Concavity, Inflection Points, and the Second Derivative Test

By the end of this lesson, you will be able to determine intervals of concavity from $f''$, find inflection points, and apply the second derivative test to classify critical points.

Concavity

Concave up ($f'' > 0$): the curve looks like a bowl (cup ∪); the tangent line is below the curve; $f'$ is increasing.

Concave down ($f'' < 0$): the curve looks like a hill (cap ∩); the tangent line is above the curve; $f'$ is decreasing.

Memory: $f'' > 0$ → accelerating → concave UP (like accelerating upward).

Inflection Points

An inflection point is a point where the concavity changes (from up to down or down to up).

Candidates: where $f''(c) = 0$ or $f''(c)$ does not exist.

Important: $f''(c) = 0$ does NOT guarantee an inflection point. The concavity must actually change.

Counterexample: $f(x) = x^4$. $f''(x) = 12x^2 = 0$ at $x = 0$, but $f'' \geq 0$ everywhere → no inflection point.

Test: make a sign chart for $f''$. If the sign changes at $c$, it's an inflection point.

Second Derivative Test for Local Extrema

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

$f''(c)$	Conclusion
$f''(c) > 0$	Local minimum (concave up at $c$ → bottom of a valley)
$f''(c) < 0$	Local maximum (concave down at $c$ → top of a hill)
$f''(c) = 0$	Inconclusive — use first derivative test

When the second derivative test is inconclusive: $f(x) = x^4$ has $f'(0) = f''(0) = 0$, but it has a minimum at $x = 0$. The second derivative test fails; use the first derivative test instead.

Example

$f(x) = x^4 - 4x^3 + 6x^2 - 4x$

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

Critical point: $x = 1$ (where $f'(1) = 0$)

$f''(x) = 12x^2 - 24x + 12 = 12(x-1)^2$

$f''(1) = 0$ → Second derivative test is inconclusive.

First derivative test: $f'(x) = 4(x-1)^3 < 0$ for $x < 1$ and $> 0$ for $x > 1$ → local at $x = 1$.