Pregunta 1

Let us start out with an

Example 3.2. For the graph pertinent to the example, see Figure 3.1. Con-

sider the function

f(x) =x2.

The line S through the points (.5, f(.5)) = (.5, .25) and (2.5, f(2.5)) = (2.5,6.25) has slope

s= f(2.5)−f(.5) 2.5−.5 =

6.25−.25 2.5−.5 = 3.

2_{This definition is technically less painful and conceptually more sensible than one}

3.2. CAUCHY’S MEAN VALUE THEOREM 139 0.5 1 1.5 2 2.5 1 2 3 4 5 6

Figure 3.1: Cauchy’s Theorem

We call the slopesof the lineSthe average slope (or average rate of change) of f over the interval [.5,2.5]. Now, remember that

f0(x) = 2x.

If c = 3/2, then f0(c) = 3. So, the tangent line L to the graph of f at (3/2, f(3/2)) has the same slope as the lineS. This means, for the number

c= 3/2, .5< c <2.5, we have that

f0(c) = f(2.5)−f(.5) 2.5−.5 .

In other words, there exists a numbercbetween the endpoints of the interval, such that the slope of the graph of f at this point equals the average slope of f over the interval. In geometric terms it means that there exists a point in the interval such that the tangent line at this point is parallel to the line

S, the secant line over the interval. ♦

The following theorem is named after Augustin-Louis Cauchy (1789– 1857). It expresses the observation which we made in the example. The average slope of a differentiable function over an interval equals the slope of the graph of the function at some point in the interval.

Theorem 3.3 (Chauchy’s Mean Value Theorem). Letf be a real val- ued function which is defined and differentiable on the interval [a, b], where

a < b.3 Then there exists a numberc∈(a, b) such that

f0(c) = f(b)−f(a)

b−a .

The following special case of the theorem, called Rolle’s theorem (named after Michel Rolle (1652–1719)), is of particular interest.

Theorem 3.4 (Rolle’s Theorem). Letf be as in Theorem 3.3. Iff(a) =

f(b), then there exists a number c between a and b (i.e., a < c < b) such that

f0(c) = 0.

We are not going to say anything about the proof of these two theorems, except that Cauchy’s theorem and Rolle’s theorem are equivalent (each is an easy consequence of the other one), and that the proof of both of them depends heavily on the completeness4 of the real numbers. We are also not interested in finding the points c, as they occur in the two theorems. We are interested in more general consequences.

Corollary 3.5. Let f be a real valued function which is defined and differ-

entiable on an interval I. If f0(x) = 0 for all x∈ I, then f is constant on this interval. In other words, there exists a numberdsuch that f(x) =dfor allx∈I.

Proof. A different formulation of the claim is that f(a) = f(b) for all a,

b ∈ I. We prove this statement using Cauchy’s theorem. If f(a) 6= f(b), thena6=b and there exists some c∈(a, b), such that

f0(c) = f(b)−f(a)

b−a 6= 0.

But this contradicts the assumption that f0(c) = 0 for all c ∈ I, and the corollary is proved.

We are going to use the following corollary frequently.

3_{More typically it is assumed that the function is continuous on [a, b] and differentiable}

on (a, b). Then the same conclusion is true. But, as we have not introduced continuous functions, we content ourselves with this more restrictive version of the theorem.

3.2. CAUCHY’S MEAN VALUE THEOREM 141

Corollary 3.6. Lethandgbe functions which are defined and differentiable

on an interval I. If h0(x) = g0(x) for all x ∈ I, then h and g differ by a constant, i.e., there exists a number dsuch that

h(x) =g(x) +d

for all x∈I.

Proof. Apply the previous corollary tof(x) =h(x)−g(x).

Uniqueness of Solutions of Some Differential Equations

Let us apply the principles which we just discussed to finding all solutions of some differential equations. You are familiar with the fact that, for given numbersb and c,c6= 0, there is exactly one numberx such that

cx=b.

(3.1)

Consideringxas the unknown, you can also express this by saying that (3.1) has a unique solution. In a differential equation the unknown is a function. We like to see to which extent some differential equations have a unique solution.

Example 3.7. Find all functionsf(x) which are defined and differentiable

on the entire real line, and for which f0(x) = 0 for all x.

Solution: We know that the derivative of a constant function vanishes

(is everywhere zero). Furthermore, Corollary 3.5 tells us that constants are the only functions with this property. So we found all functions which have the desired properties. The functions which we were looking for are the constant functions. ♦

Example 3.8. Find all functionsf(x) which are defined and differentiable

on the entire real line, and whose derivative is

f0(x) = 2x.

Solution: There is one obvious solution for the problem, the function

f(x) =x2. Corollary 3.6 says that any other solution of the problem differs from f only by a constant, so that the functions

f(x) =x2+c

are the only functions with the desired property. Here c is an arbitrary constant. ♦

We may formulate the ideas of the last two examples in a more general way.

Example 3.9. Suppose you are given a function h(x) which is defined on

an intervalI. Find all functionsf(x) which are defined on I and for which

f0(x) =h(x).

(3.2)

Solution: Find5 one functionH(x) which is defined onI, and for which

H0(x) = h(x). If there is such a function, then any solution of (3.2) is of the form

f(x) =H(x) +c,

wherecis an arbitrary constant. ♦

Exercise 99. Find all function f(x) which satisfy the equation:

(1) f0(x) = 5x2+ 7 (2)f0(x) = 3 sin 5x (3)f0(x) = sec2x.

Hint: Guess a functionH(x), such that H0(x) =f0(x).

In the following example we verify the second claim which we made in Theorem 2.12. We like to see which functions satisfy the Malthusian Law. This law was the basis for the population and radioactive decay models discussed in Section 2.7.

Example 3.10. Find all functionsf(x) which are defined and differentiable

on an interval and for which

f0(x) =af(x).

Solution: We know some functions f(x) which satisfy the differential

equation, namely all functions of the formf(x) =ceaxwherecis a constant. We want to see once again that these are all of the solutions of the differential equation.

Letf(x) be any function which satisfies the differential equation on some interval. Consider the function

h(x) =f(x)e−ax.

5_{For the time being you depend on being able to guess such a function H(x). By}

3.2. CAUCHY’S MEAN VALUE THEOREM 143 As a product of differentiable functions, his differentiable and its derivative is

h0(x) =f0(x)e−ax−af(x)e−ax=af(x)e−ax−af(x)e−ax= 0.

Corollary 3.5 tells us that h(x) is a constant function. Calling the constant

c we find that

f(x) =ceax.

This means that all solutions of the differential equation f0(x) =af(x) are of the form f(x) = ceax_{, wherec is a constant. With this we have verified}

the second claim in Theorem 2.12. ♦

Without any further information, the solutions of the differential equa- tions are not unique. In either of the above problems, we get a unique solution if we prescribe the value of the function at one point.

Example 3.11. Find all functionsf(x) which are defined and differentiable

on the entire real line and for which

f0(x) = 2f(x) and f(0) = 3.

Solution: We learned that the only functions which satisfy the differen-

tial equation f0(x) = 2f(x) are of the form f(x) =ce2x_{. Substitutingx= 0}

into this expression we see thatf(0) =ce0 =c. We conclude thatc= 3 and that f(x) = 3e2x. ♦

Remark 13. The uniqueness of the solution of an initial value problem

as in the previous example is not only of theoretical importance. Imagine you study the growth rate of a strain of bacteria. Before you can publish your result, it must be certain that your experiment can be reproduced at a different time in a different location. That is a requirement which any experiment in science must satisfy. If there is more than one mathematical solution to your problem, then you have to expect that the experiment can go either way, and this would invalidate your experiment.

Exercise 100. Find the unique solutions of the problems:

1. f0(x) = 5f(x) andf(0) = 7. 2. f0(x) = 3f(x) andf(2) = 3. 3. f0(x) = 2x2+ 3 and f(2) = 1.