## 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.