Second Derivative and Concavity
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Direct answer
Positive second derivative on an interval means the curve is convex there — bending upward, tangents below the graph; negative means concave downward, tangents above; and a point of inflexion is where concavity actually changes sign, strictly stronger than f″ = 0. The second derivative test converts concavity into classification: at a critical point with f′ = 0, f″ < 0 gives a local maximum, f″ > 0 a local minimum, and f″ = 0 says nothing — x⁴ at the origin has f″ = 0 yet a clear minimum. When the test stays silent, the first derivative's sign change decides, and it never fails. Convexity also proves inequalities: the curve lies above every tangent.
What you must remember
- Concavity test: f″ > 0 on an interval ⇒ convex (cup); f″ < 0 ⇒ concave (cap); the words reverse in some books, the geometry does not.
- Second derivative test: f′ = 0 with f″ < 0 → local maximum; f″ > 0 → local minimum; f″ = 0 → inconclusive, escalate to sign change of f′.
- Inflexion, properly: f″ changes sign at the point (equivalently, the lowest nonzero derivative there is of odd order); f″ = 0 alone certifies nothing — compare x³ against x⁴ at 0.
- Horizontal inflexion: x³ at 0 has f′ = 0 and f″ = 0 yet is neither max nor min — a favourite "which is true" item.
- Monotonicity of f′: f″ > 0 means f′ is increasing — the reading that connects this topic to curve sketching.
- Tangent inequality: on an interval where f″ > 0, f(x) ≥ f(a) + f′(a)(x − a) for every a — convexity's one-line proof of e^x ≥ 1 + x and tan x ≥ x on [0, π/2).
- Testing order: find f′ = 0 points first; f″ matters only at those points for classification.
The silence of f″ = 0
JEE Main asks the classification directly, and the option design always includes the x⁴-style trap: a point where f″ = 0 is labelled a minimum (which it may be!) when the question's function behaves like x³ (which it is not) — the only defence is actually checking the sign change. A parallel Main pattern gives a graph and asks where concavity flips; the answer is where the bending reverses, not merely where the curvature vanishes numerically. JEE Advanced prefers the higher-order language: if f′(a) = f″(a) = 0 and f‴(a) ≠ 0, the point is an inflexion; if f′(a) = f″(a) = f‴(a) = 0 with the fourth derivative positive, it is a minimum — the parity of the first nonzero derivative decides extremum versus inflexion. The inequality tier is the other Advanced face: prove tan x > x on (0, π/2) by convexity. Keep straight under pressure: f″ = 0 is a clue, the sign change is the verdict.
Frequently asked questions
What does the sign of f″ tell you?
f″ > 0 means the curve is convex (bends upward, tangents below); f″ < 0 means concave downward with tangents above the graph.
When does the second derivative test fail?
When f″ = 0 at the critical point — the point may be a maximum, a minimum, or neither, and only the sign change of f′ settles it.
What defines a point of inflexion?
A genuine change in the sign of f″ across the point; equivalently, the lowest-order nonzero derivative there is of odd order.
Can a point be both critical and inflexion?
Yes — x³ at the origin has f′ = 0 and an inflexion, the so-called horizontal inflexion that defeats careless tests.
How does convexity prove e^x ≥ 1 + x?
The curve lies above its tangent at 0, whose equation is y = 1 + x; convexity makes the inequality global.