Calculus I · Maxima, Minima, and the Shape of a Function

The second derivative: how a function curves

The derivative of the derivative, $f''$, is the rate of change of the slope — and that measures curving. Watch what its sign means:

• $f'' > 0$: the slope is increasing (getting more positive or less negative as you move right), so the curve bends upward — like a cup, or a valley. We say $f$ is concave up.
• $f'' < 0$: the slope is decreasing, so the curve bends downward — like a cap, or a hilltop. We say $f$ is concave down.

A point where the concavity switches — where $f''$ changes sign — is an inflection point, the spot where a curve stops bending one way and starts bending the other (the steepest point of an S-shaped growth curve, for instance). (Figure 8.3: a single curve colored by the sign of $f''$ — concave-up stretches as cups, concave-down as caps, with the inflection point where they meet.)

This gives a second, often faster, way to classify critical points.

The second-derivative test (and where it came from)

At a critical point $a$ — where $f'(a) = 0$ — the concavity settles the question immediately:

Second-Derivative Test. Suppose $f'(a) = 0$. - If $f''(a) > 0$, then $a$ is a local minimum (the curve cups upward there). - If $f''(a) < 0$, then $a$ is a local maximum (it caps downward). - If $f''(a) = 0$, the test is inconclusive — $a$ could be a max, a min, or neither.

Where does this come from? Straight from Chapter 7. Near a critical point, the Taylor quadratic loses its linear term (because $f'(a) = 0$) and becomes

$$f(x) \approx f(a) + \tfrac12 f''(a)\,(x-a)^2.$$

That is a parabola sitting at $(a, f(a))$. If $f''(a) > 0$ the parabola opens upward — a minimum; if $f''(a) < 0$ it opens downward — a maximum. The second-derivative test is nothing more than reading the leading Taylor term at a critical point. And the inconclusive case is honest: when $f''(a) = 0$, the parabola is flat and the next Taylor term takes over — for $x^4$ (a min), $-x^4$ (a max), and $x^3$ (neither), all have $f'(0) = f''(0) = 0$, and only the higher terms tell them apart. (Figure 8.4: the three functions $x^4$, $-x^4$, $x^3$ near $0$, all flat to second order, showing why $f''=0$ can’t decide.)

Toolkit: Classifying a critical point $a$ (where $f'(a)=0$)

1. Second-derivative test (fastest): compute $f''(a)$. Positive → min, negative → max, zero → fall back to step 2.
2. First-derivative test (always works): check the sign of $f'$ just left and just right of $a$. $+\to-$ → max, $-\to+$ → min, no change → neither.

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