Concavity, Points of Inflection, and the Second Derivative Test

Concavity and Points of Inflection

Definitions: If the graph of $f$ lies above all of its tangent lines on an open interval, the we say it is concave up on that interval. If the graph of $f$ lies below all of its tangent lines on an open interval, then we say it is concave down on that interval. A point, $P$, on a continuous curve $f(x)$ is an inflection point if $f$ changes concavity there.

When a curve is concave up, it is sort of bowl-shaped, and you can think it might hold water. When it is concave down, it is sort of upside-down-bowl-like, and water would run off of it.

How-to

The intervals of concavity can be found in the same way used to determine the intervals of increase/decrease, except that we use the second derivative instead of the first. In particular, since $(f')'=f''$, the intervals of increase/decrease for the first derivative will determine the concavity of $f$:

If possible, factor $f''$. If $f''$ is a quotient, factor the numerator and denominator (separately). Find all critical numbers $x=s$ of $f'$. These are the points where $(f')'=0$ or $(f')'$ doesn't exist (i.e., the points where $f''=0$ or where $f''$ doesn't exist). Draw a number line with tick marks at each critical number $s$. For each interval in which the function $f$ is defined, find the sign of the second derivative $f''$. If $f''(b) \gt 0$, then $f'$ is increasing on the interval containing $b$. This means that the slopes are increasing, so $f$ is concave up. Draw a right-side-up bowl over that interval on your number line. Similarly, if $f''(b) \lt 0$, draw an upside-down bowl. That's it! You can now see the intervals where $f$ is concave up or down.

The Second Derivative Test

Theorem: If $f''(x)>0$ for all $x$ in an open interval, then it is concave up on that interval. If $f ''(x) < 0$ for all $x$ in an open interval, then it is concave down on that interval.

Second Derivative Test: If $f'(c)=0$ and $f''(c) \gt0$, then there is a local minimum at $x=c$. If $f'(c)=0$ and $f''(c) \lt 0$, then there is a local maximum at $x=c$. If $f'(c)=0$ and $f''(c)=0$, or if $f''(c)$ doesn't exist, then the test is inconclusive. There might be a local maximum or minimum, or there might be a point of inflection.

The reasoning behind the test is simple: if $f''(c) \gt 0$, then $f'(x)$ is increasing near $x=c$. Since $f'(c)=0$, this means that $f'(x)$ used to be negative and is about to be positive. So the curve bottoms out at $x=c$ and then heads back up. The critical number $x=c$ is the bottom of the concave-up bowl. Likewise, if $f''(c) \lt 0$ and $f'(c)=0$, then $f'(x)$ is decreasing; it used to be positive and is about to be negative. The point $x=c$ is at the top of an upside-down bowl.

Example: Find the concavity of $f(x) = x^3 - 3x^2$. Solution: Since $f'(x)=3x^2-6x=3x(x-2)$,our two critical points for $f$ are at $x=0$ and $x=2$. Meanwhile, $f''(x)=6x-6$, so the only critical point for $f'$ is at $x=1$. It's easy to see that $f''$ is negative for $x \lt 1$ and positive for $x \gt 1$, so our curve is concave down for $x \lt 1$ and concave up for $x \gt 1$, and there is a point of inflection at $x=1$.

As for the critical points, $f''(0)=-6 \lt 0$, so we have a local maximum at $x=0$. $f''(2)=6 > 0$, so we have a local minimum at $x=2$. These results agree with what we got from the first derivative test.