Concavity and the Second Derivative Test

By the end of this lesson, you will be able to use $f''(x)$ to determine concavity and inflection points, apply the Second Derivative Test to classify critical points, and understand the geometric meaning of concavity.

Concavity

$f$ is concave up on an interval if $f''(x) > 0$ there. The function "holds water" — the slope is increasing.

$f$ is concave down on an interval if $f''(x) < 0$ there. The function is "spills water" — the slope is decreasing.

Geometric: concave up → curve bends away from any tangent (bowl shape); concave down → curve bends toward the tangent.

Inflection Points

An inflection point is where $f$ changes concavity — where $f''$ changes sign.

Finding inflection points:

1. Find where $f''(x) = 0$ or is undefined
2. Check for a sign change of $f''$ at each candidate

If $f''$ doesn't change sign, it's not an inflection point.

Second Derivative Test

At a critical point $c$ where $f'(c) = 0$:

• $f''(c) > 0$: local minimum (concave up at $c$)
• $f''(c) < 0$: local maximum (concave down at $c$)
• $f''(c) = 0$: inconclusive — use the First Derivative Test

Example 1 — Concavity Analysis

$f(x) = x^4 - 6x^2$

$f'(x) = 4x^3 - 12x$; $f''(x) = 12x^2 - 12 = 12(x^2-1) = 12(x-1)(x+1)$

$f''(x) = 0$ at $x = \pm 1$.

$f''(-2) = 12(3) > 0$ (concave up on $(-\infty,-1)$)

$f''(0) = -12 < 0$ (concave down on $(-1,1)$)

$f''(2) = 12(3) > 0$ (concave up on $(1,\infty)$)

Inflection points at $x = \pm 1$: $f(-1) = 1-6 = -5$, $f(1) = -5$.

Example 2 — Second Derivative Test

$g(x) = x^3 - 3x + 2$:

$g'(x) = 3x^2 - 3 = 0 \implies x = \pm 1$

$g''(x) = 6x$

$g''(-1) = -6 < 0$ → local max at $x = -1$; $g(-1) = 4$

$g''(1) = 6 > 0$ → local min at $x = 1$; $g(1) = 0$

Example 3 — When Second Derivative Test Fails