Concavity and Points of Inflection

The running example on this page is $f(x) = \frac{1}{2}x^4 - \frac{3}{2}x^3 - 3x^2 + 2x$, with derivative $f'(x) = 2x^3 - \frac{9}{2}x^2 - 6x + 2$. Its graph is drawn below with three tangent lines, and the graph of $f'$ is drawn under it on the same $x$-scale.

The tangent slopes at $x = -1.5$, $0.5$, and $3$ are $f'(-1.5) = -5.875$, $f'(0.5) = -1.875$, and $f'(3) = -2.5$. All three tangent lines slope down, but they don't behave alike. Left of $x = -\frac{1}{2}$ the slopes increase as $x$ increases, and the graph bends up. Between $-\frac{1}{2}$ and $2$ they decrease, and the graph bends down. Past $2$ they increase again. So the graph of $f$ is concave up, then concave down, then concave up, and it changes exactly where the graph of $f'$ turns.

Concavity is reported on open intervals, as the AP items report it.

In the figure, the tangent lines at $x = -1.5$ and $x = 3$ lie below the graph, and the tangent at $x = 0.5$ lies above it. That's true of every tangent line, and it follows from the definition.

Here's the proof for concave up. Take $x > a$ in $I$. Since $f$ is differentiable on $I$, it's continuous on $[a, x]$ and differentiable on $(a, x)$. By the Mean Value Theorem there's a $c$ in $(a, x)$ with

Since $c > a$ and $f'$ is increasing, $f'(c) > f'(a)$. Multiplying by $x - a > 0$ gives $f(x) - f(a) > f'(a)(x - a)$. For $x < a$, the $c$ lies in $(x, a)$, so $f'(c) < f'(a)$. Multiplying by $x - a < 0$ reverses that inequality, and the same conclusion follows.

The reason is that $f''$ is the derivative of $f'$. A function whose derivative is positive on an interval is increasing there, so $f'$ is increasing on $I$ when $f'' > 0$ on $I$. Similarly, $f'$ is decreasing when $f'' < 0$.

Suppose $f''(c)$ exists. Then $f'$ is differentiable at $c$, so it's continuous there. Since $f'$ is increasing on one side of $c$ and decreasing on the other, $f'$ has a relative extremum at $c$. Fermat's theorem, applied to $f'$, gives $f''(c) = 0$.

The converse is false. For $f(x) = x^4$, $f''(x) = 12x^2$, so $f''(0) = 0$. But $f'' > 0$ on both sides of $0$, the graph is concave up on both sides, and $(0, 0)$ isn't a point of inflection. A zero of $f''$, or a point where $f''$ doesn't exist, is only a candidate. It's a point of inflection when $f$ is continuous there and $f''$ changes sign there. On a free-response answer, the reason is the sign change: "$f''(x) = 0$" alone doesn't justify an inflection point.

When the graph of $f'$ is given, concavity comes from its slope. The graph of $f$ is concave up where the graph of $f'$ rises and concave down where it falls. Points of inflection are where the graph of $f'$ changes from rising to falling or from falling to rising: at its relative extrema. The sign of $f'$ says whether $f$ is increasing, and it says nothing about concavity.

A graph of $f''$ is read by its sign. The graph of $f$ is concave up where the graph of $f''$ is above the axis and concave down where it's below. The graph of $f$ has a point of inflection wherever the graph of $f''$ crosses the axis. Its turning points mark something else: where $f''$ has a relative extremum, the graph of $f'$ has a point of inflection.