Section 4.1
(continued):
Theorem 3 (
Extreme Value Theorem
): If
f
is continuous
on the closed interval [
a
,
b
], then there exist
c
and
d
in [
a
,
b
]
such that
f
(
x
)
≤
f
(
c
) for all
x
in [
a
,
b
] and
f
(
x
)
≥
f
(
d
) for all
x
in [
a
,
b
].
That is, every continuous function whose domain
is a closed interval must have a global maximum value
f
(
c
)
and a global minimum value
f
(
d
).
Last time we saw that the hypothesis of continuity cannot
be dropped.
Is the Extreme Value Theorem still true if the domain is the
open interval (
a
,
b
) instead of the closed interval [
a
,
b
]?
..?..
No; e.g.,
f
(
x
) =
x
on (–1,1), or
f
(
x
) =
x
2
on (0,1).
It’s hard to prove the Extreme Value Theorem; I may say
more about this at the end of the semester, if time permits
(and enough of you are interested).
Like the Intermediate
Value Theorem, it can be derived from the Cut Axiom.
How do we find global maxima and minima for a function
on a closed interval?
We look at endpoints and local maxima and minima.
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How do we find local maxima and minima?
We look at all the critical points
.
Definition: A
critical number
of a function
f
is a number
c
in the domain of
f
such that either
f
′
(
c
) = 0 or
f
′
(
c
) does
not exist.
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 Fall '11
 Staff
 Calculus, critical number

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