← Back to course

Concavity, Inflection Points and the Second Derivative Test

You’ll be able to

Concavity is the sign of the second derivative

f is concave up where f″ > 0 and concave down where f″ < 0. The intuition worth carrying is that concave up means f′ is increasing — the slopes are getting larger as you move right, whether they are negative and shrinking toward zero or positive and growing. Concavity is a statement about how the slope is changing, which is why it is a property of f″ and says nothing directly about whether f is rising or falling.

An inflection point requires a sign change

A point of inflection is where the concavity changes. Finding where f″ = 0 produces candidates, not answers: f(x) = x⁴ has f″(0) = 0 and no inflection point there, because f″ = 12x² is positive on both sides. Concavity can also change where f″ is undefined. So the procedure mirrors the First Derivative Test one level up: list candidates from f″ = 0 and f″ undefined, then test the sign of f″ on each side.

The Second Derivative Test, and its blind spot

At a critical point c where f′(c) = 0: if f″(c) > 0 the curve is concave up there, so c is a relative minimum; if f″(c) < 0, a relative maximum. If f″(c) = 0 the test is inconclusive and tells you nothing — fall back on the First Derivative Test. The Second Derivative Test is faster when it works, but it requires f′(c) = 0 exactly, so it cannot classify a critical point at a corner where f′ is undefined.

Concavity and the Second Derivative Test
f″ > 0 ⇒ concave up ⇒ relative min at a critical point · f″ < 0 ⇒ concave down ⇒ relative max · f″ = 0 ⇒ inconclusive
Inflection points need a sign change in f″, which f″ = 0 alone does not guarantee.
Worked example

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

  1. 1.f′(x) = 4x³ − 12x², and f″(x) = 12x² − 24x = 12x(x − 2).
  2. 2.Candidates for inflection: f″ = 0 at x = 0 and x = 2. f″ is a polynomial, so it is defined everywhere and there are no other candidates.
  3. 3.Test signs: at x = −1, f″ = 12(−1)(−3) = 36 > 0. At x = 1, f″ = 12(1)(−1) = −12 < 0. At x = 3, f″ = 12(3)(1) = 36 > 0.
  4. 4.So f is concave up on (−∞, 0), concave down on (0, 2), and concave up on (2, ∞).
  5. 5.Concavity changes at both x = 0 and x = 2, so both are inflection points.
Answer: Concave up on (−∞, 0) and (2, ∞), concave down on (0, 2); inflection points at x = 0 and x = 2, where f″ changes sign.
Tip

An inflection point is a point on the graph of f, so if a question asks for "the inflection point" as a coordinate pair, you still have to compute f at that x. Reporting only the x-value answers a slightly different question.

Checkpoint

g″(x) = (x − 3)² (x + 1). At which x does the graph of g have an inflection point?

Checkpoint

h′(2) = 0 and h″(2) = 0. What can be concluded about x = 2?

On the exam

The exam asks "for what values of x does the graph of f have a point of inflection? Justify." A justification that stops at "f″(x) = 0 there" is incomplete every time — say that f″ changes sign.

Answer the 2 checkpoints as you read.

Sign in to save your progress