Lesson 20: Derivatives and the shapes of curves

Section 4.2
Derivatives and the Shapes of Curves

V63.0121.002.2010Su, Calculus I

New York University

June 9, 2010

Announcements
No office hours tonight
Quiz 3 on Thursday (3.3, 3.4, 3.5, 3.7)
Assignment 4 due Tuesday

. . . . . .

Announcements

No office hours tonight
Quiz 3 on Thursday (3.3,
3.4, 3.5, 3.7)
Assignment 4 due Tuesday

. . . . . .

V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Shapes of Curves June 9, 2010 2 / 30

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.
Use the second derivative
of a function to determine
the intervals along which
the graph of the function is . . . . . .

concave up or concave 4.2 The Shapes of Curves
V63.0121.002.2010Su, Calculus I (NYU) Section June 9, 2010 3 / 30

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
.
= f′ (c).
b−a . .
a
.

. . . . . .



Theorem (The Mean Value Theorem)

c
..
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
.
= f′ (c).
b−a . .
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) = 0.
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 (a, b) if

f(x) < f(y)

whenever x and y are two points in (a, b) with x < y.

. . . . . .


What does it mean for a function to be increasing?

Definition
A function f is increasing on (a, b) if

f(x) < f(y)

whenever x and y are two points in (a, b) with x < y.

An increasing function “preserves order.”
Write your own definition (mutatis mutandis) of decreasing,
nonincreasing, nondecreasing

. . . . . .



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

. . . . . .



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

Proof.
It works the same as the last theorem. Pick two points x and y in (a, b)
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) > 0.
y−x
So
f(y) − f(x) = f′ (c)(y − x) > 0.

. . . . . .


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

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

. . . . . .



Example

Solution

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

Solution
1
Since f′ (x) = is always positive, f(x) is always increasing.
1 + x2

. . . . . .


Finding intervals of monotonicity II

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

. . . . . .



Example

Solution

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

. . . . . .



Example

Solution

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

. . . . . .



Example

Solution

−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.

. . . . . .



Example

Solution

−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.

So f is decreasing on (−∞, 0) and increasing on (0, ∞).

. . . . . .



Example

Solution

−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.

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

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1

The critical points are 0 and and −4/5.

−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5

. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
0
.. ×
.. .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0
.. ×
.. .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. −
. 4/5 0
. f
.(x)
. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 f
.(x)
. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
. . . . . .



Theorem (The First Derivative Test)
Let f be continuous on [a, b] and c a critical point of f in (a, b).
If f′ (x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a local
maximum.
If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local
minimum.
If f′ (x) has the same sign on (a, c) and (c, b), then c is not a local
extremum.

. . . . . .



Example

Solution

−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.

[0, ∞)
. . . . . .



Example

Solution

−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
m
. in
[0, ∞)
. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
m
. ax
. . . . . .


Example

Solution

f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 3 x−1/3 (5x + 4)
3
1


−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
5
.x+4
−
. 4/5
..
+ 0 − ×
.. . . . . ..
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
m
. ax . inm
. . . . . .


Outline


Monotonicity

Concavity
Definitions

. . . . . .


Concavity

Definition
The graph of f is called concave up on an interval I if it lies above all
its tangents on I. The graph of f is called concave down on I if it lies
below all its tangents on I.

. .

concave up concave down
We sometimes say a concave up graph “holds water” and a concave
down graph “spills water”.
. . . . . .


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 up
i
.nflection point
. .
.
concave
down

. . . . . .


Theorem (Concavity Test)

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

. . . . . .


Theorem (Concavity Test)

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

Proof.
Suppose f′′ (x) > 0 on I. 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)

f(x) − f(a)
By MVT, there exists a c between a and x with = f′ (c). So
x−a

f(x) = f(a) + f′ (c)(x − a) ≥ f(a) + f′ (a)(x − a) = L(x) .

. . . . . .


Finding Intervals of Concavity I

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

. . . . . .



Example

Solution

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

. . . . . .



Example

Solution

This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3

. . . . . .



Example

Solution

This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3
So f is concave down on (−∞, −1/3), concave up on (−1/3, ∞),
and has an inflection point at (−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

Solution
10 −1/3 4 −4/3 2 −4/3
f′′ (x) = x − x = x (5x − 2)
9 9 9

. . . . . .



Example

Solution
10 −1/3 4 −4/3 2 −4/3
f′′ (x) =x − x = x (5x − 2)
9 9 9
The second derivative f′′ (x) is not defined at 0

. . . . . .



Example

Solution
10 −1/3 4 −4/3 2 −4/3
f′′ (x) =x − x = x (5x − 2)
9 9 9
Otherwise, x−4/3 is always positive, so the concavity is
determined by the 5x − 2 factor

. . . . . .



Example

Solution
10 −1/3 4 −4/3 2 −4/3
f′′ (x) =x − x = x (5x − 2)
9 9 9
Otherwise, x−4/3 is always positive, so the concavity is
determined by the 5x − 2 factor
So f is concave down on (−∞, 0], concave down on [0, 2/5),
concave up on (2/5, ∞), and has an inflection point when x = 2/5

. . . . . .



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. Since f′′ is continuous, f′′ (x) > 0 for
all x sufficiently close to c. Since f′′ = (f′ )′ , we know f′ is increasing
near c. 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. 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.

. . . . . .


Using the Second Derivative Test I

Example
Find the local extrema of f(x) = x3 + x2 .

. . . . . .



Example

Solution

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

. . . . . .



Example

Solution

Remember f′′ (x) = 6x + 2

. . . . . .



Example

Solution

Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.

. . . . . .



Example

Solution

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

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

. . . . . .



Example

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

. . . . . .



Example

Solution
1 −1/3
3
10 −4/3
9
x = −4/5
So x = −4/5 is a local maximum.

. . . . . .



Example

Solution
1 −1/3
3
10 −4/3
9
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):
y
.

. −4/5, 1.03413)
( .
.
. 2/5, 1.30292)
(

. . x
.
. −2, 0)
( . 0, 0)
(

. . . . . .


When the second derivative is zero

At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?

. . . . . .



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) = 4x3 , f′ (0) = 0
f(x) = x4 min
f′′ (x) = 12x2 , f′′ (0) = 0 .
.
g′ (x) = −4x3 , g′ (0) = 0
g(x) = −x4 max
g′′ (x) = −12x2 , g′′ (0) = 0
h′ (x) = 3x2 , h′ (0) = 0
h(x) = x3 infl.
h′′ (x) = 6x, h′′ (0) = 0 .

. . . . . .



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 shapes of curves

Recommandé

Recommandé

Contenu connexe

Tendances

Tendances (18)

En vedette

En vedette (20)

Similaire à Lesson 20: Derivatives and the shapes of curves

Similaire à Lesson 20: Derivatives and the shapes of curves (20)

Plus de Matthew Leingang

Plus de Matthew Leingang (20)

Dernier

Dernier (20)

Lesson 20: Derivatives and the shapes of curves