Download Wednesday, March 25

Wednesday, March 25
Math 142, Section 01, Spring 2009
This lesson introduces the notions of absolute maxima and absolute minima. We discuss a general
method for finding the absolute extrema of a continuous function defined on a closed interval. We
give some examples of finding absolute extrema for a function defined on an open interval.
A function f (x) has an absolute maximum at x = c if f (x) ≤ f (c) for all x in the domain of
f (x). A function f (x) has an absolute minimum at x = c if f (x) ≥ f (c) for all x in the domain of
f (x).
The terms absolute maximum and global maximum are synonomous. Likewise, absolute minimum
and global minimum are synonomous.
A function need not have absolute extrema. The Extreme Value Theorem gives conditions on
f (x) which guarantee that f (x) has both a absolute maximum and a absolute minimum.
Extreme Value Theorem: Suppose f (x) is continuous on a closed, bounded interval [a, b] (thus,
−∞ < a < b < ∞). Then f (x) has a absolute maximum and a absolute minimum.
This theorem tells us nothing about how many absolute maxes or absolute mins f (x) may have.
Nor does it tell us how to find these extrema. It simply asserts there always is at least one absolute
max and at least one absolute minimum, provided f (x) is continuous on a closed, bounded interval.
The conclusion of the theorem may fail if f (x) is continuous on an open or infinite interval. For
example, consider the function f (x) = x1 . This function is continuous on the open interval (0, ∞),
but f (x) has no largest nor no smallest value.
If f (x) is has at most a finite number of points of nondifferentiablity on the open interval (a, b),
in addition to being continuous on the closed interval [a, b], then we can use the following strategy
to locate the absolute extrema:
(1) Find all of the critical numbers. Call them c1 , c2 , · · · , cN .
(2) Evaluate f (x) at each of the critical numbers, also evaluate f (x) at the endpoints, f (a) and
f (b).
(3) The absolute maximum is the largest of the numbers f (a), f (c1 ), . . . , f (cN ), f (b). The absolute minimum is the smallest of the numbers f (a), f (c1 ), . . . , f (cN ), f (b).
The method outlined above is sometimes refered to as the closed interval method.
Example 1: Let f (x) = x + cos (2x). Find the absolute maximum and absolute minimum of f (x)
on [0, π4 ].
Solution: First find critical points: f 0 (x) = 1 − 2 sin (2x). Notice f 0 (x) is defined everywhere so
we only need to look for zeros of the derivative when looking for critical numbers. Now, f 0 (x) = 0
if and only if sin (2x) = 12 , if and only if
2x =
π
5π
+ 2π · k or 2x =
+ 2π · k
6
6
1
2
for k a whole number. Therefore,
π
+ π · k.
12
π
The only critical value in the interval [0, π4 ] is 12
+π·0=
x=
π
.
12
Now,
√
π
π
π
3
π
+ cos (2 · ) =
+
≈ 1.1278.
f( ) =
12
12
12
12
2
Next, we check the endpoints.
f (0) = 0 + cos (2 · 0) = 1.
π
π
π
π
f ( ) = + cos (2 · ) = ≈ 0.785.
4
4
4
4
Of the three numbers 1.1278, 1, and 0.785, we see 0.785 is the smallest and 1.1278 is the largest. So
π
the absolute maximum occurs at x = 12
and
the absolute minimum occurs at x = π4 , and the value
√
π
π
) = 12
+ 23 and the value of the absolute minimum is f ( π4 ) = π4 .
of the absolute maximum is f ( 12
When locating absolute extrema we need to pay attention to both the function and the domain
the function is defined on. The next example looks at the same function as the previous example,
but considers a larger domain.
Example 2: Let f (x) = x + cos (2x). Find the absolute maximum and absolute minimum of f (x)
on [0, π].
Solution: As in example 1, the critical values are of the form
π
5π
x=
+ π · k or x =
+π·k
12
12
π
where k is a whole number. Notice that 12
and 5π
are the only critical values in the interval [0, π].
12
Now
√
π
π
3
+
≈ 1.1278.
f( ) =
12
12
2
and
√
5π
5π
5π
5π
3
f( ) =
+ cos (2 ·
)=
−
≈ 0.4430.
12
12
12
12
2
Now we check the endpoints.
f (0) = 0 + cos (2 · 0) = 1.
f (π) = π + cos (2π) = π + 1 ≈ 4.1416.
Of the four numbers 1.1278, 0.4430, 1, and 0.5708, we see 0.4430 is the smallest and 4.1416 is the
largest. So the absolute maximum occurs at x = π and the absolute minimum occurs at x = 5π
12
and the values√of the absolute maximum is f (π) = π + 1 and the value of the absolute minimum is
f ( 5π
) = 5π
− 23 .
12
12
Locating absolute extrema for functions defined on infinite intervals can be a bit more challenging.
First, we already noted that there is no guarantee that a continuous function on an infinite interval
need have absolute extrema. It is interesting to notice that the function f (x) = x + cos (2x) has
infinitely many local maxima and local minima on (−∞, ∞) but no absolute extrema. Also, the
3
function g(x) = cos x has infinitely many absolute extrema on (−∞, ∞). The absolute maximum
is 1 and the absolute minimum is −1. A absolute maximum occurs at each x = 2πk, where k is any
whole number. A absolute minimum occurs at each x = π + 2πk where k is any whole number.
To find absolute extrema of a function defined on an infinite or open interval we first locate all
critical points. We also need to consider the behavior as x goes off to infinity, or off to the edge of
the interval.
2
Example 3: Let g(x) = xe−x . Find the absolute maximum and absolute minimum (if either
exists) of g(x) on (−∞, ∞).
Solution: First we find critical values to find the local extrema.
2
2
2
g 0 (x) = e−x − 2x2 e−x = (1 − 2x2 )e−x .
−1
g 0 (x) = 0 if and only if x = ± √12 . You should check that x = √
gives a local min and x = √12 gives
2
a local max using either the first or the second derivative tests. Also,
−1
−1
g( √ ) = √
2
2·e
and
1
1
g( √ ) = √
.
2
2·e
Next we look at the behavior of g(x) as x → ±∞. We rewrite g(x) = exx2 so that we may apply
L’Hospital’s rule. The problem is that either of the local extrema may fail to be absolute extrema
since g(x) possibly could take larger and larger values as x → ∞.
1
x
lim
= 0.
2 =
x
x→±∞
e
2xex2
Thus, we see that g(x) has a horizontal asymptote along y = 0 as x → ±∞. Thus, the local max
and local min of g(x) are indeed a absolute max and a absolute min.
There is at least one situation in which we can immediately claim that a local extrema actually
gives rise to an absolute extrema when f (x) is defined on either an infinite or open interval. We
state the test for local / absolute minima. An analogous test holds for maxima.
Suppose f (x) is continuous on an interval (possibly open, possibly infinite). Suppose that f (x)
has a local minimum at x = c. Suppose further that f (x) has no critical values other than x = c.
Then f (x) has a absolute minimum at x = c.
lim
x→±∞
Example 4: Let L(x) = x − ln x. Find the local maxima and local minima, if either exists.
Solution: The natural domain of definition of L(x) is (0, ∞). We find L0 (x) = 1 −
critical value is x = 1. Now,
1
L00 (x) = 2
x
so
L00 (1) > 0
1
x
so the only
4
and so the second derivative test tells us that x = 1 gives a local minimum for L(x). Furthermore,
since x = 1 is the only critical value of L(x), we can assert that x = 1 gives a absolute minimum.
Example 5: The force between two atoms in a molecule is given by
A
B
F (r) = − 2 + 3
r
r
where r is the distance between the centers of the atoms and A and B are positive constants. At
what distance of seperation is the force a minimum?
Solution: For physical reasons, we restrict the domain of F (r) to (0, ∞). We find that F (r) has a
single critical point:
2A 3B
2Ar − 3B
F 0 (r) = 3 − 4 =
.
r
r
r4
F 0 (r) = 0 if and only if the numerator is zero, so r = 3B
is the only critical point. Next,
2A
−6Ar + 12B
6 · (2B − Ar)
−6A 12B
+ 5 =
=
,
4
5
r
r
r
r5
5
6 · (2B − A 3B
)
3B
3B
00 3B
2A
F ( )=
·
=
> 0.
3B 5
2A
2
2A
( 2A )
F 00 (r) =
so
gives an absolute minimum since it gives a local minimum and it is the only critical point.
r = 3B
2A
This minimal force is
3B
3B
3B
3B
3B
−A 2A 2
F ( ) = −A · ( )−2 + B · ( )−3 = (−A + B · ( )−1 )( )−2 =
·( ) .
2A
2A
2A
2A
2A
3
3B
The minimal force is negative, meaning the force is attractive when it is minimal.
Notice that limr→0 F (r) = ∞ and limr→0 F (r) = 0. To verify these, we rewrite F (r) = B−Ar
.
r3
Then
B − Ar
=0
lim
r→∞
r3
since the denominator has higher degree (you could also use L’Hospital’s rule here.)
B − Ar
lim
=∞
r→0
r3
since the numerator approaches the nonzero number B while the denominator gets smaller and
smaller.
It is useful to think about these limits physically as well. As the distance that separates the
atoms shrinks, the repulsive force between the positively charged nuclei overpowers any attracitive
forces between the atoms, resulting in a very large force. For the other limit, the attractive force
between the atoms diminishes as the distance grows without bound.