Mean Value Theorem

In document Calculus One (Page 174-179)

on the interval(a, b)and is continuous on the interval[a, b]. Then

f′(c)=f(b)−f(a)

ba

for somea < c < b.

a c b x

y

Figure 10.11: A geometric interpretation of the Mean Value Theorem

Proof Let

m=f(b)−f(a)

ba ,

and consider a new function g(x) = f(x)m(xa)f(a). We know that g(x) has a derivative on [a, b], since g′(x) = f′(x)m. We can compute g(a)=f(a)m(aa)f(a)=0and

g(b)=f(b)m(ba)f(a)=f(b)f(b)−f(a)

ba (ba)−f(a)

=f(b)(f(b)f(a))f(a) =0.

So g(a)=g(b)=0. Now by Rolle’s Theorem, that at some c, g(c)=0 for some a < c < b.

But we know that g(c)=f(c)m, so

0=f′(c)m=f′(c)f(b)−f(a)

Hence

f(c)=f(b)−f(a)

ba .

We can now answer our second question above.

Example 10.3.4 Suppose you drive a car from toll booth on a toll road to another toll booth 30 miles away in half of an hour. Must you have been driving at 60 miles per hour at some point?

Solution If p(t)is the position of the car at time t, and0hours is the starting

time with1/2hours being the final time, the Mean Value Theorem states there is a time c

p′(c)=30−0

1/2 =60 where0< c <1/2.

Since the derivative of position is velocity, this says that the car must have been driving at60miles per hour at some point.

Now we will address the unthinkable, could there be a function f(x) whose derivative is zero on an interval that is not constant? As we will see, the answer is “no.”

Theorem 10.3.5 Iff(x)=0for allx in an intervalI, thenf(x)is constant on

I.

Proof Let a < b be two points in I. By the Mean Value Theorem we know

f(b)f(a)

ba =f

(c)

for some c in the interval(a, b). Since f′(c)=0we see that f(b)=f(a). Moreover, since a and b were arbitrarily chosen, f(x)must be the constant function.

Example 10.3.6 Suppose two different functions have the same derivative. What can you say about the relationship between the two functions?

Solution Set h(x)=f(x)g(x), so h(x)=f(x)g(x). Now h(x)=0on the

interval(a, b). This means that h(x)=k where k is some constant. Hence g(x)=f(x)+k.

Example 10.3.7 Describe all functions whose derivative issin(x).

Solution One such function is −cos(x), so all such functions have the form

−cos(x)+k, see Figure10.12.

1 2 3 4 5 6 −4 −2 2 4 x y

Figure 10.12: Functions of the form−cos(x)+k, each of whose derivative issin(x).

Exercises for Section 10.3

(1) Letf(x)=x2. Find a valuec(−1,2)so thatf(c)equals the slope between the endpoints off(x)on[−1,2].

(2) Verify thatf(x)=x/(x+2)satisfies the hypotheses of the Mean Value Theorem on the interval[1,4]and then find all of the values,c, that satisfy the conclusion of the theorem.

(3) Verify thatf(x) =3x/(x+7)satisfies the hypotheses of the Mean Value Theorem on the interval[−2,6]and then find all of the values,c, that satisfy the conclusion of the theorem.

(4) Letf(x)=tan(x). Show thatf(π)=f(2π)=0 but there is no numberc(π,2π)such thatf(c)=0. Why does this not contradict Rolle’s theorem?

(5) Letf(x) =(x3)−2. Show that there is no valuec (1,4)such thatf(c) =(f(4)

f(1))/(4−1). Why is this not a contradiction of the Mean Value Theorem?

(6) Describe all functions with derivativex2+47x5.

(7) Describe all functions with derivative 1

1+x2.

(8) Describe all functions with derivativex3−1

x.

(9) Describe all functions with derivativesin(2x).

(10) Show that the equation 6x4−7x+1=0 does not have more than two distinct real roots.

(11) Letf(x)be differentiable onR. Suppose thatf(x),0 for everyx. Prove thatf has at most one real root.

11.1

Basic Antiderivatives

Computing derivatives is not too difficult. At this point, you should be able to take the derivative of almost any function you can write down. However, undoing derivatives is much harder. This process of undoing a derivative is called taking an

antiderivative.

Definition A functionF(x)is called anantiderivativeoff(x)on an interval if

F′(x)=f(x) for allx in the interval.

We have special notation for the antiderivative:

Definition The antiderivative is denoted by Z

f(x)dx=F(x)+C,

wheredxidentifiesxas the variable andCis a constant indicating that there are many possible antiderivatives, each varying by the addition of a constant. This is often called theindefinite integral.

Here are the basic antiderivatives. Note each of these examples comes directly from our knowledge of basic derivatives.

Theorem 11.1.1 (Basic Antiderivatives) • Z k dx=kx+C. • Z xndx= x n+1 n+1 +C (n,−1). • Z exdx =ex+C. • Z axdx= a x ln(a)+C. • Z 1 x dx=ln|x|+C. • Z cos(x)dx =sin(x)+C. • Z sin(x)dx=−cos(x)+C. • Z tan(x)dx=−ln|cos(x)|+C. • Z sec2(x)dx=tan(x)+C. • Z csc2(x)dx=−cot(x)+C. • Z

sec(x) tan(x)dx=sec(x)+C.

• Z csc(x) cot(x)dx=−csc(x)+C. • Z 1 x2+1dx=arctan(x)+C. • Z 1 √ 1−x2dx=arcsin(x)+C.

It may seem that one could simply memorize these antiderivatives and antidifferen- tiating would be as easy as differentiating. This isnotthe case. The issue comes up when trying to combine these functions. When taking derivatives we have the

product ruleand thechain rule. The analogues of these two rules are much more difficult to deal with when taking antiderivatives. However, not all is lost. We have the following analogue of the Sum Rule for derivatives, Theorem3.2.6.

Theorem 11.1.2 (The Sum Rule for Antiderivatives) Given two functions

f(x)andg(x)wherek is a constant:

• Z kf(x)dx=kF(x)+C. • Z (f(x)+g(x))dx=F(x)+G(x)+C.

Let’s put this rule and our knowledge of basic derivatives to work. Example 11.1.3 Compute

Z

In document Calculus One (Page 174-179)