James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

290 Chapter 4 Applications of Differentiation
y
O
FIGURE 6
y=˛- x
B
c 2
x
Figure 6 illustrates this calculation: The tangent line at this value of c is parallel to the
secant line OB.
n
ExamplE 4 If an object moves in a straight line with position function s − f std, then
the average velocity between t − a and t − b is
f sbd 2 f sad
b 2 a
and the velocity at t − c is f 9scd. Thus the Mean Value Theorem (in the form of Equation
1) tells us that at some time t − c between a and b the instantaneous velocity f 9scd
is equal to that average velocity. For instance, if a car traveled 180 km in 2 hours, then
the speedometer must have read 90 kmyh at least once.
In general, the Mean Value Theorem can be interpreted as saying that there is a
number at which the instantaneous rate of change is equal to the average rate of change
over an interval.
n
The main significance of the Mean Value Theorem is that it enables us to obtain information
about a function from information about its derivative. The next example provides
an instance of this principle.
ExamplE 5 Suppose that f s0d − 23 and f 9sxd < 5 for all values of x. How large can
f s2d possibly be?
SOLUtion We are given that f is differentiable (and therefore continuous) everywhere.
In particular, we can apply the Mean Value Theorem on the interval f0, 2g. There exists
a number c such that
f s2d 2 f s0d − f 9scds2 2 0d
so
f s2d − f s0d 1 2f 9scd − 23 1 2f 9scd
We are given that f 9sxd < 5 for all x, so in particular we know that f 9scd < 5. Multiplying
both sides of this inequality by 2, we have 2f 9scd < 10, so
f s2d − 23 1 2f 9scd < 23 1 10 − 7
The largest possible value for f s2d is 7.
n
The Mean Value Theorem can be used to establish some of the basic facts of differential
calculus. One of these basic facts is the following theorem. Others will be found
in the following sections.
5 Theorem If f 9sxd − 0 for all x in an interval sa, bd, then f is constant on sa, bd.
Proof Let x 1 and x 2 be any two numbers in sa, bd with x 1 , x 2 . Since f is differentiable
on sa, bd, it must be differentiable on sx 1 , x 2 d and continuous on fx 1 , x 2 g. By
applying the Mean Value Theorem to f on the interval fx 1 , x 2 g, we get a number c such
that x 1 , c , x 2 and
6 f sx 2 d 2 f sx 1 d − f 9scdsx 2 2 x 1 d
Since f 9sxd − 0 for all x, we have f 9scd − 0, and so Equation 6 becomes
f sx 2 d 2 f sx 1 d − 0 or f sx 2 d − f sx 1 d
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.