MATH 150A: Calculus I Study Notes
Topic: Definition and Determination of Extrema: The Second Derivative Test (The
Second Sufficient Condition)
CSU Northridge (Northridge, CA)
Part I: Personal Insight and Conceptual Bridge
The Second Derivative Test (SDT) represents a significant conceptual leap forward
from the First Derivative Test (FDT). While the FDT requires extensive analysis of
intervals around critical points—often necessitating tedious substitution into a
factored derivative—the SDT offers an elegant, instantaneous classification based
on a single number: the sign of the second derivative evaluated at the critical point.
My key learning insight is that the SDT is fundamentally a test of concavity at a
location where the function is momentarily flat. A local extremum occurs when the
derivative,
f′(x)
, equals zero. This point, where the tangent line is perfectly
horizontal, is known as a stationary point (a special type of critical number). The
question then becomes: What happens immediately after this moment of flatness?
The second derivative,
f″(x)
, answers this by quantifying the rate of change of the
slope.
If
f″(x0)
is positive at the stationary point
x0
, it means the slope (
f′(x)
) is increasing
at that exact moment. Since the slope is zero at
x0
, it must have been negative just
before
x0
and will become positive just after
x0
. A function whose slope goes from
negative (decreasing) to positive (increasing) must "hold water," forming a concave
up shape, which mathematically forces the critical point to be the bottom of a bowl
—a Local Minimum.
If
f″(x0)
is negative, the slope (
f′(x)
) is decreasing. Since the slope is zero at
x0
, it
must have been positive just before
x0
and will become negative just after
x0
. A
function whose slope goes from positive (increasing) to negative (decreasing) must
"spill water," forming a concave down shape, which mathematically forces the
critical point to be the top of a hill—a Local Maximum.
This relationship between the second derivatives sign and the geometric concavity
provides the necessary and sufficient condition to classify the extremum
instantaneously, making the SDT a powerful shortcut, provided, of course, that the
second derivative exists and is non-zero at that point. We will explore the rigorous
derivation of this logic below.
Part II: Formal Definitions and Theoretical Foundation
Before stating the Second Derivative Test formally, we must establish the precise
definitions of the concepts upon which it relies.
1. Definition of Local Extrema
For a function
f(x)
defined on an interval
I
containing
c
:
Local Maximum:
f
has a local maximum at
c
if
f(x)≤ f (c)
for all
x
in some open
interval around
c
. Geometrically, it is the peak of a small hill in its immediate
vicinity.
Local Minimum:
f
has a local minimum at
c
if
f(x)≥ f (c)
for all
x
in some open
interval around
c
. Geometrically, it is the lowest point of a small valley in its
immediate vicinity.
Critical Number (Recap): A value
c
in the domain of
f
where
f′(c)=0
(stationary
point) or
f′(c)
does not exist (DNE). Crucially, the Second Derivative Test only
applies to stationary points.
2. Definition of Concavity
Concavity is the geometric property that describes the way a curve bends, and it is
directly controlled by the second derivative.
Concave Upward: A function
f
is concave upward on an interval
I
if its graph lies
above all its tangent lines on
I
.
Test: If
f″(x)>0
for all
x
in
I
, then
f
is concave upward on
I
. The rate of change of the
slope is increasing. The curve is bending up like a U-shape.
Concave Downward: A function
f
is concave downward on an interval
I
if its graph
lies below all its tangent lines on
I
.
Test: If
f″(x)<0
for all
x
in
I
, then
f
is concave downward on
I
. The rate of change of
the slope is decreasing. The curve is bending down like an inverted U-shape.
3. The Second Derivative Test (SDT)
Let
f
be a function such that
f′(c)=0
(i.e.,
c
is a stationary point), and assume that
f″(x)
is continuous in an open interval containing
c
.
The Three Cases of the Second Derivative Test:
Case 1: Classification as a Local Minimum
If
f″(c)>0
, then
f
has a Local Minimum at
c
.
Geometric Interpretation: Positive second derivative implies concave up. A
stationary point on a concave-up curve must be the minimum value in that local
region.
Case 2: Classification as a Local Maximum
If
f″(c)<0
, then
f
has a Local Maximum at
c
.
Geometric Interpretation: Negative second derivative implies concave down. A
stationary point on a concave-down curve must be the maximum value in that local
region.
Case 3: Inconclusive Test (The Failure Condition)
If
f″(c)=0
or
f″(c)
DNE, the test is Inconclusive.
Action Required: When this failure condition occurs, the student must revert
immediately to the more fundamental and powerful First Derivative Test to classify
the critical point. The point could be a maximum, a minimum, or neither (an
inflection point, such as with
f(x)=x3
at
x=0
, where
f′(0)=0
and
f″(0)=0
).
Part III: The Rigorous Proof and Derivation of the SDT
The power of the Second Derivative Test lies in its direct link between the local
concavity and the required sign change of the first derivative. We can rigorously
derive both classification cases using the definition of the second derivative and the
Monotonicity Test.
1. Proof of Case 1:
f″(c)>0⇒
Local Minimum
We begin with the premise that
c
is a stationary point, meaning
f′(c)=0
.
Step 1: The Definition of
f″(c)
The second derivative
f″(c)
is the derivative of the first derivative,
f′(x)
, evaluated
at
c
. By definition, the derivative is a limit:
f″(c)=
lim
x →c
f′(x)− f ′(c)
x − c
Since we are given that
f′(c)=0
, this simplifies to:
f″(c)=
lim
x →c
f′(x)
x −c
Step 2: Using the Premise
f″(c)>0
If
f″(c)>0
, it means the limit we calculated in Step 1 is a positive number. By the
definition of a limit, if the limit of a function is positive, the function itself must be
positive for all
x
in a small open interval
(a , b)
around
c
(excluding
c
itself).
f′(x)
x −c >0 for x∈(a , b), x ≠ c
Step 3: Analyzing the Interval to the Right of
c
(
x>c
)
Consider
x
values just to the right of
c
. In this case,
x>c
, which implies that the
denominator,
x − c
, is positive (
x − c >0
).
For the fraction
f′(x)
x −c
to be positive (as established in Step 2), the numerator,
f′(x)
,
must also be positive.
If x>c(so x − c>0¿,then f′(x)must be>0 for the ratio to be>0.
Conclusion:
f′(x)>0
immediately to the right of
c
. The function
f
is increasing here.
Step 4: Analyzing the Interval to the Left of
c
(
x<c
)
Now consider
x
values just to the left of
c
. In this case,
x<c
, which implies that the
denominator,
x − c
, is negative (
x− c<0
).
For the fraction
f′(x)
x −c
to be positive (as established in Step 2), the numerator,
f′(x)
,
must be negative (since a negative divided by a negative yields a positive).
If x<c(so x − c<0¿,then f′(x)must be<0 for the ratio to be>0.
Conclusion:
f′(x)<0
immediately to the left of
c
. The function
f
is decreasing here.
Step 5: Final Classification
We have demonstrated that the first derivative changes sign from negative to
positive at the stationary point
c
. By the First Derivative Test (Monotonicity Test), a
function that switches from decreasing to increasing forms a valley.
f′(x): Negative (Decreasing)→0→Positive (Increasing)
Therefore,
f(c)
is a Local Minimum. This rigorously proves the theorems first
sufficient condition.
2. Proof of Case 2:
f″(c)<0⇒
Local Maximum
Again, we start with the premise that
c
is a stationary point,
f′(c)=0
.
Step 1: The Simplified Limit
As before, the second derivative at
c
is the limit:
f″(c)=
lim
x →c
f′(x)
x −c
Step 2: Using the Premise
f″(c)<0
If
f″(c)<0
, the limit is a negative number. Thus, the fraction
f′(x)
x −c
must be negative
for all
x
in a small open interval
(a , b)
around
c
.
f′(x)
x −c <0 for x∈(a , b), x ≠ c
Step 3: Analyzing the Interval to the Right of
c
(
x>c
)
Consider
x
values just to the right of
c
, where the denominator
x − c
is positive (
x − c >0
).
For the fraction
f′(x)
x −c
to be negative (as established in Step 2), the numerator,
f′(x)
,
must be negative (since a negative divided by a positive yields a negative).
If x>c(so x − c>0¿,then f′(x)must be<0 for the ratio to be<0.
Conclusion:
f′(x)<0
immediately to the right of
c
. The function
f
is decreasing here.
Step 4: Analyzing the Interval to the Left of
c
(
x<c
)
Consider
x
values just to the left of
c
, where the denominator
x − c
is negative (
x − c <0
).
For the fraction
f′(x)
x −c
to be negative (as established in Step 2), the numerator,
f′(x)
,
must be positive (since a positive divided by a negative yields a negative).
If x<c(so x − c<0¿,then f′(x)must be>0 for the ratio to be<0.
Conclusion:
f′(x)>0
immediately to the left of
c
. The function
f
is increasing here.
Step 5: Final Classification
We have demonstrated that the first derivative changes sign from positive to
negative at the stationary point
c
. By the First Derivative Test, a function that
switches from increasing to decreasing forms a peak.
f′(x): Positive (Increasing)→0→Negative (Decreasing)
Therefore,
f(c)
is a Local Maximum. This completes the rigorous proof for the
second sufficient condition.
Part IV: The Practical Algorithm for Classification (No Tables)
The following sequence of steps provides a structured, prose-based method for
applying the Second Derivative Test to find and classify local extrema.
Step-by-Step Procedure for the Second Derivative Test
Define Domain and Continuity: The first step is always to establish the functions
domain. The SDT requires the function to be twice differentiable around the
stationary points, meaning
f′(x)
and
f″(x)
must exist in that neighborhood.
Polynomials and combinations of transcendental functions (like
ex
,
sin x
,
cos x
) are
usually safe.
Calculate the First Derivative (
f′(x)
): Compute the first derivative of the function,
f′(x)
. This derivative is necessary to find the locations where the tangent line is
horizontal.
Identify Stationary Points: Set the first derivative equal to zero,
f′(x)=0
, and solve
the resulting equation for all real values of
x
. These values are the stationary points
—the candidates for local extrema that the SDT can analyze. We explicitly exclude
critical numbers where
f′(x)
DNE, as the SDT cannot be used on those.
Calculate the Second Derivative (
f″(x)
): Compute the second derivative by
differentiating
f′(x)
. This expression will quantify the concavity and the rate of
change of the slope across the functions domain.
Evaluate Concavity at Stationary Points (The Sufficient Condition Check): This is the
core of the test. For each stationary point
c
found in Step 3, substitute
c
into the
second derivative expression,
f″(x)
, to find
f″(c)
.
If
f″(c)
is calculated and the result is a positive number (
f″(c)>0
): We immediately
conclude that the function is concave up at
c
. The Second Derivative Test confirms
that
f
has a Local Minimum at
c
.
If
f″(c)
is calculated and the result is a negative number (
f″(c)<0
): We immediately
conclude that the function is concave down at
c
. The Second Derivative Test
confirms that
f
has a Local Maximum at
c
.
If
f″(c)
is calculated and the result is exactly zero (
f″(c)=0
): The test provides no
information regarding the classification. We must state clearly that the test is
inconclusive and proceed to the comprehensive sign analysis of the First Derivative
Test for this specific point.
Determine the Extreme Values: For every stationary point confirmed to be an
extremum (i.e., where
f″(c)≠0
), substitute the
x
-coordinate back into the original
function,
f(x)
, to find the corresponding
y
-coordinate, which is the actual extreme
value. The classification is then summarized as a pair
(c , f (c))
.
Part V: Comprehensive Example 1: Polynomial Function
Let us apply the Second Derivative Test to the polynomial function
f(x)=x4−8x3+18 x2−1
. This function is a quartic, and we anticipate up to three
potential extrema.
Domain and Continuity:
f(x)
is a polynomial, defined and infinitely differentiable for
all real numbers,
(− ∞ , ∞ )
.
Calculate the First Derivative (
f′(x)
):
f′(x)= d
dx (x4−8x3+18 x2−1)=4x3−24 x2+36 x
Identify Stationary Points: Set
f′(x)=0
and solve. We factor the resulting cubic
equation:
4x3−24 x2+36 x=0
We factor out the greatest common factor,
4x
:
4x(x2−6x+9)=0
Recognizing the perfect square trinomial inside the parenthesis, we factor further:
4x¿
This equation yields two distinct stationary points:
x1=0
x2=3
(This root has multiplicity two, which often suggests the point may not be an
extremum, but we must use the SDT to confirm.)
Calculate the Second Derivative (
f″(x)
): We differentiate the expression for
f′(x)
from Step 2:
f″(x)= d
dx (4x3−24 x2+36 x)=12 x2−48 x+36
Evaluate Concavity at Stationary Points (The Sufficient Condition Check): We test
our two stationary points in the
f″(x)
expression.
Test Stationary Point
x1=0
:
We substitute
x=0
into
f″(x)
:
f″(0)=12¿
Since the result,
f″(0)=36
, is a positive number (
f″(0)>0
), the function is concave
upward at
x=0
.
Conclusion for
x=0
: The Second Derivative Test classifies this stationary point as a
Local Minimum.
Test Stationary Point
x2=3
:
We substitute
x=3
into
f″(x)
:
f″(3)=12¿
Since the result,
f″(3)=0
, the Second Derivative Test is Inconclusive for the
stationary point
x=3
. We must revert to the First Derivative Test for classification at
this point.
Reverting to FDT for
x=3
(Required due to SDT Inconclusive Case):
We use the factored form of the first derivative:
f′(x)=4x¿
. We test the intervals
immediately surrounding
x=3
.
Test
x=2.5
(Left of 3):
f′(2.5)=4(2.5)¿
. This is positive (
f
is increasing).
Test
x=3.5
(Right of 3):
f′(3.5)=4(3.5)¿
. This is positive (
f
is increasing).
Since
f′(x)
changes from positive to positive at
x=3
, there is no sign change in the
first derivative.
Conclusion for
x=3
: This is Not a Local Extremum; it is a horizontal point of
inflection.
Determine the Extreme Values: We only found one local extremum at
x=0
. We
calculate the extreme value using the original function
f(x)=x4−8x3+18 x2−1
.
Local Minimum Value (at
x=0
):
f(0)=¿
Final Result Summary: The Second Derivative Test efficiently classified
x=0
as a
local minimum. It failed at
x=3
, which then required the First Derivative Test to
correctly classify it as a point that is neither a local maximum nor a local minimum.
The single local extremum is the Local Minimum at
(0, −1)
.
Part VI: Comprehensive Example 2: Transcendental Function
Let us apply the Second Derivative Test to a transcendental function, which often
requires careful differentiation of products and chains, such as
g(x)=x⋅e− x /2
.
Domain and Continuity: The domain of
g(x)
is
(− ∞ , ∞)
, as both
x
and
e− x/2
are
defined for all real numbers. It is infinitely differentiable.
Calculate the First Derivative (
g′(x)
): We apply the Product Rule:
g′(x)=f′h+f h′
.
g′(x)=(1)e− x /2+x⋅
(
−1
2e− x/2
)
Factor out the common term
e− x/2
:
g′(x)=e− x/2
(
1−1
2x
)
Identify Stationary Points: Set
g′(x)=0
. Since
e− x/2
is always positive and never zero,
we only need to set the factor in the parenthesis to zero:
1−1
2x=0⇒1
2x=1⇒x=2
The only stationary point is
c=2
.
Calculate the Second Derivative (
g″(x)
): We must differentiate
g′(x)=e− x/2
(
1−1
2x
)
using the Product Rule again.
g″(x)= d
dx
[
e− x /2
]
⋅
(
1−1
2x
)
+e− x/2⋅d
dx
[
1−1
2x
]
g″(x)=
(
−1
2e− x/2
)(
1−1
2x
)
+e− x /2
(
−1
2
)
Factor out the common term
−1
2e− x/2
:
g″(x)=−1
2e− x/2
[
(
1−1
2x
)
+1
]
Simplify the terms inside the brackets:
g″(x)=−1
2e− x/2
(
2−1
2x
)
Evaluate Concavity at Stationary Point (The Sufficient Condition Check): We test our
only stationary point
x=2
in the
g″(x)
expression.
g″(2)=−1
2e−2/2
(
2−1
2(2)
)
g″(2)=−1
2e−1(2−1)g″(2)=−1
2e−1(1)=−1
2e
Since
e ≈2.718
is positive,
2e
is positive. Therefore, the result
g″(2)=−1
2e
is a
negative number (
g″(2)<0
).
Conclusion for
x=2
: The function is concave downward at
x=2
. The Second
Derivative Test classifies this stationary point as a Local Maximum.
Determine the Extreme Values: We calculate the extreme value using the original
function
g(x)=x⋅e− x /2
.
Local Maximum Value (at
x=2
):
g(2)=2⋅e−2/2=2e−1=2
e
Final Result Summary: The Second Derivative Test successfully classified the
stationary point
x=2
. The function
g(x)
has a single local extremum, a Local
Maximum at
(
2,2
e
)
.
Part VII: The Crucial Failure Case and Necessity of FDT Backup
The Second Derivative Test is an excellent tool due to its speed and simplicity, but
its limitations are critical to understand in MATH 150A. It only works when
f″(c)≠0
.
The Inconclusive Condition:
f″(c)=0
When
f″(c)=0
, the Second Derivative Test is inconclusive because the sign of the
concavity is changing or momentarily flat at the stationary point
c
. This zero result
means that the concavity switches at
c
, which defines an inflection point. However, a
stationary point that is also an inflection point can be one of three things:
A Point of Inflection (Not an Extremum): The most common case, such as
f(x)=x3
at
x=0
.
f′(x)=3x2
, so
f′(0)=0
. (Stationary)
f″(x)=6x
, so
f″(0)=0
. (SDT Inconclusive)
Using FDT:
f′(x)
is positive on both sides of
x=0
. Conclusion: No extremum.
A Local Minimum: For example,
f(x)=x4
at
x=0
.
f′(x)=4x3
, so
f′(0)=0
. (Stationary)
f″(x)=12 x2
, so
f″(0)=0
. (SDT Inconclusive)
Using FDT:
f′(x)
changes from negative (for
x<0
) to positive (for
x>0
). Conclusion:
Local Minimum.
A Local Maximum: For example,
f(x)=− x4
at
x=0
.
f′(x)=−4x3
, so
f′(0)=0
. (Stationary)
f″(x)=−12 x2
, so
f″(0)=0
. (SDT Inconclusive)
Using FDT:
f′(x)
changes from positive (for
x<0
) to negative (for
x>0
). Conclusion:
Local Maximum.
The Inconclusive Condition:
f′(c)
DNE (Not Twice Differentiable)
The Second Derivative Test also fails entirely when the function is not twice
differentiable at the critical number
c
, such as at a cusp or corner (where
f′(c)
DNE).
For instance, with
h(x)=¿x∨¿
at
x=0
,
h′(0)
DNE, so we cannot even calculate
h″(0)
.
The Overarching Rule: The First Derivative Test is the master test. It relies on the
most fundamental property—the change in direction—and is robust enough to
classify every single critical number, regardless of whether
f′(c)=0
or
f′(c)
DNE,
and regardless of the value of
f″(c)
. The Second Derivative Test is merely a highly
efficient shortcut for the specific scenario where
f′(c)=0
and concavity is non-zero.
When the shortcut fails, we must always return to the foundational analysis of the
First Derivative Test. This understanding is non-negotiable for success in MATH
150A.
Part VIII: Summary, Comparison, and Context for MATH 150A
In conclusion, the Second Derivative Test is a powerful, elegant, and sufficient
condition for classifying stationary points by utilizing the concept of concavity. Its
primary advantage is efficiency; its primary disadvantage is its limited scope.
Comparison Table in Prose
Feature
First Derivative Test (FDT)
Second Derivative Test (SDT)
Applicability
Universal. Applies to all critical numbers:
f′(c)=0
or
f′(c)
DNE.
Limited. Only applies to stationary points where
f′(c)=0
and
f″(c)≠0
.
Criterion
Sign change of
f′(x)
across the critical number.
Sign of
f″(x)
evaluated at the stationary point.
Concept
Monotonicity (Increasing
→
Decreasing or vice-versa).
Concavity (Concave Up
→
Minimum, Concave Down
→
Maximum).
Advantage
Always works; the gold standard. Reveals horizontal inflection points like
x3
.
Fastest classification when it works. Avoids interval testing.
Disadvantage
Requires testing multiple intervals (time-consuming).
Fails when
f″(c)=0
or
f′(c)
DNE (forces switch to FDT).
Final Study Recommendations
As a student navigating MATH 150A, your strategy should be:
Always calculate
f′(x)
and
f″(x)
first. Having both available saves time regardless of
the test you choose.
Always use the SDT first for stationary points (
f′(c)=0
). This is the most efficient
way to classify them, provided the differentiation is not overly complex.
Be prepared to switch. If
f″(c)=0
or if the critical number arose because
f′(c)
DNE,
immediately and confidently transition to the rigorous sign-analysis procedure of
the First Derivative Test. This flexibility proves a deep understanding of the
underlying principles of local extrema. The mastery of calculus lies in knowing not
just how to apply a test, but also when to rely on a different, more fundamental tool.
Practice the Inconclusive Cases. Spending time on functions like
x4
and
x3
at the
origin is crucial to internalize why
f″(c)=0
is a genuine failure and not just a
complication. The distinction between a local minimum, a local maximum, and a
horizontal inflection point where
f″(c)=0
is a common point of confusion that only
the FDT can resolve.
