MATH 136
The Extreme Value Theorem
Let f be a continuous function on the interval [ a , b ]. The Extreme Value Theorem (EVT)
states that f obtains maximum value and a minimum value on [ a , b ]. These extreme
values occur at either an endpoint or at a critical point within [ a , b ].
€
Recall: A critical point is a value x in the €
domain of f for which f ′(x) = 0 or for which
€ €
f ′(x) is undefined.
€ €
Example 1. Let f ( x) = (€x 2 − 9x)1/3 on [–4, 8]. Find all critical points. Then find the
maximum value and the minimum value of f on the given interval and state where
these extreme values occur.
Solution. We first evaluate f ′ (x ) . Then we find the critical points by (i) solving
f ′ (x ) = 0 and (ii) determining where f ′ (x ) is undefined but f ( x) is defined.
1 2
2x − 9
−2 /3
× (2x − 9) =
By the Chain Rule, f ′ (x ) = (x − 9x )
. (i) Then
2
3
3(x − 9x )2 /3
f ′ (x ) = 0 when 2x − 9 = 0 , giving x = 4.5 as a critical point.
(ii) f ′ (x ) is undefined when its denominator is 0; that is, x 2 − 9x = x( x − 9) = 0 , giving
x = 0 and
€ x = 9 . Because€ f (0) and f (9) are defined, x = 0 and x = 9 are critical points
of f as well (even though x = 9 is not in the interval under consideration.)
€
By graphing f on [–4, 8], we see that the
€
€
min occurs at the critical point x = 4.5 and the
minimum value €is f (4.5) ≈ −2.725681 . The
max occurs at the left endpoint x = –4 and the
maximum value is f (−4 ) ≈ 3. 7325112 .
€
€
€
10 x
Example 2. Find the critical points of f ( x) =
2 and find the extreme values of f
4−x
on the interval [3, 6].
10(4 − x 2 ) − 10 x(−2x ) 40 + 10 x 2
=
Solution. By the Quotient Rule, f ′ (x ) =
. (i) Then
(4 − x 2 )2
(4 − x 2 )2
2
f ′ (x ) = 0 when 40 + 10x 2 = 0 , giving 10 x = − 40 , which has no real solutions. Thus,
f ′ (x ) = 0 yields no critical points.
(ii) f ′ (x€) is undefined when 4 − x 2 = 0 , giving x = ± 2 . But f itself is not defined at
x = ± 2 , so these values are not critical points. Thus, f has no critical points.
f is not continuous at x = ± 2 , but f is
€ the EVT applies. From
continuous on [3, 6], so
the graph, we see that the min occurs at the
left endpoint x = 3 and the minimum value is
The max occurs at the right
f (3) = −6 .
endpoint x = 6 and the maximum value is
f (6) = −1.875 .
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€
The Extreme Value Theorem does not necessarily apply if we do not have a
continuous function, or if the function is defined only on an open interval. We illustrate
with several examples.
1
on (0, ∞). Then f is continuous, but f has no maximum value on the
x
1
= +∞ . And athough f (x) > 0 for all x in (0, ∞) , f
open interval (0, ∞) because lim
x → 0+ x
€
€
€has no minimum value in this interval. The actual function values get smaller and
smaller as x increases, and there is always another function value that is smaller than
€
any previous function value.
€
€
€
€
(a) Let f (x) =
(b)€ Let f (x) = x 2 on [2, 5). Then f is continuous and f has a minimum value at the
closed left endpoint of the interval [2, 5) and this minimum value is f (2) = 4 . But f has
2
no maximum value on [2, 5). Even though lim x = 25 , there is no function value that
x → 5−
€
€
€equals 25 because the domain
is [2, 5) and the range is [4, 25). As x increases to 5
€ that is greater
€
without equaling 5, there is always another function value
than any
previous function value.
€
€
tan x if 0 ≤ x < π /2
(c) On the interval [0, π], let f (x) = 
. Then f is defined on a closed
if π/2 ≤ x < π
 x
interval [0, π], but f is not continuous at the point x = π /2 within this interval because
lim f (x) ≠ lim f (x) . And f has no maximum value in this interval because
π−
π+
€
x→
x→
2
2 €
€
lim f (x)€= lim tan x = +∞ .
π−
π−
€
x→
x→
2
€
€
2