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Reconstructing f from f′ and f″

You’ll be able to

The two-row sign chart

Almost every Unit 5 question reduces to a table with two rows. The f′ row controls direction: f′ > 0 means f is increasing, f′ < 0 means decreasing, and a sign change in f′ marks a local extremum — positive-to-negative gives a maximum, negative-to-positive gives a minimum. The f″ row controls curvature: f″ > 0 means f is concave up, f″ < 0 means concave down, and a sign change in f″ marks an inflection point. Build both rows across the same number line, mark every point where either derivative is zero or undefined, and the shape of f is fully determined up to vertical position. That last qualification matters: sign information gives shape but never the actual height, which is why these questions supply one value of f when they want a specific curve.

Four shapes, and the vocabulary for each

The four sign combinations describe four visually distinct behaviors, and being able to name them prevents most errors. f′ > 0, f″ > 0: increasing and concave up — increasing at an increasing rate, the shape of accelerating growth. f′ > 0, f″ < 0: increasing and concave down — increasing at a decreasing rate, growth that is levelling off, which is the most common applied case and the one described in words most often. f′ < 0, f″ < 0: decreasing at an increasing rate, falling ever faster. f′ < 0, f″ > 0: decreasing at a decreasing rate, a decline that is slowing — the shape of cooling coffee or of a decaying quantity approaching an asymptote. Learn the third-column phrasings, because applied problems state conditions in exactly that language.

Inflection points require a change, not just a zero

An inflection point occurs where f″ changes sign — where concavity actually switches. Two failure modes the exam exploits. First, f″(c) = 0 is not sufficient: for f(x) = x⁴, f″(x) = 12x² is zero at x = 0 but is positive on both sides, so the function is concave up throughout and there is no inflection point. Second, an inflection point can occur where f″ is undefined, provided the sign changes there and f itself is continuous. So the procedure is: find candidates where f″ = 0 or f″ is undefined, then test the sign on each side, and report only those where it flips. A response that lists the zeros of f″ as inflection points has skipped the only step that matters.

Second derivative test
If f′(c) = 0 and f″(c) > 0, f has a local minimum at c; if f′(c) = 0 and f″(c) < 0, a local maximum
If f″(c) = 0 the test is inconclusive — fall back to the first derivative test, which always works.
Worked example

For f(x) = x⁴ − 4x³, find the intervals of concavity and all inflection points.

  1. 1.First derivative: f′(x) = 4x³ − 12x².
  2. 2.Second derivative: f″(x) = 12x² − 24x = 12x(x − 2). Candidates where f″ = 0 are x = 0 and x = 2.
  3. 3.Test the sign of f″ on each of the three intervals. At x = −1: 12(−1)(−3) = 36 > 0, so concave up on (−∞, 0).
  4. 4.At x = 1: 12(1)(−1) = −12 < 0, so concave down on (0, 2).
  5. 5.At x = 3: 12(3)(1) = 36 > 0, so concave up on (2, ∞).
  6. 6.The sign changes at both x = 0 and x = 2, so both are inflection points. Their heights: f(0) = 0, and f(2) = 16 − 32 = −16.
Answer: Concave up on (−∞, 0) and (2, ∞), concave down on (0, 2). Inflection points at (0, 0) and (2, −16), since concavity changes at each.
Watch out

f″(c) = 0 alone never establishes an inflection point — x⁴ at the origin is the counterexample to memorize. And when asked for the inflection point, give both coordinates; a bare x value answers a different question than the one asked.

Checkpoint

A quantity is described as "increasing at a decreasing rate." This means

Checkpoint

For which function does f″(0) = 0 without x = 0 being an inflection point?

Answer the 2 checkpoints as you read.

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