The Second Derivative Test

Previously, we determined whether a critical number $c$ of a function $f$ indicated a local extreme value of $f$ by looking at the sign of $f'$ on either side of $c$.  If $f'$ changed sign at $c$, there was a local extreme value there, since $f'$ changes sign exactly where $f$ changes direction.  This is the first derivative test, where we use the first derivative to determine where we have local extrema. 

On this slide we use another method that might allow us to determine whether there is a local extreme value at $x=c$.  It uses the second derivative as well as the first, so we call it the second derivative test.

First, we formalize the concavity concepts from our previous work:

Concavity Test: If $f''(x)>0$ for all $x$ in an open interval, then f is concave up on that interval; i.e. $$\large f''>0\Longleftrightarrow f'\uparrow\Longleftrightarrow f \cup$$ If $f ''(x) < 0$ for all $x$ in an open interval, then f is concave down on that interval; i.e. $$\large f''<0\Longleftrightarrow f'\downarrow\Longleftrightarrow f \cap$$

The reasoning behind the test is simple:  If a point is a max, at the top of a hill, it must be on the (concave down) top of the upside-down bowl.  Similarly, if a point is a min, at the bottom of a valley, it must be on the (concave up) bottom of the right-side-up bowl.  So $f$ concave down at a critical number $c$ indicates a max, and $f$ concave up at $c$ indicates a min.

Another perspective:  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.

Notice:  As is indicated in the third option of the test, if a critical number $c$ is also a subcritical number, then the second derivative test cannot help determine whether or not there is a max or min at $c$.

Example:  Find the concavity of $f(x) = x^3 - 3x^2$ using the second derivative test. 

DO:  Try this before reading the solution, using the process above.

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 subcritical number for $f$ is at $x=1$.  As we saw in our previous work on this problem, $f''$ 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 $x=1$.

To use the second derivative test, we check the concavity of $f$ at the critical numbers.  We see that at $x=0$, $x<1$ so $f$ is concave down there.  Thus we have a local maximum at $x=0$.  At $x=2$, since $x>1$ $f$ is concave up there, so we have a local minimum at $x=2$.  These results agree with what we got from the first derivative test.