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

Section 4.2 The Mean Value Theorem 289
y
A
ƒ
FIGURE 5
h(x)
y=ƒ
0 a x
b x
f(b)-f(a)
f(a)+ (x-a)
b-a
Lagrange and the
Mean Value Theorem
The Mean Value Theorem was first
formulated by Joseph-Louis Lagrange
(1736–1813), born in Italy of a French
father and an Italian mother. He was a
child prodigy and became a professor in
Turin at the tender age of 19. Lagrange
made great contributions to number
theory, theory of functions, theory of
equations, and analytical and celestial
mechanics. In particular, he applied
calculus to the analysis of the stability
of the solar system. At the invitation
of Frederick the Great, he succeeded
Euler at the Berlin Academy and, when
Frederick died, Lagrange accepted King
Louis XVI’s invitation to Paris, where he
was given apartments in the Louvre
and became a professor at the Ecole
Polytechnique. Despite all the trappings
of luxury and fame, he was a kind and
quiet man, living only for science.
B
Proof We apply Rolle’s Theorem to a new function h defined as the difference
between f and the function whose graph is the secant line AB. Using Equation 3
and the point-slope equation of a line, we see that the equation of the line AB can
be written as
f sbd 2 f sad
y 2 f sad − sx 2 ad
b 2 a
or as
So, as shown in Figure 5,
y − f sad 1
4 hsxd − f sxd 2 f sad 2
f sbd 2 f sad
b 2 a
sx 2 ad
f sbd 2 f sad
b 2 a
sx 2 ad
First we must verify that h satisfies the three hypotheses of Rolle’s Theorem.
1. The function h is continuous on fa, bg because it is the sum of f and a first-degree
polynomial, both of which are continuous.
2. The function h is differentiable on sa, bd because both f and the first-degree polynomial
are differentiable. In fact, we can compute h9 directly from Equation 4:
f sbd 2 f sad
h9sxd − f 9sxd 2
b 2 a
(Note that f sad and f f sbd 2 f sadgysb 2 ad are constants.)
3. hsad − f sad 2 f sad 2
Therefore hsad − hsbd.
hsbd − f sbd 2 f sad 2
f sbd 2 f sad
b 2 a
f sbd 2 f sad
b 2 a
sa 2 ad − 0
sb 2 ad
− f sbd 2 f sad 2 f f sbd 2 f sadg − 0
Since h satisfies the hypotheses of Rolle’s Theorem, that theorem says there is a number
c in sa, bd such that h9scd − 0. Therefore
and so
0 − h9scd − f 9scd 2
f 9scd −
f sbd 2 f sad
b 2 a
f sbd 2 f sad
b 2 a
n
ExamplE 3 To illustrate the Mean Value Theorem with a specific function, let’s
consider f sxd − x 3 2 x, a − 0, b − 2. Since f is a polynomial, it is continuous and
differentiable for all x, so it is certainly continuous on f0, 2g and differentiable on s0, 2d.
Therefore, by the Mean Value Theorem, there is a number c in s0, 2d such that
f s2d 2 f s0d − f 9scds2 2 0d
Now f s2d − 6, f s0d − 0, and f 9sxd − 3x 2 2 1, so this equation becomes
6 − s3c 2 2 1d2 − 6c 2 2 2
which gives c 2 − 4 3 , that is, c − 62ys3 . But c must lie in s0, 2d, so c − 2ys3 .
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