The Mean Value Theorem and Its Consequences (Part II)

The Mean Value Theorem & Its Consequences (Part II)

Let us consider the following example to illustrate the use of the Mean Value Theorem. In a 100-meter dash race, a very fast athlete takes only 9.6 seconds to finish the race; thus his average velocity during this race is 100 meters divided by 9.6 seconds, or 10.4 meters per second.

We know that the velocity of an athlete throughout a race is not constant. The initial velocity is usually lower, and then he picks up speed toward the end of the race.

So this graph represents an athlete's speed over time during a race.

The average speed of the athlete is 10.4 meters per second, as shown by this red line. Therefore, according to the Mean Value Theorem, we know that at a certain point during the race, the instantaneous velocity of this athlete is exactly equal to 10.4 meters per second, which is the average speed.

So in fact, before he reached this point, his speed was somewhat lower than it was after he passed through it.

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From now on, his speed would be higher. So we will now look at some consequences of the Mean Value Theorem. The first consequence is that functions with no derivatives at all are constant functions.

So, if f(x) equals zero for every value of x in the open interval a < b, then f(x) must be a constant function in that interval. Here is the proof: Let us fix any point x 0 in this open interval and consider any other point x in the same interval.

So by the Mean Value Theorem, this ratio here--f(x) minus f(x zero) divided by x minus x zero--is equal to f dash c for some point c in the interval. However, it is assumed that f dash c is zero for all values of x in a b. In other words, this ratio is always equal to zero. That means the numerator must be zero for all values of x in a b. That is to say, f equals constant number f(x-naught).

Another corollary is the following. Corollary two, functions with the same derivative differ by a constant. Suppose we have two functions f and g. If their derivatives are equal at every point x in an open interval [a, b], then f - g must be a constant function in that interval.

In other words, there exists a constant a such that f(x) equals to g(x) plus a. However, the proof is as follows: You let h be f minus g. h dash equals to f dash minus g dash equals zero as given. Therefore by corollary one, since h dash always equal zero, h is a constant. f minus g is equal to the constant a. This is about anti-derivative.

If we have a function f defined on a certain interval, its anti-derivative is another function F. If differentiating F gives back f, then F is an anti-derivative of f. Starting from f, you find the anti-derivative F whose derivative is equal to f.

And we have a theorem that tells us:

If f is an anti-derivative of another function F, then any other anti-derivative of F can be expressed as some particular capital letter F plus a constant.

If you recall, when you learned integration in high school, you always found one function whose derivative equaled the given function and then used this fact to prove that all other functions were equal to this particular one plus a constant.

This is it. All the anti-derivative of a given small f is equal to one of them plus a constant.

Now, the proof is like this. Suppose you have this function F and its derivative of small f, and suppose another function G is also the derivative of small f. That means capital F equals to small f, and also equal to G dash. Therefore, F and G are always identical. According to corollary two, G minus capital F is equal to C which is a constant, so G equals just F plus C.