Extreme Values of Functions
Chapter 5.1
Absolute (Global) Extreme Values
• Up to now we have used the derivative in applications to find rates of
change
• However, we are not limited to the rate-of-change interpretation of the
derivative
• In this section you will learn how we can use derivatives to find
extreme values of functions (that is maximum or minimum values)
2
Definition of Extreme Values on an Interval
DEFINITION:
Let 𝑓 be defined on an interval 𝐼 containing 𝑐.
a) 𝑓(𝑐) is the minimum of 𝒇 on 𝑰 if 𝑓 𝑐 ≤ 𝑓(𝑥) for all 𝑥 in 𝐼
b) 𝑓(𝑐) is the maximum of 𝒇 on 𝑰 if 𝑓 𝑐 ≥ 𝑓 𝑥 for all 𝑥 in 𝐼
The minimum and maximum of a function on an interval are the extreme values or
extrema (plural of extremum), of the function on the interval. The minimum and
maximum of a function on an interval are also called the absolute minimum and
absolute maximum. They are also sometimes called the global minimum and
global maximum.
3
Definition of Extreme Values on an Interval
• The definition given is slightly different than the one given in your
textbook
• Most of the time we are only concerned with extreme values in an
interval, rather than on the entire domain (many functions have neither
absolute maxima nor absolute minima over the entire domain)
• Over an interval, extrema can occur either in the interior or at the
endpoints
• It is possible for a function to have no extrema on an interval
4
Definition of Extreme Values on an Interval
5
Definition of Extreme Values on an Interval
6
Definition of Extreme Values on an Interval
7
Example 1: Exploring Extreme Values
Use the graphs of 𝑓 𝑥 = cos 𝑥 and 𝑔 𝑥 = sin 𝑥 on
determine absolute maxima and minima (if any).
𝜋 𝜋
− ,
2 2
to
8
Example 1: Exploring Extreme Values
9
The Extreme Value Theorem
THEOREM:
If 𝑓 is continuous on a closed interval [𝑎, 𝑏], then 𝑓 has both a maximum value and
a minimum value on the interval.
10
The Extreme Value Theorem
• The proof of this theorem requires more advanced calculus, so we will
take the theorem as given
• Note that continuity is a requirement of the proof; if we know that the
function is not continuous on a given interval, then we cannot use the
theorem
• In plain words, this tells us that a function is guaranteed to have both a
maximum and a minimum value on a closed interval (if continuous)
• These extrema may be either at the endpoints of the interval or the
interior of the interval
11
The Extreme Value Theorem
12
The Extreme Value Theorem
13
The Extreme Value Theorem
14
The Extreme Value Theorem
15
Local (Relative) Extrema
• In addition to absolute (or global) extrema, which are always the
greatest/least function value on an interval, we will want to define
relative (local) extrema
• These occur when “nearby” values are all less (for a relative
maximum) or greater (for a relative minimum)
• Relative extrema may also be absolute extrema, but not all relative
extrema are absolute extrema (but all absolute extrema are also
relative extrema)
16
Local (Relative) Extrema
DEFINITION:
Let 𝑐 be an interior point of the domain of the function 𝑓. Then 𝑓(𝑐) is a
a) local maximum value at 𝑐 if and only if 𝑓 𝑥 ≤ 𝑓(𝑐) for all 𝑥 in some open
interval containing 𝑐
b) local minimum value at 𝑐 if and only if 𝑓 𝑥 ≥ 𝑓(𝑐) for all 𝑥 in some open
interval containing 𝑐
c) A function 𝑓 has a local maximum or local minimum at an endpoint 𝑐 if the
appropriate inequality holds for all 𝑥 in some half-open interval containing 𝑐
17
Local (Relative) Extrema
18
Local (Relative) Extrema
• Relative extrema in the interior of an interval occur at points where the
graph of a function changes direction (from increasing to decreasing,
or vice versa)
• We would like to be able to find both absolute and relative extrema for
a function over a closed interval
• To narrow down the possibilities for the interior of an interval, we can
ask, “is there anything about relative extrema by which we can identity
them?”
19
Critical Point
DEFINITION:
Let 𝑓 be a function defined over some interval 𝐼. A point 𝑐, 𝑓 𝑐 , where 𝑐 is in the
interior of 𝐼, at which 𝑓 ′ 𝑐 = 0 or 𝑓 is not differentiable is called a critical point
of 𝑓.
The number 𝑐 in the interval is called a critical number of 𝑓.
20
Critical Point
• Note that critical points occur where either
• 𝑓′ is zero
• 𝑓 is not differentiable
• The next theorem (not in your text and presented without proof) gives
us an answer to our previous question
21
Relative Extrema Occur Only at Critical
Numbers
THEOREM:
If 𝑓 has a relative minimum or relative maximum at 𝑥 = 𝑐, then 𝑐 is a critical
number of 𝑓 (and 𝑐, 𝑓 𝑐 is a critical point of 𝑓).
22
Relative Extrema Occur Only at Critical
Numbers
• This theorem joins the definition of a critical number with the
definition of relative extrema
• Specifically, it says that relative extrema occur only at critical
numbers
• Therefore, to find relative extrema, we need only find critical
numbers, which occur where either 𝑓′ is zero or where 𝑓 is not
differentiable
• However, we must be clear about what it does not say: it does not say
that, if we find a critical number, then we have found a relative
extremum
23
Relative Extrema Occur Only at Critical
Numbers
24
Finding Absolute Extrema
• We now have enough understanding to be able to find the absolute
extrema on a closed interval
• If we have a closed interval, the Extreme Value Theorem guarantees that there
exist both an absolute maximum and absolute minimum in the interval
• These absolute extrema may occur at either the endpoints or in the interior
• If absolute extrema occur in the interior, then they occur at relative extrema
(which, in turn, occur at critical numbers)
• To find absolute extrema, we will first find relative extrema in the
interior (i.e., find critical numbers), evaluate the function at this
25
Finding Absolute Extrema
• To find absolute extrema
1.
2.
3.
4.
We will first find relative extrema in the interior (i.e., find critical numbers)
Evaluate the function at all critical numbers
Evaluate the function at its endpoints
Compare these values; the largest of these is the absolute maximum and the
smallest is the absolute minimum
26
Example 3: Finding Absolute Extrema
2
3
Find the absolute maximum and minimum values of 𝑓 𝑥 = 𝑥 on the
interval [−2,3].
27
Example 3: Finding Absolute Extrema
2
3
Find the absolute maximum and minimum values of 𝑓 𝑥 = 𝑥 on the
interval [−2,3].
′
Find the critical numbers by taking the derivative: 𝑓 𝑥 =
2 −1
𝑥 3
3
=
2
33 𝑥
Critical numbers occur where 𝑓′ is zero or where 𝑓 is not differentiable. Note that
𝑓′ is never zero. However, 𝑓′ is not defined at 𝑥 = 0, which means that 𝑓 is not
differentiable at 𝑥 = 0, by our definition, 𝑥 = 0 is a critical number (and 0,0 is a
critical point). Now, check the function values at 𝑥 = −2, 𝑥 = 0, and 𝑥 = 3:
𝑓 −2 ≈ 1.587, 𝑓 0 = 0, 𝑓 3 ≈ 2.08
The maximum is approximately 2.08 and occurs at 𝑥 = 3; the minimum is 0 and
occurs at 𝑥 = 0
28
Example 3: Finding Absolute Extrema
29
Example 4: Finding Extreme Values
Find the extreme values of 𝑓 𝑥 =
1
4−𝑥 2
.
30
Example 4: Finding Extreme Values
Find the extreme values of 𝑓 𝑥 =
1
4−𝑥 2
.
First note that no interval is provided. From the function it is clear that the domain
is −2,2 . Since this is not a closed interval, we cannot conclude that the function
has both an absolute minimum and absolute maximum. We differentiate to find
critical numbers:
−1
1
1
𝑥
−
−1
2
′
2
𝑓 𝑥 = 4 − 𝑥 2 ⟹ 𝑓 𝑥 = − ⋅ −2𝑥 ⋅ 4 − 𝑥 2 =
3
2
4 − 𝑥2 2
We have that 𝑓 ′ 𝑥 = 0 if 𝑥 = 0 and this is the only critical number because critical
1
numbers are only interior values of an interval. The critical point occurs at 0,
2
31
Example 4: Finding Extreme Values
Find the extreme values of 𝑓 𝑥 =
1
4−𝑥 2
.
How can we know whether 𝑓(0) is a maximum, minimum, or neither? If 𝑥 > 0,
then the denominator of 𝑓 decreases, so its reciprocal increases. The same is true if
𝑥 < 0. This means that, for all 𝑥 ≠ 0, 𝑓 𝑥 > 𝑓(0) and by definition this means
that we have an absolute minimum at 𝑥 = 0.
Is there an absolute maximum? As 𝑥 approaches 2 from the left, the denominator
approaches zero, so the function approaches infinity. The same is true as 𝑥
approaches −2 from the right. So this function has no maximum value.
32
Example 4: Finding Extreme Values
33
Example 5: Finding Local Extrema
Find the local extrema of
5 − 2𝑥 2 , if 𝑥 ≤ 1
𝑓 𝑥 =
𝑥 + 2, if 𝑥 > 1
34
Example 5: Finding Local Extrema
Find the local extrema of
5 − 2𝑥 2 , if 𝑥 ≤ 1
𝑓 𝑥 =
𝑥 + 2, if 𝑥 > 1
This function cannot have absolute extrema since function values continue
increasing to the right of 1, and continue decreasing to the left of 1. What about
local (relative) extrema? We differentiate 𝑓 by differentiating the two pieces of the
function. But we must determine whether the function is differentiable at 𝑥 = 1. To
do this, find the left- and right-hand derivatives:
𝑓 1+ℎ −𝑓 1
1 + ℎ + 2 − (1 + 2)
ℎ
lim
= lim+
= lim+ = 1
ℎ→0+
ℎ→0
ℎ→0 ℎ
ℎ
ℎ
35
Example 5: Finding Local Extrema
Find the local extrema of
5 − 2𝑥 2 , if 𝑥 ≤ 1
𝑓 𝑥 =
𝑥 + 2, if 𝑥 > 1
𝑓 1+ℎ −𝑓 1
5−2 1+ℎ 2− 5−2 1
lim−
= lim−
ℎ→0
ℎ→0
ℎ
ℎ
2
−4ℎ − 2ℎ2
= lim−
= lim−(−4 − 2ℎ) = −4
ℎ→0
ℎ→0
ℎ
The derivatives are different, so 𝑓 is not differentiable at 𝑥 = 1.
36
Example 5: Finding Local Extrema
Find the local extrema of
5 − 2𝑥 2 , if 𝑥 ≤ 1
𝑓 𝑥 =
𝑥 + 2, if 𝑥 > 1
So our derivative is
𝑓′
−4𝑥, if 𝑥 < 1
𝑥 =
1, if 𝑥 > 1
Therefore we have critical points at 𝑥 = 0 and 𝑥 = 1. Since the function piece
defined for 𝑥 ≤ 1 is a parabola that opens down, then 𝑓(0) must be a local
maximum.
37
Example 5: Finding Local Extrema
We can determine how to classify the critical point (1,3) by examining what
happens to the function values at 𝑥 on either side of (but near) 𝑥 = 1. To the
left of 𝑥 = 1, note that 𝑓 .9 = 3.38, 𝑓 0.8 = 3.72, 𝑓 0.7 = 4.02; in general,
these nearby values are greater than 3. To the right of 𝑥 = 1: 𝑓 1.1 = 3.1,
𝑓 1.2 = 3.2, 𝑓 1.3 = 3.3; in general, these nearby values are greater than 3.
So we have, that 𝑓 𝑥 > 𝑓(1) for values near 𝑥 = 1 so by definition, 1,3
occurs at a local minimum.
38
Example 5: Finding Local Extrema
39
Example 6: Finding Local Extrema
Find the local extrema of 𝑓 𝑥 = ln
𝑥
1+𝑥 2
.
40
Example 6: Finding Local Extrema
Find the local extrema of 𝑓 𝑥 = ln
𝑥
1+𝑥 2
.
The textbook asks you to use a calculator, but differentiating this function is not
beyond your ability if you first note the following:
Suppose that 𝑢 is a differentiable function of 𝑥. Then 𝑦 = ln 𝑢 can be written as a
piecewise defined function
ln 𝑢 , if 𝑢 > 0
𝑦=
ln(−𝑢) , if 𝑢 < 0
𝑑𝑦
1
𝑑𝑦
1
1
Then we get = 𝑢′ ⋅ if 𝑢 > 0 and = −𝑢′ ⋅
= 𝑢′ ⋅ , if 𝑢 < 0. We can
𝑑𝑥
𝑢
𝑑𝑥
−𝑢
𝑢
ignore the absolute value sign!
41
Example 6: Finding Local Extrema
Find the local extrema of 𝑓 𝑥 = ln
Now, if 𝑢 =
𝑥
,
1+𝑥 2
𝑥
1+𝑥 2
.
then
′
𝑢
𝑓 𝑥 = ln 𝑢 ⟹ 𝑓 ′ 𝑥 =
𝑢
Find 𝑢′ (with respect to 𝑥):
2 ⋅ 1 − 𝑥 2𝑥
2
1
+
𝑥
1
−
𝑥
𝑢′ =
=
2
2
1+𝑥
1 + 𝑥2 2
Therefore,
2
2
2
1
−
𝑥
1
+
𝑥
1
−
𝑥
𝑓′ 𝑥 =
⋅
=
2
2
1+𝑥
𝑥
𝑥 1 + 𝑥2
42
Example 6: Finding Local Extrema
Find the local extrema of 𝑓 𝑥 = ln
𝑥
1+𝑥 2
.
2
2
2
1
−
𝑥
1
+
𝑥
1
−
𝑥
𝑓′ 𝑥 =
⋅
=
2
2
1+𝑥
𝑥
𝑥 1 + 𝑥2
Find where 𝑓′ is zero and where 𝑓 is not differentiable. The function is not
differentiable at 𝑥 = 0, but this doesn’t count because ln 0 is not defined (so zero is
not in the domain of 𝑓). We have 1 − 𝑥 2 = 0 ⟹ 𝑥 = ±1. The function values for
1
these critical numbers are 𝑓 1 = ln
= 𝑓 −1 ≈ −0.693
2
43
Example 6: Finding Local Extrema
Find the local extrema of 𝑓 𝑥 = ln
𝑥
1+𝑥 2
.
Finally, check some nearby function values to determine whether these are
local maxima, minima, or neither.
At 𝑥 = −1: 𝑓 −1.1 ≈ −0.698, 𝑓 −0.9 ≈ −0.699; this appears to be a local
maximum
At 𝑥 = 1: 𝑓 1.1 ≈ −0.698, 𝑓 0.9 ≈ −0.699; this, too, appears to be a local
maximum.
44
Example 6: Finding Local Extrema
45
Exercise 5.1
46