Lesson 20: Derivatives and the Shapes of Curves

Section 4.3
Derivatives and the Shapes of
Curves

V63.0121, Calculus I

March 25-26, 2009

.

.
Image credit: cobalt123
. . . . . .

Outline

Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test

Concavity
Deﬁnitions
Testing for Concavity
The Second Derivative Test

. . . . . .


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:
.′
− f
+
. .
0
..
0
.

. . . . . .


Example

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

. . . . . .


Example

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

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

. . . . . .


Example

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

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

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

. . . . . .

Example
Solution

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


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

. . . . . .

Example
Solution

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


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

. . . . . .

The First Derivative Test

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

. . . . . .

Example
Solution

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


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

Deﬁnition
The graph of f is called concave up on and 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”.

. . . . . .

Definition
A point P on a curve y = f(x) is called an inflection point if f is
continuous there 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 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 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)
= f′ (b). So
By MVT, there exists a b between a and x with
x−a

f(x) = f(a) + f′ (b)(x − a) ≥ f(a) + f′ (a)(x − a) = L(x)

. . . . . .

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)

. . . . . .

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 (5x − 2)
=x
x
9 9 9

. . . . . .

Example

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

. . . . . .

The Second Derivative Test

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 f(c) is a local maximum.
If f′′ (c) > 0, then f(c) is a local minimum.

If f′′ (c) = 0, the second derivative test is inconclusive (this does not
mean c is neither; we just don’t know yet).

. . . . . .

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.

. . . . . .

Example
Find the local extrema of f(x) = x2/3 (x + 2)

. . . . . .

Example

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

. . . . . .

Example

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

. . . . . .

Example

Solution
1 −1/3
x
3
10 −4/3
x
9
x=− 4/5

So x = −4/5 is a local maximum.

. . . . . .

Example

Solution
1 −1/3
x
3
10 −4/3
x
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.

. . . . . .

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:

g(x) = −x4
f(x) = x4 h(x) = x3

. . . . . .


Consider these examples:

g(x) = −x4
f(x) = x4 h(x) = x3

All of them have f′′ (0) = 0. 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

. . . . . .

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
Next week: Graphing functions

. . . . . .

Lesson 20: Derivatives and the Shapes of Curves

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