CSUN Math 📚150A Study Notes: The Second Derivative and
Concavity
I. Student Insights and Learning Takeaways
🧐 The "Why" Behind the Second Derivative
When I first learned about the first derivative (
f′(x)
), it was clear—it tells us where
the function
f(x)
is increasing or decreasing. It describes the direction of the
movement.
However, the second derivative (
f″(x)
) is where things get really interesting. It tells
us how the direction is changing—it describes the shape or curvature of the graph.
Its the difference between driving uphill at a constant speed (increasing, constant
f′
)
and accelerating uphill (increasing, positive
f″
).
Key Insight: The second derivative is essentially the rate of change of the
slope.
oIf the slopes are getting larger (more positive or less negative), the
function is bending up.
oIf the slopes are getting smaller (less positive or more negative), the
function is bending down.
Analogy: If
f(t)
is position,
f′(t)
is velocity, and
f″(t)
is acceleration.
Positive acceleration means your velocity is increasing; positive
f″(x)
means
your slope (
f′(x)
) is increasing. This direct link makes the concept more
intuitive.
Connecting 💡
f′(x)
and
f″(x)
Understanding the relationship between the first and second derivatives is crucial
for a complete curve sketch:
f′(x)
Sign
f″(x)
Sign Function Behavior Mnemonic
+
(Increasing)
+ (Concave Up) Increasing at an increasing
rate.
Speeding Up
+
(Increasing)
- (Concave
Down)
Increasing at a decreasing rate. Slowing
Down
-
(Decreasing)
+ (Concave Up) Decreasing at a decreasing
rate.
Slowing
Down
-
(Decreasing)
- (Concave
Down)
Decreasing at an increasing
rate.
Speeding Up
Strategic Problem Solving🛠️
A common pitfall is stopping after finding the points where
f″(x)=0
. Always
remember:
1. A potential inflection point must be a point where
f″(x)=0
or
f″(x)
is
undefined.
2. Crucially, the concavity must change sign across that point for it to be a true
Point of Inflection (P.I.). The function
f(x)=x4
is a classic counterexample:
f″(0)=0
, but
f″(x)=12 x2≥0
everywhere, so there is no change in concavity,
and thus, no P.I. at
x=0
.
II. Knowledge Point Summary
1. The Second Derivative Definition
The second derivative of a function
f
is the derivative of its first derivative,
f′
.
Notations: f″(x),d2y
d x2, y″, Dx
2y
Formal Definition:
f″(x)= d
dx
[
f′(x)
]
=d
dx
[
dy
dx
]
2. Concavity Defined
Concavity describes the way the graph of a function bends or curves.
Definition of Concave Up
A function
f
is concave up on an open interval
(a , b)
if its graph lies above all of its
tangent lines on that interval.
Calculus Test:
f
is concave up on
(a , b)
if and only if the second derivative is
positive for every
x
in
(a , b)
.
f″(x)>0
Geometric Implication: The slope of the tangent line,
f′(x)
, is increasing on
the interval. (It opens "like a cup").
Definition of Concave Down
A function
f
is concave down on an open interval
(a , b)
if its graph lies below all of
its tangent lines on that interval.
Calculus Test:
f
is concave down on
(a , b)
if and only if the second derivative
is negative for every
x
in
(a , b)
.
f″(x)<0
Geometric Implication: The slope of the tangent line,
f′(x)
, is decreasing
on the interval. (It opens "like a frown").
3. Points of Inflection
A Point of Inflection (P.I.) is a point on the graph of a function where the concavity
changes (from up to down or from down to up).
Necessary Condition: If
(c , f (c))
is a point of inflection, then either:
f″(c)=0
f″(c)is undefined
The
x
-values where this occurs are called Hypercritical Numbers (or
possible inflection points).
Sufficient Condition (The Inflection Test): A hypercritical number
c
is a
point of inflection if and only if
f″(x)
changes sign as
x
moves across
c
.
4. The Second Derivative Test for Local Extrema
The Second Derivative Test provides an alternative way to classify critical points
(where
f′(c)=0
) as local maxima or minima, often more efficiently than the First
Derivative Test.
Suppose
f′(c)=0
and
f″(x)
exists on an open interval containing
c
.
Condition Conclusion Concavity Visual
f″(c)>0
Local Minimum at
x=c
.
Concave Up (holds water).
f″(c)<0
Local Maximum at
x=c
.
Concave Down (spills water).
f″(c)=0
Test is Inconclusive. Use the First Derivative Test.
III. Step-by-Step Procedure for Concavity Analysis
To find the intervals of concavity and the points of inflection for a function
f(x)
:
Step 1: Find the First and Second Derivatives
Calculate
f′(x)
and
f″(x)
.
Step 2: Find Hypercritical Numbers
Set
f″(x)=0
and solve for
x
. Also, find any
x
-values where
f″(x)
is undefined. These
are your hypercritical numbers (possible Points of Inflection).
Step 3: Create a Sign Chart for
f″(x)
Use the hypercritical numbers to divide the domain of
f
into test intervals. Pick a
test value in each interval and plug it into
f″(x)
to determine its sign.
Interva
l
Test Value
c
Sign of
f″(c)
Conclusion about
f(x)
(− ∞ , a)
c1
+Concave Up (
∪
)
(a , b)
c2
-Concave Down (
∩
)
Step 4: State Intervals of Concavity
Based on the sign chart:
f(x)
is Concave Up where
f″(x)>0
.
f(x)
is Concave Down where
f″(x)<0
.
Step 5: Identify Points of Inflection (P.I.)
A hypercritical number
c
corresponds to a P.I. if
f″(x)
changes sign at
c
. Find the
corresponding
y
-coordinate by evaluating
f(c)
in the original function.
IV. Detailed Examples (Pages 11-15)
Example 1: Basic Polynomial Function
Let
f(x)=x4−4x3
. Find the intervals of concavity and the Points of Inflection.
Step 1: Find Derivatives
f′(x)=4x3−12 x2f″(x)=12 x2−24 x
Step 2: Find Hypercritical Numbers
Set
f″(x)=0
:
12 x2−24 x=012 x(x − 2)=0
The hypercritical numbers are
x=0
and
x=2
.
Step 3: Create a Sign Chart for
f″(x)
Test intervals:
(− ∞ , 0),(0,2),(2, ∞)
.
Interva
l
Test Value
c
f″(c)=12 c(c −2)
Sign of
f″(c)
Concavity
(− ∞ , 0)
c=−1
12(−1)(−3)=36
+Concave Up (
∪
)
(0,2)
c=1
12(1)(−1)=−12
-Concave Down (
∩
)
(2, ∞)
c=3
12(3)(1)=36
+Concave Up (
∪
)
Step 4: State Intervals of Concavity
Concave Up:
(− ∞ , 0)∪(2, ∞)
Concave Down:
(0,2)
Step 5: Identify Points of Inflection
The concavity changes at both
x=0
and
x=2
.
At
x=0
:
f(0)=04−4¿
. P.I. is
(0,0)
.
At
x=2
:
f(2)=24−4¿
. P.I. is
(2, −16)
.
Example 2: Application of the Second Derivative Test
Let
g(x)=x3−6x2+5
. Use the Second Derivative Test to classify its local extrema.
Step 1: Find Derivatives
g′(x)=3x2−12 x g″(x)=6x−12
Step 2: Find Critical Points
Set
g′(x)=0
:
3x2−12 x=03 x(x− 4)=0
Critical points are
x=0
and
x=4
.
Step 3: Apply the Second Derivative Test
Evaluate
g″(x)
at each critical point.
Test
x=0
:
g″(0)=6(0)−12=−12
Since
g′(0)=0
and
g″(0)<0
, there is a Local Maximum at
x=0
.
g(0)=03−6¿
. Local Maximum at
(0,5)
.
Test
x=4
:
g″(4)=6(4)−12=24 −12=12
Since
g′(4)=0
and
g″(4)>0
, there is a Local Minimum at
x=4
.
g(4)=43−6¿
. Local Minimum at
(4, −27)
.
Example 3: Function with Undefined Second Derivative
Let
h(x)=x1/3
. Find the concavity and any P.I.s.
Step 1: Find Derivatives
h′(x)= 1
3x−2/3h″(x)= 1
3⋅(−2
3)x−5/3=−2
9x−5/3=−2
9x5/3
Step 2: Find Hypercritical Numbers
Set
h″(x)=0
:
$ − 2
9x5/3=0$
. This has no solution (the numerator is never
zero).
Find where
h″(x)
is undefined:
h″(x)
is undefined when the denominator is
zero, so
9x5/3=0⇒x=0
.
The only hypercritical number is
x=0
.
Step 3: Create a Sign Chart for
h″(x)
Test intervals:
(− ∞, 0),(0, ∞)
.
Interva
l
Test Value
c
h″(c)=−2
9c5/3
Sign of
h″(c)
Concavity
(− ∞ , 0)
c=−1
−2
9(−1)=2
9
+Concave Up (
∪
)
(0, ∞)
c=1
−2
9(1)=−2
9
-Concave Down (
∩
)
Step 4 & 5: Conclusion
The function
h(x)
is continuous at
x=0
and the concavity changes.
Concave Up:
(− ∞ , 0)
Concave Down:
(0, ∞)
Point of Inflection:
h(0)=01/3=0
. The P.I. is
(0,0)
.
