Inflection points & curvature

In an epidemic there comes a day when the number of new cases stops rising and for the first time is smaller than the day before. The total keeps growing, but more slowly. Mathematically that day is exactly an inflection point, and in the news it is the moment everybody is waiting for. An inflection point says nothing about high or low, it says something about a change in curvature.

Curving left and right

Imagine cycling along the graph from left to right. Sometimes you have to hold the handlebars to the left, sometimes to the right. On a parabola opening upwards you steer left the whole time, it curves left everywhere. On one opening downwards you steer right everywhere. Only when you have to change the steering on the way is there an inflection point, exactly at the place where you change.

The second derivative

The first derivative $f'$ measures the slope. Differentiate it once more and you get $f''$, which measures how the slope changes. If $f''$ is positive, the slope is growing and the curve bends left. If $f''$ is negative, the slope is shrinking and the curve bends right. And a switch between the two means $f''$ changes sign, so it passes through zero on the way.

How to calculate it

An example: $f(x) = x^3 - 3x^2$. First differentiate twice: $f'(x) = 3x^2 - 6x$ and $f''(x) = 6x - 6$. Then set $f''$ to zero: $6x - 6 = 0$ gives $x = 1$. To confirm, differentiate once more: $f'''(x) = 6$, which is not zero. So at $x = 1$ there really is an inflection point. As always, the height comes from the original function: $f(1) = 1 - 3 = -2$, so the inflection point is $(1|-2)$.

The procedure is the same as for maxima and minima, just one derivative further along. There you set $f'$ to zero and check with $f''$. Here you set $f''$ to zero and check with $f'''$. If you remember this parallel, you only need one scheme instead of two.