terrace point. (Why? The function climbs through 0 uninterrupted.)
Your turn: f(x) = −x³ — classify x = 0.
Answer: Neither — a terrace point.
f′(x) = −3x² is − on both sides of 0: no sign change, no verdict. The function falls straight through.
Example 3 — the cusp: f(x) = x2/3
- Critical points. f′(x) = (2/3)x−1/3 = 2/(3∛√x) — never 0, undefined at x = 0. So x = 0 is critical (f(0) = 0 is defined).
- Sign chart. For x < 0, ∛√x < 0 → f′ < 0. For x > 0, f′ > 0. So − left, + right.
- Verdict. −→+ → local min, f(0) = 0 — a sharp cusp valley. (The second derivative test can’t touch this one; the first test handles it.)
Your turn: f(x) = x4/3 — classify x = 0.
Answer: Local min, f(0) = 0.
f′(x) = (4/3)x1/3: for x < 0, x1/3 < 0 so f′ < 0; for x > 0, f′ > 0. −→+ gives a min.
Example 4 — the repeated factor: f(x) = 3x4 − 4x³
- Critical points. f′(x) = 12x³ − 12x² = 12x²(x−1) = 0 at x = 0, 1.
- Sign chart. 12x² ≥ 0 always — the sign comes entirely from (x−1): − for x < 1 (x ≠ 0), + for x > 1. (Why? An even power can’t flip a sign.)
- Verdicts. At x = 0: −→−, no change → neither. At x = 1: −→+ → local min, f(1) = 3 − 4 = −1.
Your turn: f(x) = x⁴ − 4x³ — classify each critical point.
Answer: Neither at x = 0; local min at x = 3 (f = −27).
f′ = 4x³ − 12x² = 4x²(x − 3) = 0 at x = 0, 3. 4x² ≥ 0 never flips, so the sign comes from (x − 3): − left of 3 (x ≠ 0), + right. At 0: −→−, neither; at 3: −→+, min, f(3) = 81 − 108 = −27 ✓.
Memorization tips
- The trail chant: “up then down, peak; down then up, valley.” +→− max, −→+ min.
- Suspects, not verdicts: f′(c) = 0 only earns c a trial. The sign change convicts or acquits.
- Undefined counts: cusps and corners are critical points — x2/3 at 0 is the poster child.
- Even powers don’t flip: x², x4 keep their sign through their zero — only odd powers change the verdict.
- Where, then what: the test gives the location c; f(c) gives the max/min value. Report both.
- Silent test = answer: no sign change means “neither” — the terrace point is a verdict too.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the first derivative test 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
What is the first derivative test?
At a critical point c: if f′ changes from + to −, f has a local max at c; if f′ changes from − to +, a local min; if f′ doesn’t change sign, neither (a terrace point like x³ at 0).
Does f′(c) = 0 mean c is a max or min?
Not by itself. f(x) = x³ has f′(0) = 0 but 0 is neither a max nor a min — the sign of f′ is + on both sides. Only a sign change delivers the verdict.
Can the first derivative test handle critical points where f′ is undefined?
Yes — that’s its advantage over the second derivative test. f(x) = x2/3 has an undefined derivative at 0, yet f′ changes − to + there, so 0 is a local min (a cusp).
First vs. second derivative test — which to use?
The first derivative test always works (it just needs a sign chart) but takes longer. The second derivative test is faster when f″(c) ≠ 0, but goes silent when f″(c) = 0 or f′ is undefined — then fall back to the first.
Why does a + to − sign change give a local max?
Left of c, f′ > 0 so f climbs toward c; right of c, f′ < 0 so f falls away. Climbing in, falling away — c is a peak, i.e. f(c) ≥ nearby values.
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