Given: f(x) is a continuous function on a closed
interval [a,b].
f(x)
is differentiable on the open interval (a,b).
f(a)
= f(b)
Then: f '(c) = 0 for some c in (a,b)
Proof: Let f(a) = f(b) = d
I. Suppose
f(x) = d for all x in [a,b].
Then f(x) is a constant function and f '(x)
=
0 for all x.
II. Suppose
f(x) > d for some x in [a,b].
The
extreme value theorem says f(x) has a maximum at some c
in the interval.
Since
f(c) > d, the maximum is not an endpoint.
Therefore
f(c) is a maximum and c is a critical
number.
Since
f(x) is differentiable, f '(c) = 0.
III. Suppose f(x)
< d. The proof is similar to part II.
Given: f(x) is a continuous function on a
closed
interval [a,b].
f(x) is differentiable on the open interval (a,b).
Then: There exists a c in (a,b)
such that f ' (c) = f(b) - f(a)
b
- a
Proof: The equation of the secant line through a
and b is y = f(b) - f(a) (x - a)
+ f(a)
b
- a
Let
g(x) be f(x) - y
Then
g(x) = f(x) - [f(b) - f(a)] (x - a) - f(a)
b
- a
Now
g(a) = 0 = g(b) by putting a and b
into g(x).
Also,
since f(x) is differentiable, g is
differentiable
and Rolle's Theorem applies.
Thus,
g'(c) = 0 for some c in (a,b).
And
0 = g'(c) = f '(c) - f(b) - f(a)
b
- a
And
f '(c) = f(b) - f(a)
b
- a