## Wednesday, December 3, 2008

### Values, Equations, and Theorems

There is some confusion about theorems. For example, consider the Mean Value Theorem: If f is continuous on [a,b] and differentiable on (a,b), then there is some value c∈(a,b) so that f'(c)=(f(b)-f(a))/(b-a).

Some students think that the ratio (f(b)-f(a))/(b-a) is the Mean Value Theorem. But it is not; it is just a value that is called the average rate of change of f between a and b. You can compute this value as long as both f(a) and f(b) exist. It has nothing to do with derivatives or continuity.

Other students think that the equation f'(c)=(f(b)-f(a))/(b-a). This is closer to the truth, but still is incorrect. First of all, what is c? Second, this statement may not be true. For example, suppose that f(x)=-1 if x<0 and f(x)=1 if x>0. Suppose that a=-2 and b=+2. Then the ratio (f(b)-f(a))/(b-a) is equal to 1/2. But f'(x)=0 everywhere except at x=0, where f'(0) does not exist.

Even closer is to say that f'(c)=(f(b)-f(a))/(b-a) for some c between a and b. This is actually the conclusion of the Mean Value Theorem. It requires the entire statement, particularly the statement that the equation is true for some c and that the value c must be between a and b.

But my example above provides an example where the conclusion of the Mean Value Theorem is false. That does not mean that the Mean Value Theorem itself is false. After all, it is a theorem, and that means that it has been proved to be true always. The part that is missing is the hypothesis for the theorem. The conclusion can only be guaranteed to be true using the theorem if the hypotheses are all satisfied. In this case, you must also check (or give a reason why) the function f is continuous and differentiable on the interval from a to b, including the endpoints for continuity.

Similarly, ∫ab f(x) dx/(b-a) computes the average value of a function f on an interval [a,b]. The value can be computed anytime the function is integrable over the interval [a,b]. The Mean Value Theorem for Integrals has nothing to do (in principle) with this calculation.

However, if f is continuous on [a,b], then
f(c)=∫ab f(x) dx/(b-a)
for some c∈(a,b). This entire statement comprises the Mean Value Theorem for Integrals. The hypothesis that must be verified to use the theorem is that f is continuous on [a,b]. The conclusion is that you are guaranteed that
f(c)=∫ab f(x) dx / (b-a)
for at least one value c between a and b.

## Tuesday, December 2, 2008

### Limit or Function Evaluation?

I've noticed that some students are perplexed about when they use a limit or function evaluation. I presume that the cause of this confusion is that students have learned that you evaluate a limit by plugging in a value. But this is only because nearly all functions that they work with are continuous.

You use a limit evaluation when you need to know what the value of the function should be by using information from the side of the point of interest. When using a limit, you must use limit notation: limx→c f(x). Then you use the appropriate rules of limits to evaluate (and hopefully, the function is continuous).

You use a function evaluation when you need to know the value of the function at an actual point. There is no limit involved, just function evaluation. You just use function notation, say f(c), and compute the value defined by the function.

For example, suppose you are calculating an instantaneous rate of change as the limit of an average rate of change. The average rate of change only makes sense when the interval of interest includes two points (endpoints of an interval). The instantaneous rate of change is found by seeing what the value of the average rate of change does when the two points move closer to each other, or more particularly, as the second point approaches the first point.

On the other hand, suppose that you know the derivative, which is itself a function. Then the instantaneous rate of change is calculated by function evaluation using the derivative function. (The limit was already used to create this new function.)

In particular, I have noticed this problem when dealing with finding extreme values of a function. When the interval of interest is an open interval, we are acting as though the domain does not include the endpoints. So, with this restricted domain, evaluation of the function is not possible (since the points are not in the domain). So we must use the information about the function immediately adjacent to the endpoints, and this is with a limit. In this context, the value found in the limit is not achieved at the endpoint, although it might be achieved somewhere else in the domain.

On the other hand, if the interval is a closed interval, then the endpoints are included in the restricted domain. If the function is continuous at these points, then evaluating the function directly is appropriate, since the point is in the domain.

Final remark, if the function is discontinuous at some point in the interval, you must also check the limits at that point for consideration when looking for extreme values.