The Second Derivative Test

We have been concentrating on the problem of finding the overall maximum and minimum values of a function--the global extrema.

We quite often want to know about local extrema as well, especially when trying to decide how a function behaves (for example when graphing it). In this case it would be nice to know whether a given critical point (derivative zero) is a local maximum, a local minimum or neither.

Often, the easiest approach is to look at the sign of the derivative near the point.

At a local maximum the derivative goes from positive to zero to negative as you pass the point in the direction of increasing x (or, more precisely, it is not negative just to the left and is not positive just to the right--it could just be flat).

At a local minimum the derivative goes from negative to zero to positive as you pass through the point. You can see this in Fig. 2.5.

Figure 2.5: The slope of the crurve near a critical point.

I would encourage you to use this method quite a lot because it gets you used to the idea of analysing what is happening to a function.

There is another, more routine, method known as the Second Derivative Test. I will state the test, show you a difficulty and then try to explain where the test comes from.

Theorem 2.12 (2nd derivative test)   Suppose that the function f (x) can be differentiated twice. Suppose that x = a is a critical point (so f'(a) = 0). Then if f''(a) < 0 the point a is a local maximum and if f''(a) > 0 the point a is a local minimum. If f''(a) = 0 then we can draw no conclusions.

For example, consider the function f (x) = x³ - 3x + 1. Then, as we have seen, f'(x) = 3x² - 3 and f''(x) = 6x. The critical points are x = - 1 and x = 1. f''(- 1) = - 6 < 0, so x = - 1 is a local maximum. f''(1) = 6 > 0, so x = 1 is a local minimum.

The method looks very simple and is--so long as it is easy to work out and evaluate the second derivative. Sometimes the function is so messy that it is easier to look at the signs of f'(x) than it is to calculate f''(x).

The `difficulty' I mentioned is really a warning. The statement of the test said quite clearly that if the second derivative is zero at the critical point then no conclusions can be drawn about the nature of the critical point. Let me illustrate this with some simple examples. Consider the three functions f (x) = x⁴, g(x) = x³ and h(x) = - x⁴. You should check that each of these functions has precisely one critical point x = 0 and that at this point the second derivative is zero in all three cases. Now look at the graphs:

Figure 2.6: figure

You can see that f has a local minimum at 0, g has an inflexion at 0 and h has a local maximum at 0.

Now for some examples.