Lesson 21: Derivatives and the Shapes of Curves

Section 4.2
Derivatives and the Shapes of Curves

V63.0121.034, Calculus I

November 11, 2009

Announcements
Final Exam Friday, December 18, 2:00–3:50pm
Wednesday, November 25 is a regular class day

.
.
Image credit: cobalt123
. . . . . .

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 .
b
.
f(b) − f(a) . .
= f′ (c). a
.
b−a

. . . . . .


Theorem (The Mean Value c
..
Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that .
b
.
f(b) − f(a) . .
= f′ (c). a
.
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) = 0.
y−x

So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.

. . . . . .

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

. . . . . .


Example

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

. . . . . .


Example

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

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

. . . . . .


Example

Solution
is.
−
. 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 = 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
0
.. ×
.. .′ (x)
f
−
. 4/5 0
. f
.(x)

. . . . . .

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

. . . . . .

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

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 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).
x−a
So

f(x) = f(a) + f′ (c)(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 − 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 f(c) is a local maximum.
If f′′ (c) > 0, then f(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 f(c) is a local maximum.
If f′′ (c) > 0, then f(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.

. . . . . .

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) = x (5x + 4) which is zero when
3
x = −4/5

. . . . . .

Example

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

. . . . . .

Example

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

. . . . . .

Example

Solution
1 −1/3
3
x = −4/5
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.

. . . . . .

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 ) = x 4 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 ) = x 4 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 ) = x 4 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.

. . . . . .

What have we learned today?

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

. . . . . .

What have we learned today?

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 21: Derivatives and the Shapes of Curves

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