MATH 150A: Calculus I Study Notes
Topic: Finding Absolute Maximum and Minimum Values on a Closed Interval
CSU Northridge (Northridge, CA)
Introduction: The Shift from Local to Absolute
The initial phase of differential calculus focuses on local extrema, where we analyze
a functions behavior in small, immediate neighborhoods using the First and Second
Derivative Tests. However, in practical applications, such as maximizing profit,
minimizing cost, or determining the absolute highest point on a suspension bridge
cable, we are not interested in a momentary peak or valley, but rather the Absolute
Maximum and Absolute Minimum values a function attains across its entire domain
or a specific, bounded portion of it.
This shift in focus—from local behavior to global behavior within a constrained
domain—is mathematically formalized and guaranteed by the Extreme Value
Theorem (EVT). My personal insight into this topic is the profound simplicity and
necessity of the theorems conditions. The calculus only provides the candidates for
extrema (the critical numbers), but it is the geometry of a closed interval that forces
one of those candidates, or a boundary point, to be the absolute winner. The
algorithm we follow is merely a systematic way of applying the EVTs guarantee:
checking the only three places an absolute extremum can possibly occur.
The requirement that we check endpoints, stationary points, and non-differentiable
points is not just a rule; it’s a comprehensive checklist derived directly from the
mathematical proof that these are the only possibilities. If an absolute extremum
exists, it must be one of these points. Our job is simply to find the complete list of
candidates and compare the functions value at each one.
Part I: The Theoretical Foundation – The Extreme Value Theorem
The entire process of finding absolute extrema on a closed interval is underpinned
by a single, powerful existence theorem from advanced calculus. We do not need to
prove the theorem in MATH 150A, but we must understand the precise conditions
required for its guarantee to hold.
1. Statement of the Extreme Value Theorem (EVT)
The Extreme Value Theorem is stated as follows:
If a function
f(x)
is continuous on a closed interval
[a , b]
, then
f(x)
attains an
absolute maximum value
f(c)
and an absolute minimum value
f(d)
at some
numbers
c
and
d
in the interval
[a , b]
.
This statement is profound because it is an existence theorem. It doesnt tell us how
to find these absolute extrema, but it guarantees that they must exist, provided two
conditions are met:
A. The Necessity of Continuity
The function
f(x)
must be continuous over the entire interval
[a , b]
. If a function has
a discontinuity, the theorem fails, and an absolute extremum might not exist.
Counterexample 1: Discontinuity.
Consider the piecewise function:
f(x)=
{
xif 0 ≤ x <1
0 if x=1
on the closed interval
[0,1]
. This function is discontinuous at
x=1
.
The absolute minimum is
f(1)=0
.
As
x
approaches 1 from the left,
f(x)
approaches
1
. However, the function never
actually reaches
1
. It gets arbitrarily close to
1
(e.g.,
0.9 ,0.99 ,0.999 ,…
) but never
attains that value.
Therefore, the function fails to attain an absolute maximum on the interval
[0,1]
,
precisely because the condition of continuity was violated. The theorems guarantee
is lost.
B. The Necessity of a Closed Interval
The interval must be closed, meaning it includes its endpoints, denoted by the
square brackets:
[a , b]
. If the interval is open,
(a , b)
, or half-open,
¿
or
¿
, the theorem
fails, and the function might never attain its maximum or minimum value.
Counterexample 2: Open Interval.
Consider the simple function
f(x)=x
on the open interval
(0,1)
.
As
x
approaches
1
from the left,
f(x)
approaches
1
.
As
x
approaches
0
from the right,
f(x)
approaches
0
.
However, because the interval is open,
x
can never equal
0
or
1
. The function value
gets arbitrarily close to
1
, but never reaches it, so there is no absolute maximum.
Similarly, there is no absolute minimum.
The EVTs guarantee is essential because the process of "checking the endpoints"
only makes sense if the endpoints
a
and
b
are actually part of the domain, which is
the definition of a closed interval.
2. Definition of the Candidates: Critical Numbers
The second crucial theoretical component is the Critical Number. Critical numbers
are the only places in the interior of the interval where a local extremum, and thus a
potential absolute extremum, can occur.
Formal Definition of a Critical Number
A critical number
c
is a number in the domain of
f
such that:
f′(c)=0
(Stationary Point): This is where the tangent line is perfectly horizontal. The
slope momentarily stops changing direction. These are typically the peaks and
valleys of smooth curves.
f′(c)
does not exist (DNE) (Singular Point): This is where the graph has a sharp
point, a corner, or a cusp, or a vertical tangent. The function is still continuous at this
point, but it is not differentiable. These points, though technically non-smooth, can
and often do contain absolute extrema.
The Geometric Necessity of Critical Numbers
If an absolute extremum occurs at some interior point
c
(where
a<c<b
), then by
definition,
f(c)
must be a local extremum. Fermats Theorem states that if
f(c)
is a
local extremum and
f′(c)
exists, then
f′(c)
must be equal to zero. If
f′(c)
does not
exist, then
c
must be a singular point. This means that every interior absolute
extremum must be a critical number.
3. Absolute Extrema vs. Local Extrema
It is vital to distinguish between these concepts:
Local Extrema: Determined by the sign change of
f′(x)
. They only consider the
immediate neighborhood. A function can have many local maxima and minima.
Absolute Extrema: Determined by comparison of function values over the entire
domain
[a , b]
. There can only be one absolute maximum value and one absolute
minimum value (though they might occur at multiple
x
-values).
The Role of the Endpoints
The functions overall shape is what matters for absolute extrema. Imagine a
function that is always increasing on
[a , b]
. It will have no local extrema. However,
the EVT still guarantees an absolute minimum at
x=a
and an absolute maximum at
x=b
. This is why the endpoints are essential candidates; they act as the "fences" that
define the restricted domain, and the absolute extrema will often be found right on
these boundaries, independent of the interior calculus.
Part II: The Complete Method and Algorithm
The Extreme Value Theorem and the concept of Critical Numbers lead directly to the
formal, three-step algorithm for finding the absolute extrema of a continuous
function
f
on a closed interval
[a , b]
. This is often called The Closed Interval Method.
The Closed Interval Method Algorithm
Step 1: Locate All Candidate Points (Endpoints and Critical Numbers)
The absolute extremum must occur at one of two types of points:
Endpoints: The boundaries of the interval,
x=a
and
x=b
.
Critical Numbers: Any value
c
in the open interior
(a , b)
such that
f′(c)=0
or
f′(c)
DNE.
To execute this step:
Identify
a
and
b
.
Compute the first derivative,
f′(x)
.
Solve
f′(x)=0
for
x
. These are the stationary critical numbers.
Identify where
f′(x)
DNE. These are the singular critical numbers.
Crucially: Discard any critical numbers found that lie outside the interval
[a , b]
.
Step 2: Evaluate the Function at All Candidate Points
Create a complete list of all candidate
x
-values from Step 1 (the two endpoints and
all interior critical numbers). For each candidate
x
-value, substitute it into the
original function,
f(x)
, to determine the corresponding
y
-value.
f(a)
f(b)
f(c1)
,
f(c2)
,
…
(for all interior critical numbers)
This step produces a finite set of
y
-values.
Step 3: Compare and Conclude
The final set of
y
-values contains the absolute maximum and absolute minimum
values guaranteed by the Extreme Value Theorem.
Absolute Maximum Value: The largest value in the set from Step 2.
Absolute Minimum Value: The smallest value in the set from Step 2.
The conclusion must be stated clearly, identifying both the value (the
y
-coordinate)
and the location (the
x
-coordinate). For example: "The absolute maximum value is
Y
at
x=X
."
Part III: Comprehensive Example 1: A Polynomial Function
We will apply the Closed Interval Method in painstaking detail to the function
f(x)=x3−6x2+5
on the closed interval
[−1,5]
.
A. Theoretical Verification
Continuity: The function
f(x)=x3−6x2+5
is a polynomial. All polynomials are
continuous everywhere, and therefore
f(x)
is continuous on the interval
[−1,5]
.
Closed Interval: The given interval is
[−1,5]
, which is a closed interval.
EVT Guarantee: Since both conditions (continuity and closed interval) are satisfied,
the Extreme Value Theorem guarantees that an absolute maximum and an absolute
minimum value must exist within the interval.
B. Step 1: Locating All Candidate Points
1. Endpoints
The endpoints are
a=−1
and
b=5
. These are the first two candidates.
2. Critical Numbers
i. Find the First Derivative
f′(x)= d
dx (x3−6x2+5)=3x2−12 x
Since
f′(x)
is a polynomial, it exists for all
x
. Therefore, there are no critical numbers
where
f′(x)
DNE.
ii. Find Stationary Points (
f′(x)=0
)
We set the derivative equal to zero and solve:
3x2−12 x=0
Factor the expression:
3x(x − 4)=0
This yields two stationary points:
xc=0
xc=4
3. Filtering Critical Numbers
We must check if these critical numbers lie within the open interior
(−1,5)
:
x=0
: Since
−1<0<5
,
x=0
is an interior critical number. Candidate.
x=4
: Since
−1<4<5
,
x=4
is an interior critical number. Candidate.
The complete list of candidate
x
-values is
{−1,5,0,4}
.
C. Step 2: Evaluating the Function at All Candidate Points
We substitute each candidate
x
-value back into the original function
f(x)=x3−6x2+5
.
1. Evaluate Endpoints
At
x=−1
(Left Endpoint):
f(−1)=¿
The value is
−2
.
At
x=5
(Right Endpoint):
f(5)=¿
The value is
−20
.
2. Evaluate Critical Numbers
At
x=0
(Stationary Point):
f(0)=¿
The value is
5
. (This value is a local maximum by the SDT, since
f″(0)=6(0)−12=−12<0
, confirming its local nature, but we must compare it
globally.)
At
x=4
(Stationary Point):
f(4)=¿
The value is
−27
. (This value is a local minimum by the SDT, since
f″(4)=6(4)−12=12>0
, confirming its local nature, but again, global comparison is
required.)
The complete set of
y
-values is
{−2,− 20 ,5, −27 }
.
D. Step 3: Comparing and Concluding
We examine the set of values
{−2,− 20 ,5, −27 }
:
Largest Value: The largest value is
5
.
Smallest Value: The smallest value is
−27
.
Formal Conclusion:
The absolute maximum value of
f(x)
on the interval
[−1,5]
is
5
, which occurs at the
stationary point
x=0
.
The absolute minimum value of
f(x)
on the interval
[−1,5]
is
−27
, which occurs at
the stationary point
x=4
.
Commentary on the Example: This example perfectly illustrates the necessity of the
full method. The absolute maximum was a local maximum (
x=0
), but the absolute
minimum was also a local minimum (
x=4
). Crucially, the endpoints, despite not
being the absolute extrema themselves, represented other local bounds that had to
be considered to ensure the global comparison was complete. For instance, the
function value at the right endpoint,
f(5)=−20
, was significantly larger than the
absolute minimum, demonstrating how the interval boundary can cut off the
function before it returns to a higher value.
Part IV: Comprehensive Example 2: Non-Differentiable Singular Points
The second necessary type of example is one that involves the critical number
where the derivative does not exist. This is a critical trap for students who rely
solely on finding stationary points. We will apply the Closed Interval Method to the
function
g(x)=x2/3(x− 10)
on the closed interval
[−1,8]
.
A. Theoretical Verification
Continuity: The function
g(x)=x2/3(x − 10)=x5/3−10 x2/3
involves fractional
exponents, but since the rational powers have odd denominators (3), the function is
defined and continuous for all real numbers,
(− ∞ , ∞ )
. Thus,
g(x)
is continuous on
[−1,8]
.
Closed Interval: The interval is
[−1,8]
.
EVT Guarantee: The EVT guarantees the existence of absolute extrema.
B. Step 1: Locating All Candidate Points
1. Endpoints
The endpoints are
a=−1
and
b=8
.
2. Critical Numbers
i. Find the First Derivative
First, rewrite the function to facilitate differentiation:
g(x)=x5/3−10 x2/3
.
g′(x)= d
dx (x5/3−10 x2/3)g′(x)= 5
3x5/3−1−10
(
2
3
)
x2/3−1g′(x)= 5
3x2/3−20
3x−1/3
ii. Find Singular Points (
g′(x)
DNE)
The derivative expression contains
x−1/3
, which is
1
x1/3
. This term is undefined when
x=0
. Since
g(x)
is defined at
x=0
,
x=0
is a critical number where the derivative
does not exist (a vertical tangent).
xc=0
. Since
−1<0<8
,
x=0
is an interior singular point. Candidate.
iii. Find Stationary Points (
g′(x)=0
)
We set the derivative equal to zero and solve:
5
3x2/3−20
3x−1/3=0
To simplify, multiply the entire equation by
3
:
5x2/3−20 x−1/3=0
Factor out the lowest power of
x
, which is
5x−1/3
:
5x−1/3
(
x2/3−(−1/3)−4
)
=05 x−1/3
(
x3/3−4
)
=05 x−1/3(x − 4)=0
For the product to be zero, one of the factors must be zero.
5x−1/3=0
: This implies
5
x1/3=0
, which has no solution (as a fraction is zero only if the
numerator is zero). Note that setting the
x−1/3
term to zero would only yield
x=0
again, but
x=0
is already correctly classified as a DNE point, not a stationary point.
x − 4=0
: This yields
xc=4
.
Since
−1<4<8
,
x=4
is an interior stationary point. Candidate.
3. Filtering Critical Numbers
The complete list of candidate
x
-values is
{−1,8,0,4}
.
C. Step 2: Evaluating the Function at All Candidate Points
We substitute each candidate
x
-value back into the original function
g(x)=x2/3(x − 10)
.
1. Evaluate Endpoints
At
x=−1
(Left Endpoint):
g(−1)=¿
Since
¿
:
g(−1)=(1)(−11)=−11
The value is
−11
.
At
x=8
(Right Endpoint):
g(8)=¿
Since
¿
:
g(8)=(4)(−2)=−8
The value is
−8
.
2. Evaluate Critical Numbers
At
x=0
(Singular Point):
g(0)=¿
The value is
0
.
At
x=4
(Stationary Point):
g(4)=¿
Since
¿
:
g(4)= 3
√
16(−6)=−63
√
16
We can simplify
3
√
16=3
√
8⋅2=23
√
2
.
g(4)=−6(23
√
2)=−12 3
√
2
To compare this value numerically, we estimate
3
√
2≈1.2599
.
−12 3
√
2≈ − 12(1.26)≈ −15.12
The exact value is
−12 3
√
2
.
The complete set of exact
y
-values is
{−11 ,− 8,0, −12 3
√
2}
.
D. Step 3: Comparing and Concluding
We compare the set of values
{−11 ,− 8,0, −12 3
√
2≈ −15.12 }
.
Largest Value: The largest value is
0
.
Smallest Value: The smallest (most negative) value is
−12 3
√
2
.
Formal Conclusion:
The absolute maximum value of
g(x)
on the interval
[−1,8]
is
0
, which occurs at the
singular critical point
x=0
.
The absolute minimum value of
g(x)
on the interval
[−1,8]
is
−12 3
√
2
, which occurs
at the stationary point
x=4
.
Commentary on the Example: This powerful example underscores the absolute
necessity of identifying and checking both types of critical numbers. The absolute
maximum was found at
x=0
, a point where the derivative DNE (a cusp). Had we
only searched for stationary points by solving
g′(x)=0
, we would have entirely
missed the location of the highest point on the graph within the specified domain.
This failure to check singular points is a common and costly error, and the
comprehensive Closed Interval Method is designed precisely to prevent it.
Part V: Contextualizing the Extreme Value Theorem in Optimization
In the context of MATH 150A, the ultimate goal of learning the Closed Interval
Method is to solve real-world Optimization Problems. These problems always follow
a common structural pattern, and understanding the EVT is the key to framing the
solution.
The Optimization Framework
Objective Function (The Function
f(x)
): The quantity to be maximized or minimized
(e.g., area, volume, profit, time) must be expressed as a function of a single variable,
f(x)
.
Constraint Equation (Defining the Interval
[a , b]
): All real-world problems have
constraints (e.g., "use exactly 100 feet of fencing," or "the value of
x
must be
positive"). These constraints determine the domain, and in almost all MATH 150A
problems, this domain will be a closed interval
[a , b]
, either explicitly given or
implicitly determined (e.g.,
x ≥ 0
and
x
must be less than some maximum value).
The Guaranteed Search (Applying the EVT): Once the problem is reduced to finding
the absolute extremum of a continuous function on a closed interval, the search is
guaranteed to succeed. We no longer need to worry about the First or Second
Derivative Tests for classification; the comparison of candidates is sufficient.
The Final Answer: The conclusion must always address the original question in the
context of the problem, often requiring units (e.g., "The maximum area is 1,250
square feet," not just "The maximum value is 1,250.").
Rationale for the EVT’s Role in Optimization
The reason the Extreme Value Theorem is so central to optimization is that it allows
us to stop worrying about end behavior. In many optimization problems, the
"endpoints" represent physically nonsensical or trivial situations (e.g.,
x=0
often
means "no material used," which yields zero profit or zero area). However,
mathematically, they serve a vital purpose: they guarantee that we have
encompassed all possible scenarios, and the global optimum must occur somewhere
between these bounds. By checking the boundary conditions and all the points of
horizontal or non-existent slope in the interior, we ensure that no potential
maximum or minimum is overlooked.
Conclusion and Final Remarks
The Closed Interval Method, which is the practical application of the Extreme Value
Theorem, represents one of the most powerful and reliable tools we acquire in
MATH 150A. Its rigor stems from the fact that we are not simply guessing; we are
methodically checking the only possible locations where an absolute extremum can
reside.
Remember the non-negotiable checklist for every problem of this type:
Verify Continuity on a Closed Interval: This confirms the EVT applies.
Find the Full List of Candidates: This includes the endpoints and all critical numbers
(both
f′(c)=0
and
f′(c)
DNE) that fall within the interior.
Plug Candidates ONLY into the Original Function
f(x)
: The value of the derivative is
only used to find the candidates; the value of the function is what determines the
actual maximum and minimum values.
Compare and State the Absolute Conclusion: State the highest and lowest values
found in the comparison.
