10/19/2010
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4.2 THE MEAN VALUE THEOREM
f’c = 0
ROLLE’S THEOREM
Let f be a function that satisfies the following three conditions:
i. f is continuous on [a, b]
ii. f is differentiable on the open interval (a, b)
iii. f(a) = f(b)
Rolle’s Theorem says that a differentiable curve has at least one horizontal tangent line between any 2 point where it crosses a horizontal line.
x
a
Then is a number c in (a, b) such that f’c = 0.
c
b
y
x
a
[0,2]
c2
b
THE MEAN VALUE THEOREM:
Suppose y = f(x) is continuous on a closed interval [a, b] and differentiable on the interval’s interior (a, b). Then there is at
Ex. 1: Verify that the function satisfies the three
hypotheses of Rolle’s Theorem on the given interval.
Then find all numbers c that satisfy the conclusion of
Rolles’ Theorem.
f(x) = x3 -3x2 + 2x + 5,
c1
least one point c in (a, b) at which f (b ) − f (a ) = f '(c )
b−a
slope =f’(c1)
y
Geometrically, the Mean Value Theorem says that somewhere b t
between A & B, the curve has at least A & B th
h
tl t
one tangent parallel to chord AB.
B
f is continuous on [0,2]
[0 2] and differentiable on (0
(0, 2)
2).
Also, f(0) = f(2) = 5.
⇔ 3c 2 − 6 c + 2 = 0 ⇔ c =
6 ±
36 − 24 6 ± 2 3
=
6
6
3± 3
⇔
. N ote : both of these values are in
3
(0, 2)
Ex. 1: Show that the function f(x)= x3 + x -1 satisfies the
hypotheses of the Mean Value Theorem on the interval
[0, 2]. Then find all numbers c that satisfy the conclusion
of the Mean Value Theorem.
f(x) is a polynomial function that is continuous on [0,2]
and differentiable on (0, 2).
f (2) − f (0)
f (b) − f (a)
f '(c) =
2−0
b−a
9 − (−1)
3c 2 + 1 =
⇔ 3c 2 + 1 = 5 ⇔ 3c 2 = 4
2−0
2
2
⇔c =±
, but only
is in ( 0, 2 ) .
3
3
f '(c) =
slope =
A
Slope = f’(c2)
a
c1
f ( b ) − f (a )
b−a
x
c2
b
Ex. 2 Show that 1 + 2x + x3+4x5 = 0 has exactly one real root.
Let f(x) = 1 +2x + x3 + 4x2. f(0) =1>0 and
f(-1) =-6 < 0.
By the Intermediate Value Theorem, there is at least one
zero in (-1, 0).
Suppose that there are two distinct real roots a and b
with a<b
a<b. Then f(a) and f(b) =0.
=0
f(x) is continuous on [a, b] and differentiable on (a, b). Then
by Rolle’s Theorem, there exists at least one number c in
(a,b) such that f’(c) =0.
But f’(x) = 2 +3x2 + 20x4, which can never =0
This contradiction shows that there cannot be 2 distinct
roots. So there is exactly one real root.
1