Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)

..
Section 4.2
Derivatives and the Shapes of Curves
V63.0121.041, Calculus I
New York University
November 15, 2010
Announcements
Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7
There is class on November 24
. . . . . .

. . . . . .
Announcements
Quiz 4 this week in
recitation on 3.3, 3.4, 3.5,
3.7
There is class on
November 24
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 2 / 32

. . . . . .
Objectives
Use the derivative of a
function to determine the
intervals along which the
function is increasing or
decreasing (The
Increasing/Decreasing
Test)
Use the First Derivative
Test to classify critical
points of a function as local
maxima, local minima, or
neither.

. . . . . .
Objectives
Use the second derivative
of a function to determine
the intervals along which
the graph of the function is
concave up or concave
down (The Concavity Test)
Use the first and second
derivative of a function to
classify critical points as
local maxima or local
minima, when applicable
(The Second Derivative
Test)

. . . . . .
Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test

. . . . . .
Theorem (The Mean Value Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that
f(b) − f(a)
b − a
= f′
(c). ...
a
..
b

. . . . . .
(a, b) such that
f(b) − f(a)
b − a
= f′
(c). ...
a
..
b
..
c

. . . . . .
(a, b) such that
f(b) − f(a)
b − a
= f′
(c). ...
a
..
b
..
c
Another way to put this is that there exists a point c such that
f(b) = f(a) + f′
(c)(b − a)

. . . . . .
Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′
= 0 on an interval (a, b). Then f is constant on (a, b).
Proof.
Pick any points x and y in (a, b) with x < y. Then f is continuous on
[x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y)
such that
f(y) = f(x) + f′
(z)(y − x)
So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.

. . . . . .
Outline
Monotonicity
Concavity
Definitions

. . . . . .
What does it mean for a function to be increasing?
Definition
A function f is increasing on the interval I if
f(x) < f(y)
whenever x and y are two points in I with x < y.

. . . . . .
What does it mean for a function to be increasing?
Definition
A function f is increasing on the interval I if
f(x) < f(y)
whenever x and y are two points in I with x < y.
An increasing function “preserves order.”
I could be bounded or infinite, open, closed, or
half-open/half-closed.
Write your own definition (mutatis mutandis) of decreasing,
nonincreasing, nondecreasing

. . . . . .
Theorem (The Increasing/Decreasing Test)
If f′
> 0 on an interval, then f is increasing on that interval. If f′
< 0 on
an interval, then f is decreasing on that interval.

. . . . . .
Theorem (The Increasing/Decreasing Test)
If f′
> 0 on an interval, then f is increasing on that interval. If f′
< 0 on
an interval, then f is decreasing on that interval.
Proof.
It works the same as the last theorem. Assume f′
(x) > 0 on an interval
I. Pick two points x and y in I with x < y. We must show f(x) < f(y). By
MVT there exists a point c in (x, y) such that
f(y) − f(x) = f′
(c)(y − x) > 0.
So f(y) > f(x).

. . . . . .
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.

. . . . . .
Example
Solution
f′
(x) = 2 is always positive, so f is increasing on (−∞, ∞).

. . . . . .
Example
Solution
f′
Example
Describe the monotonicity of f(x) = arctan(x).

. . . . . .
Example
Solution
f′
Example
Describe the monotonicity of f(x) = arctan(x).
Solution
Since f′
(x) =
1
1 + x2
is always positive, f(x) is always increasing.

. . . . . .
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2
− 1.

. . . . . .
Example
− 1.
Solution
f′
(x) = 2x, which is positive when x > 0 and negative when x is.

. . . . . .
Example
− 1.
Solution
f′
We can draw a number line:
.. f′
.− ..
0
.0. +

. . . . . .
Example
− 1.
Solution
f′
.. f′
.
f
.− .
↘
..
0
.0. +.
↗

. . . . . .
Example
− 1.
Solution
f′
.. f′
.
f
.− .
↘
..
0
.0. +.
↗
So f is decreasing on (−∞, 0) and increasing on (0, ∞).

. . . . . .
Example
− 1.
Solution
f′
.. f′
.
f
.− .
↘
..
0
.0. +.
↗
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)

. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3
(x + 2).

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
The critical points are 0 and and −4/5.
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−
..
+

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
..
+

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗

. . . . . .
Theorem (The First Derivative Test)
Let f be continuous on [a, b] and c a critical point of f in (a, b).
If f′
changes from positive to negative at c, then c is a local
maximum.
If f′
changes from negative to positive at c, then c is a local
minimum.
If f′
(x) has the same sign on either side of c, then c is not a local
extremum.

. . . . . .
Example
− 1.
Solution
f′
.. f′
.
f
.− .
↘
..
0
.0. +.
↗
[0, ∞)

. . . . . .
Example
− 1.
Solution
f′
.. f′
.
f
.− .
↘
..
0
.0. +.
↗
.
min
[0, ∞)

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max

. . . . . .
Example
(x + 2).
Solution
f′
(x) = 2
3 x−1/3
(x + 2) + x2/3
= 1
3 x−1/3
(5x + 4)
.. x−1/3..
0
.×.− . +.
5x + 4
..
−4/5
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min

. . . . . .
Outline
Monotonicity
Concavity
Definitions

. . . . . .
Concavity
Definition
The graph of f is called concave upwards on an interval if it lies above
all its tangents on that interval. The graph of f is called concave
downwards on an interval if it lies below all its tangents on that
interval.
.
concave up
.
concave down
We sometimes say a concave up graph “holds water” and a concave
down graph “spills water”.

. . . . . .
Synonyms for concavity
Remark
“concave up” = “concave upwards” = “convex”
“concave down” = “concave downwards” = “concave”

. . . . . .
Inflection points indicate a change in concavity
Definition
A point P on a curve y = f(x) is called an inflection point if f is
continuous at P and the curve changes from concave upward to
concave downward at P (or vice versa).
..
concave
down
.
concave up
..
inflection point

. . . . . .
Theorem (Concavity Test)
If f′′
(x) > 0 for all x in an interval, then the graph of f is concave
upward on that interval.
If f′′
(x) < 0 for all x in an interval, then the graph of f is concave
downward on that interval.

. . . . . .
Theorem (Concavity Test)
If f′′
(x) > 0 for all x in an interval, then the graph of f is concave
upward on that interval.
If f′′
(x) < 0 for all x in an interval, then the graph of f is concave
downward on that interval.
Proof.
Suppose f′′
(x) > 0 on the interval I (which could be infinite). This
means f′
is increasing on I. Let a and x be in I. The tangent line
through (a, f(a)) is the graph of
L(x) = f(a) + f′
(a)(x − a)
By MVT, there exists a c between a and x with
f(x) = f(a) + f′
(c)(x − a)
Since f′
is increasing, f(x) > L(x).

. . . . . .
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3
+ x2
.

. . . . . .
Example
+ x2
.
Solution
We have f′
(x) = 3x2
+ 2x, so f′′
(x) = 6x + 2.

. . . . . .
Example
+ x2
.
Solution
We have f′
(x) = 3x2
+ 2x, so f′′
(x) = 6x + 2.
This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3

. . . . . .
Example
+ x2
.
Solution
We have f′
(x) = 3x2
+ 2x, so f′′
(x) = 6x + 2.
This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3
So f is concave down on the open interval (−∞, −1/3), concave up
on the open interval (−1/3, ∞), and has an inflection point at the
point (−1/3, 2/27)

. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3
(x + 2).

. . . . . .
Example
(x + 2).
Solution
We have f′′
(x) =
10
9
x−1/3
−
4
9
x−4/3
=
2
9
x−4/3
(5x − 2).
.. x−4/3..
0
.×.+ . +.
5x − 2
..
2/5
.
0
.
−
.
+

. . . . . .
Example
(x + 2).
Solution
We have f′′
(x) =
10
9
x−1/3
−
4
9
x−4/3
=
2
9
x−4/3
(5x − 2).
.. x−4/3..
0
.×.+ . +.
5x − 2
..
2/5
.
0
.
−
.
+
.
f′′
(x)
.
f(x)
..
2/5
.
0
..
0
.
×

. . . . . .
Example
(x + 2).
Solution
We have f′′
(x) =
10
9
x−1/3
−
4
9
x−4/3
=
2
9
x−4/3
(5x − 2).
.. x−4/3..
0
.×.+ . +.
5x − 2
..
2/5
.
0
.
−
.
+
.
f′′
(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−

. . . . . .
Example
(x + 2).
Solution
We have f′′
(x) =
10
9
x−1/3
−
4
9
x−4/3
=
2
9
x−4/3
(5x − 2).
.. x−4/3..
0
.×.+ . +.
5x − 2
..
2/5
.
0
.
−
.
+
.
f′′
(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−

. . . . . .
Example
(x + 2).
Solution
We have f′′
(x) =
10
9
x−1/3
−
4
9
x−4/3
=
2
9
x−4/3
(5x − 2).
.. x−4/3..
0
.×.+ . +.
5x − 2
..
2/5
.
0
.
−
.
+
.
f′′
(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++

. . . . . .
Example
(x + 2).
Solution
We have f′′
(x) =
10
9
x−1/3
−
4
9
x−4/3
=
2
9
x−4/3
(5x − 2).
.. x−4/3..
0
.×.+ . +.
5x − 2
..
2/5
.
0
.
−
.
+
.
f′′
(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢

. . . . . .
Example
(x + 2).
Solution
We have f′′
(x) =
10
9
x−1/3
−
4
9
x−4/3
=
2
9
x−4/3
(5x − 2).
.. x−4/3..
0
.×.+ . +.
5x − 2
..
2/5
.
0
.
−
.
+
.
f′′
(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢

. . . . . .
Example
(x + 2).
Solution
We have f′′
(x) =
10
9
x−1/3
−
4
9
x−4/3
=
2
9
x−4/3
(5x − 2).
.. x−4/3..
0
.×.+ . +.
5x − 2
..
2/5
.
0
.
−
.
+
.
f′′
(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣

. . . . . .
Example
(x + 2).
Solution
We have f′′
(x) =
10
9
x−1/3
−
4
9
x−4/3
=
2
9
x−4/3
(5x − 2).
.. x−4/3..
0
.×.+ . +.
5x − 2
..
2/5
.
0
.
−
.
+
.
f′′
(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP

. . . . . .
Theorem (The Second Derivative Test)
Let f, f′
, and f′′
be continuous on [a, b]. Let c be be a point in (a, b) with
f′
(c) = 0.
If f′′
(c) < 0, then c is a local maximum.
If f′′
(c) > 0, then c is a local minimum.

. . . . . .
Theorem (The Second Derivative Test)
Let f, f′
, and f′′
be continuous on [a, b]. Let c be be a point in (a, b) with
f′
(c) = 0.
If f′′
(c) < 0, then c is a local maximum.
If f′′
(c) > 0, then c is a local minimum.
Remarks
If f′′
(c) = 0, the second derivative test is inconclusive (this does
not mean c is neither; we just don’t know yet).
We look for zeroes of f′
and plug them into f′′
to determine if their f
values are local extreme values.

. . . . . .
Proof of the Second Derivative Test
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
..
f′′
= (f′
)′
.
f′
...
c
.+.
f′
.
f
...
c
.
0

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
sufficiently close to c.
..
f′′
= (f′
)′
.
f′
...
c
.+.
f′
.
f
...
c
.
0

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .
f′
.
f
...
c
.
0

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
f′
.
f
...
c
.
0

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
is increasing near c.
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
f′
.
f
...
c
.
0

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
↗
.
f′
.
f
...
c
.
0

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
↗
.
↗
.
f′
.
f
...
c
.
0

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
↗
.
↗
.
f′
.
f
...
c
.
0
Since f′
(c) = 0 and f′
is increasing, f′
(x) < 0 for x close to c and
less than c, and f′
(x) > 0 for x close to c and more than c.

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
↗
.
↗
.
f′
.
f
...
c
.
0
..
−
Since f′
(c) = 0 and f′
is increasing, f′

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
↗
.
↗
.
f′
.
f
...
c
.
0
..
−
..
+
Since f′
(c) = 0 and f′
is increasing, f′

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
↗
.
↗
.
f′
.
f
...
c
.
0
..
−
..
+
Since f′
(c) = 0 and f′
is increasing, f′
This means f′
changes sign from negative to positive at c, which
means (by the First Derivative Test) that f has a local minimum
at c.

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
↗
.
↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
Since f′
(c) = 0 and f′
is increasing, f′
This means f′
at c.

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
↗
.
↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
Since f′
(c) = 0 and f′
is increasing, f′
This means f′
at c.

. . . . . .
Proof.
Suppose f′
(c) = 0 and f′′
(c) > 0.
Since f′′
is continuous,
f′′
(x) > 0 for all x
Since f′′
= (f′
)′
, we know f′
..
f′′
= (f′
)′
.
f′
...
c
.+..+ .. +.
↗
.
↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
Since f′
(c) = 0 and f′
is increasing, f′
This means f′
at c.

. . . . . .
Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3
+ x2
.

. . . . . .
Example
+ x2
.
Solution
f′
(x) = 3x2
+ 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.

. . . . . .
Example
+ x2
.
Solution
f′
(x) = 3x2
+ 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′
(x) = 6x + 2

. . . . . .
Example
+ x2
.
Solution
f′
(x) = 3x2
+ 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′
(x) = 6x + 2
Since f′′
(−2/3) = −2 < 0, −2/3 is a local maximum.

. . . . . .
Example
+ x2
.
Solution
f′
(x) = 3x2
+ 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′
(x) = 6x + 2
Since f′′
(−2/3) = −2 < 0, −2/3 is a local maximum.
Since f′′
(0) = 2 > 0, 0 is a local minimum.

. . . . . .
Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3
(x + 2)

. . . . . .
Example
(x + 2)
Solution
Remember f′
(x) =
1
3
x−1/3
(5x + 4) which is zero when x = −4/5

. . . . . .
Example
(x + 2)
Solution
Remember f′
(x) =
1
3
x−1/3
Remember f′′
(x) =
10
9
x−4/3
(5x − 2), which is negative when
x = −4/5

. . . . . .
Example
(x + 2)
Solution
Remember f′
(x) =
1
3
x−1/3
Remember f′′
(x) =
10
9
x−4/3
x = −4/5
So x = −4/5 is a local maximum.

. . . . . .
Example
(x + 2)
Solution
Remember f′
(x) =
1
3
x−1/3
Remember f′′
(x) =
10
9
x−4/3
x = −4/5
So x = −4/5 is a local maximum.
Notice the Second Derivative Test doesn’t catch the local
minimum x = 0 since f is not differentiable there.

. . . . . .
Using the Second Derivative Test II: Graph
Graph of f(x) = x2/3
(x + 2):
.. x.
y
..
(−4/5, 1.03413)
..
(0, 0)
..
(2/5, 1.30292)
..
(−2, 0)

. . . . . .
When the second derivative is zero
Remark
At inflection points c, if f′
is differentiable at c, then f′′
(c) = 0
If f′′
(c) = 0, must f have an inflection point at c?

. . . . . .
Remark
(c) = 0
If f′′
Consider these examples:
f(x) = x4
g(x) = −x4
h(x) = x3

. . . . . .
When first and second derivative are zero
function derivatives graph type
f(x) = x4
f′
(x) = 4x3, f′
(0) = 0
.
min
f′′
(x) = 12x2, f′′
(0) = 0
g(x) = −x4
g′(x) = −4x3, g′(0) = 0
.
max
g′′(x) = −12x2, g′′(0) = 0
h(x) = x3
h′
(x) = 3x2, h′
(0) = 0
.
infl.
h′′
(x) = 6x, h′′
(0) = 0

. . . . . .
Remark
(c) = 0
If f′′
Consider these examples:
f(x) = x4
g(x) = −x4
h(x) = x3
All of them have critical points at zero with a second derivative of zero.
But the first has a local min at 0, the second has a local max at 0, and
the third has an inflection point at 0. This is why we say 2DT has
nothing to say when f′′
(c) = 0.

. . . . . .
Summary
Concepts: Mean Value Theorem, monotonicity, concavity
Facts: derivatives can detect monotonicity and concavity
Techniques for drawing curves: the Increasing/Decreasing Test
and the Concavity Test
Techniques for finding extrema: the First Derivative Test and the
Second Derivative Test

Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)

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