The Unapologetic Mathematician

Distinguishing Maxima and Minima

From Heine-Borel we know that a continuous function $f$ on a closed interval $\left[a,b\right]$ takes a global maximum and a minimum. From Fermat we know that any local (and in particular any global) extremum occurs at a critical points — a point where $f'(x)=0$, or $f$ has no derivative at all. But once we find these critical points how can we tell maxima from minima?

The biggest value of $f$ at a critical point is clearly the global maximum, and the smallest is just as clearly the minimum. But what about all the ones in between? Here’s where those consequences of the mean value theorem come in handy. For simplicity, let’s assume that the critical points are isolated. That is, each one has a neighborhood in which it’s the only critical point. Further, let’s assume that $f'(x)$ is continuous wherever it exists.

Now, to the left of any critical point we’ll have a stretch where $f$ is differentiable (or else there would be another critical point there) and $f'(x)$ is nonzero (ditto). Since the derivative is continuous, it must either be always positive or always negative on this stretch, because if it was sometimes positive and sometimes negative the intermediate value theorem would give us a point where it’s zero. If the derivative is positive, our corollaries of the mean value theorem tell us that $f$ increases as we move in towards the point, while if the derivative is negative it decreases into the critical point. Similarly, on the right we’ll have another such stretch telling us that $f$ either increases or decreases as we move away from the critical point.

So what’s a local maximum? It’s a critical point where the function increases moving into the critical point and decreases moving away! That is, if near the critical point the derivative is positive on the left and negative on the right, we’ve got ourselves a local maximum. If the derivative is positive on the right and negative on the left, it’s a local minimum. And if we find the same sign on either side, it’s neither! Notice that this is exactly what happens with the function $f(x)=x^3$ at its critical point. Also, we don’t have to worry about where to test the sight of the derivative, because we know that it can only change signs at a critical point.

In fact, if we add a bit more to our assumptions we can get an even nicer test. Let’s assume that the function is “twice-differentiable” — that $f'(x)$ is itself a differentiable function — on our interval. Then all the critical points happen where $f'(x)=0$. Even better now, if it changes signs as we pass through the critical point (indicating a local extremum) it’s either increasing or decreasing, and this will be reflected in its derivative $f''(x)$ at the critical point. If $f''(x)>0$ then our sign changes from negative to positive and we must be looking at a local minimum. On the other hand, if $f''(x)<0$ then we’ve got a local maximum. Unfortunately, if $f''(x)=0$ we don’t really get any information from this test and we have to fall back on the previous one.

January 25, 2008 Posted by | Analysis, Calculus | 2 Comments