Curve Sketching with Derivatives

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Skill: curve-sketching

Curve Sketching with Derivatives

Read a graph from its derivatives: f′ tells you where the curve climbs and falls, f″ tells you how it bends. Find critical points, classify them, locate inflection points, and sketch with confidence.

1 f′: where the curve climbs and falls

The sign of f′ is the whole story of up vs. down: f′ > 0 means f is increasing, f′ < 0 means decreasing. The x-values where f′ = 0 (or f′ is undefined) are the critical points — the only places a max or min can hide.

First-derivative test
f′ changes − → + : local min   |   f′ changes + → − : local max

If f′ doesn’t change sign at a critical point, it’s neither — just a flat spot (think x3 at x = 0).

Example f(x) = x3 − 3x2 − 9x + 12
  1. f′(x) = 3x2 − 6x − 9 = 3(x − 3)(x + 1) → critical points x = −1, 3.
  2. Sign chart: f′ > 0 on (−∞, −1), f′ < 0 on (−1, 3), f′ > 0 on (3, ∞).
  3. So f increases, then decreases, then increases: local max at x = −1, local min at x = 3.

2 f′′: how the curve bends

The second derivative controls the bend: f′′ > 0 means concave up (cup shape), f′′ < 0 means concave down (cap shape). Where the concavity flips — with f′′ = 0 and a genuine sign change — sits an inflection point.

Second-derivative test (for classifying critical points)
f′(c) = 0 and f′′(c) > 0 → local min   |   f′(c) = 0 and f′′(c) < 0 → local max

If f′′(c) = 0 the test says nothing — fall back to the first-derivative sign chart.

Example Same f(x) = x3 − 3x2 − 9x + 12
  1. f′′(x) = 6x − 6 = 0 at x = 1; f′′ changes − → + there.
  2. So concave down on (−∞, 1), concave up on (1, ∞), inflection point at x = 1.
  3. Check the classification: f′′(−1) = −12 < 0 → max at x = −1; f′′(3) = 12 > 0 → min at x = 3. Matches the sign chart.
local max local min inflection point concave down (cap) concave up (cup)
A full sketch reads off f′ (up/down) and f′′ (bend): max, min, and the inflection point where the bend flips.

3 The sketching checklist

For any differentiable function, in order:

  1. Domain — where is f defined? (Watch denominators and even roots.)
  2. Intercepts — f(0) and the x-intercepts if they’re easy.
  3. f′: critical points, intervals of increase/decrease, local max/min.
  4. f′′: concavity intervals and inflection points.
  5. Assemble: plot the special points, then connect them respecting up/down and the bend.
Example Quick classify: f(x) = x3

f′(x) = 3x2 = 0 at x = 0 — a critical point. But f′ ≥ 0 everywhere, so there’s no sign change: x = 0 is neither a max nor a min (it’s an inflection point with a flat tangent).

4 Common mistakes

Three traps, each with the wrong version and the fix.

1. Calling every critical point a max or min

Wrong: “f′(0) = 0 for f(x) = x3, so there’s a max/min at x = 0.”    Right: critical points include where f′ = 0 or is undefined — test each one. x3 just flattens at 0.

2. Confusing concavity with increasing/decreasing

Wrong: “f′′ > 0 means the function is increasing.”    Right: f′ controls up/down; f′′ controls the bend. A function can decrease while concave up (right half of a valley).

3. Inflection without a sign change

Wrong: “f′′(2) = 0, so x = 2 is an inflection point.”    Right: f′′ = 0 is necessary but not sufficient — the concavity must actually flip sides.

Key vocabulary

  • Critical point — x = c where f′(c) = 0 or f′(c) is undefined (and c is in the domain).
  • Increasing / decreasing — decided by the sign of f′ on an interval.
  • Concave up / down — decided by the sign of f′′; cup vs. cap shape.
  • Inflection point — where concavity changes (f′′ = 0 plus a sign change).
  • First/second derivative test — two ways to classify a critical point: sign change of f′, or sign of f′′.

5 Quick checks

Cover the answers, decide, then reveal.

1. f′ > 0 on (0, 2) and f′ < 0 on (2, 5). What happens at x = 2?
Local maximum. The function climbs into x = 2 and falls after — a peak.
2. f′(3) = 0 and f′′(3) = −4. Classify x = 3.
Local maximum. Second-derivative test: f′′ < 0 means concave down there, so the critical point is a peak.
3. True or false: if f′′(1) = 0 then x = 1 is an inflection point.
False. You also need the concavity to change sign (e.g. f(x) = x4 has f′′(0) = 0 but stays concave up).
Practice now Try the curve analyzer Open the sign-chart lab
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