Knowing where a graph rises and falls (Increasing, Decreasing and the First Derivative Test) is not the whole story: a rising graph can bend upward like a cup or downward like a cap. The second derivative from Higher Derivatives measures the bending.
Key idea
$f$ is concave up where $f'' > 0$ (slopes increasing, graph bends up) and concave down where $f'' < 0$. An inflection point is where the concavity changes. Second derivative test: if $f'(c) = 0$ and $f''(c) > 0$, local min; if $f''(c) < 0$, local max; if $f''(c) = 0$, the test says nothing.
Why it works
$f''$ is the derivative of $f'$, so $f'' > 0$ means the slope is increasing. Walking left to right, the tangent line keeps turning counterclockwise, and the graph curves upward, lying above its tangent lines. At a flat point ($f'(c) = 0$) with $f''(c) > 0$, the slope goes from negative to positive, which is the first derivative test's local min.
Worked example: Concavity and an inflection point
For $f(x) = x^3 - 6x^2 + 5$, find the intervals of concavity and the inflection point.
- $f'(x) = 3x^2 - 12x$, $f''(x) = 6x - 12$.
- $f'' < 0$ for $x < 2$: concave down. $f'' > 0$ for $x > 2$: concave up.
- The concavity changes at $x = 2$, where $f(2) = 8 - 24 + 5 = -11$.
Answer: Concave down on $(-\infty, 2)$, up on $(2, \infty)$; inflection point $(2, -11)$.
Worked example: The second derivative test
Classify the critical numbers of $f(x) = x^3 - 3x$.
- $f'(x) = 3x^2 - 3 = 0$ at $x = \pm 1$. $f''(x) = 6x$.
- $f''(1) = 6 > 0$: local min. $f''(-1) = -6 < 0$: local max.
Answer: Local max at $x = -1$, local min at $x = 1$.
Common mistake
Calling every point with $f''(c) = 0$ an inflection point. For $f(x) = x^4$, $f''(x) = 12x^2$ is 0 at $x = 0$, but it is positive on both sides: the graph is a cup all the way through. An inflection point needs a sign change of $f''$. Check the sign on both sides, just as with the first derivative test.
When $f''(c) = 0$ at a critical number, fall back on the first derivative test. $x^4$ (min), $-x^4$ (max) and $x^3$ (neither) all have $f'(0) = f''(0) = 0$.