Topic 5.6 Determining Concavity of Functions over Their Domains: use the second derivative to find concavity and points of inflection.

What this topic is asking

The College Board (Topic 5.6) connects the sign of the second derivative to concavity. Where $f''(x) > 0$ the graph is concave up; where $f''(x) < 0$ it is concave down. A point of inflection is where concavity changes. You must build a sign chart for $f''$ and justify inflection points by a sign change.

Concavity and the second derivative

A worked concavity analysis

Why a zero of f'' is not enough

The trap is assuming $f''(x) = 0$ guarantees an inflection point. For $f(x) = x^4$, $f''(x) = 12x^2$ is zero at $x = 0$ but is positive on both sides, so there is no sign change and no inflection point: the graph stays concave up throughout. An inflection point requires the concavity to actually change, which means $f''$ must change sign. Always verify the sign change, exactly as the First Derivative Test requires a sign change of $f'$ for an extremum.

The link to the shape of the graph

Concavity refines the picture that increasing/decreasing gives. A function can be increasing while concave down (rising but levelling off) or increasing while concave up (rising and accelerating); the four combinations of increasing/decreasing with concave up/down describe the four basic curve shapes. This is why the exam often pairs a concavity question with an increasing/decreasing question on the same function: together they pin down the shape well enough to sketch. On free-response questions the inflection-point justification must name the sign change of $f''$, not merely that $f'' = 0$; the sign change is the load-bearing reason.

Concavity as the rate of change of slope

Another way to read concavity is in terms of the slope. The first derivative is the slope of the curve, and the second derivative is the rate at which that slope is changing. Where $f'' > 0$, the slope is increasing, so the curve bends to the left as you trace it, which is concave up; where $f'' < 0$, the slope is decreasing, bending right, which is concave down. This interpretation is what lets you analyze the concavity of a function from the graph of its first derivative: $f$ is concave up exactly where the graph of $f'$ is rising, and concave down where the graph of $f'$ is falling. An inflection point of $f$ then corresponds to a local maximum or minimum of $f'$, the place where $f'$ stops rising and starts falling or vice versa. Reading concavity as the slope of the slope unifies the formula approach with the graph-of-derivative approach the exam frequently tests.

Concavity in applied and motion contexts

Concavity carries meaning in applied problems too. For a quantity changing over time, $f'' > 0$ means the rate of change is itself increasing, so the quantity is growing at an accelerating pace, while $f'' < 0$ means the growth is slowing. In motion, the second derivative of position is acceleration, so concavity of the position graph reports whether the object is speeding up or slowing down in the direction of motion. Recognizing concavity as "the rate is increasing or decreasing" lets you interpret an inflection point in context as the moment when growth shifts from accelerating to decelerating, a description the exam sometimes asks for in words rather than as a bare $x$-value.