Lesson20 -derivatives_and_the_shape_of_curves_021_slides

..
Section 4.2
Derivatives and the Shapes of Curves
V63.0121.021, Calculus I
New York University
November 16, 2010
Announcements
Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7
There is class on November 23. The homework is due on
November 24. Turn in homework to my mailbox or bring to class
on November 23.
. . . . . .

. . . . . .
Announcements
Quiz 4 this week in
recitation on 3.3, 3.4, 3.5,
3.7
There is class on
November 23. The
homework is due on
November 24. Turn in
homework to my mailbox
or bring to class on
November 23.
V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 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

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