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IU CalculusI Chap3

Chapter 3 discusses applications of differentiation in calculus, focusing on optimization problems such as minimizing manufacturing costs and maximizing acceleration. It defines maximum and minimum values of functions, introduces the Extreme Value Theorem, and presents Fermat's Theorem regarding local extrema. The chapter emphasizes the importance of differential calculus in solving real-world optimization challenges.

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0% found this document useful (0 votes)

2 views405 pages

IU CalculusI Chap3

Uploaded by

Phương Uyên

We take content rights seriously. If you suspect this is your content, claim it here.

Available Formats

Download as PDF, TXT or read online on Scribd

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Chapter 3: APPLICATIONS of DIFFERENTIATION

Duong T. PHAM

CALCULUS I

Duong T. PHAM 1 / 53
Outline

1 Maximum and minimum values

2 Derivative and shape of a graph

3 Indeterminate forms and L’Hopital Rule

4 Optimization problems

5 Newton’s method

6 Antiderivatives

Duong T. PHAM 2 / 53
Applications in Optimization

Duong T. PHAM 3 / 53
Applications in Optimization

Important optimization problems require differential calculus , e.g.

Duong T. PHAM 3 / 53
Applications in Optimization

Important optimization problems require differential calculus , e.g.

What is the shape of a can that minimizes manufacturing costs?

Duong T. PHAM 3 / 53
Applications in Optimization

Important optimization problems require differential calculus , e.g.

What is the shape of a can that minimizes manufacturing costs?

What is the maximum acceleration of a space shuttle?

Duong T. PHAM 3 / 53
Applications in Optimization

Important optimization problems require differential calculus , e.g.

What is the shape of a can that minimizes manufacturing costs?

What is the maximum acceleration of a space shuttle?

What is the interest rate that the banks earn most in the market?

Duong T. PHAM 3 / 53
Applications in Optimization

Important optimization problems require differential calculus , e.g.

What is the shape of a can that minimizes manufacturing costs?

What is the maximum acceleration of a space shuttle?

What is the interest rate that the banks earn most in the market?

What is the shape of a racing car or an aircraft to minimize drag?

Duong T. PHAM 3 / 53
Applications in Optimization

Important optimization problems require differential calculus , e.g.

What is the shape of a can that minimizes manufacturing costs?

What is the maximum acceleration of a space shuttle?

What is the interest rate that the banks earn most in the market?

What is the shape of a racing car or an aircraft to minimize drag?

etc.

Duong T. PHAM 3 / 53
Applications in Optimization

Important optimization problems require differential calculus , e.g.

What is the shape of a can that minimizes manufacturing costs?

What is the maximum acceleration of a space shuttle?

What is the interest rate that the banks earn most in the market?

What is the shape of a racing car or an aircraft to minimize drag?

etc.
=⇒ finding the minimum and maximum of a function .

Duong T. PHAM 3 / 53
Maximum and minimum of a function

Duong T. PHAM 4 / 53
Maximum and minimum of a function

Definition.
Let f : D → R be a function and let c ∈ D.

Duong T. PHAM 4 / 53
Maximum and minimum of a function

Definition.
Let f : D → R be a function and let c ∈ D.
f has an absolute maximum (global maximum) at c if
f (x) ≤ f (c) ∀x ∈ D.

Then, f (c) is called the maximum value of f .

Duong T. PHAM 4 / 53
Maximum and minimum of a function

Definition.
Let f : D → R be a function and let c ∈ D.
f has an absolute maximum (global maximum) at c if
f (x) ≤ f (c) ∀x ∈ D.

Then, f (c) is called the maximum value of f .

f has an absolute minimum (global minimum) at c if
f (x) ≥ f (c) ∀x ∈ D.

Then, f (c) is called the minimum value of f .

Duong T. PHAM 4 / 53
Maximum and minimum of a function

Definition.
Let f : D → R be a function and let c ∈ D.
f has an absolute maximum (global maximum) at c if
f (x) ≤ f (c) ∀x ∈ D.

Then, f (c) is called the maximum value of f .

f has an absolute minimum (global minimum) at c if
f (x) ≥ f (c) ∀x ∈ D.

Then, f (c) is called the minimum value of f .

The maximum and minimum values of f are called the extreme
values of f .

Duong T. PHAM 4 / 53
Maximum and minimum of a function

Duong T. PHAM 5 / 53
Maximum and minimum of a function

y = f (x)

Duong T. PHAM 5 / 53
Maximum and minimum of a function

y = f (x)

a b c d e x

Duong T. PHAM 5 / 53
Maximum and minimum of a function

f (d)
y = f (x)

f (a)

a b c d e x

Duong T. PHAM 5 / 53
Maximum and minimum of a function

f (d)
y = f (x)

f (a)

a b c d e x

f (a) is the absolute minimum of f ;

Duong T. PHAM 5 / 53
Maximum and minimum of a function

f (d)
y = f (x)

f (a)

a b c d e x

f (a) is the absolute minimum of f ;

f (d) is the absolute maximum of f

Duong T. PHAM 5 / 53
Maximum and minimum of a function

f (d)
y = f (x)

f (c)

f (a)

a b c1 c c2 d e x

f (a) is the absolute minimum of f ;

f (d) is the absolute maximum of f

Duong T. PHAM 5 / 53
Maximum and minimum of a function

f (d)
y = f (x)

f (c)

f (a)

a b c1 c c2 d e x

f (a) is the absolute minimum of f ;

f (d) is the absolute maximum of f
f (c) ≤ f (x) ∀x ∈ (c1 , c2 )

Duong T. PHAM 5 / 53
Maximum and minimum of a function

f (d)
y = f (x)

f (c)

f (a)

a b c1 c c2 d e x

f (a) is the absolute minimum of f ;

f (d) is the absolute maximum of f
f (c) ≤ f (x) ∀x ∈ (c1 , c2 ) =⇒ f (c) is a local minimum of f

Duong T. PHAM 5 / 53
Local maximum and minimum

Duong T. PHAM 6 / 53
Local maximum and minimum

Definition.
Let f : D → R be a function and let c ∈ D.

Duong T. PHAM 6 / 53
Local maximum and minimum

Definition.
Let f : D → R be a function and let c ∈ D.
f has an local maximum (relative maximum) at c if there exists an
interval (c1 , c2 ) ⊂ D such that c ∈ (c1 , c2 ) and
f (x) ≤ f (c) ∀x ∈ (c1 , c2 ).

Duong T. PHAM 6 / 53
Local maximum and minimum

f has an local minimum (relative minimum) at c if if there exists an

interval (c1 , c2 ) ⊂ D such that c ∈ (c1 , c2 ) and
f (x) ≥ f (c) ∀x ∈ (c1 , c2 ).

Duong T. PHAM 6 / 53
Local maximum and local minimum

Duong T. PHAM 7 / 53
Local maximum and local minimum

f (d)
y = f (x)

f (c)

f (a)

a b c1 c c2 d e x

Duong T. PHAM 7 / 53
Local maximum and local minimum

f (d)
y = f (x)

f (c)

f (a)

a b c1 c c2 d e x

f (c) is a local minimum

Duong T. PHAM 7 / 53
Local maximum and local minimum

f (d)
y = f (x)

f (c)

f (a)

a b c1 c c2 d e x

f (c) is a local minimum

f (b) is the local maximum of f ;

Duong T. PHAM 7 / 53
Local maximum and local minimum

f (d)
y = f (x)

f (c)

f (a)

a b c1 c c2 d e x

f (c) is a local minimum

f (b) is the local maximum of f ;
f (d) is the local maximum of f (also an absolute maximum)

Duong T. PHAM 7 / 53
Local maximum and local minimum

f (d)
y = f (x)

f (c)

f (a)

a b c1 c c2 d e x

f (c) is a local minimum

f (b) is the local maximum of f ;
f (d) is the local maximum of f (also an absolute maximum)
f (e) is a local minimum of f
Duong T. PHAM 7 / 53
The Extreme value Theorem

Duong T. PHAM 8 / 53
The Extreme value Theorem

Theorem.
Let f : [a, b] → R be a continuous function on [a, b]. Then f attains an
absolute maximum value f (c) and an absolute minimum value f (d) at
some numbers c and d in [a, b].

Duong T. PHAM 8 / 53
The Extreme value Theorem

Theorem.
Let f : [a, b] → R be a continuous function on [a, b]. Then f attains an
absolute maximum value f (c) and an absolute minimum value f (d) at
some numbers c and d in [a, b].

Remark: The theorem is not true if we replace the closed interval [a, b]
by the open interval (a, b) or other interval (a, b] or [a, b).

Duong T. PHAM 8 / 53
The Extreme value Theorem

Theorem.
Let f : [a, b] → R be a continuous function on [a, b]. Then f attains an
absolute maximum value f (c) and an absolute minimum value f (d) at
some numbers c and d in [a, b].

Remark: The theorem is not true if we replace the closed interval [a, b]
by the open interval (a, b) or other interval (a, b] or [a, b).

1
Ex: The function f (x) = does not attain an absolute maximum value
x
on (0, 5].

Duong T. PHAM 8 / 53
The Extreme value Theorem

Duong T. PHAM 9 / 53
The Extreme value Theorem

y
19
18
17
16
15
14
13
12
11 1
10 y=
9 x
8
7
6
5
4
3
2
1

1 2 3 4 x

Duong T. PHAM 9 / 53
The Fermat’s Theorem

Duong T. PHAM 10 / 53
The Fermat’s Theorem
y

y = f (x)

Duong T. PHAM 10 / 53
The Fermat’s Theorem
y
(c, f (c)

y = f (x)

c x

Duong T. PHAM 10 / 53
The Fermat’s Theorem
y
(c, f (c)

y = f (x)

(d, f (d)

c d x

Duong T. PHAM 10 / 53
The Fermat’s Theorem
y
(c, f (c)

y = f (x)

(d, f (d)

c d x

Duong T. PHAM 10 / 53
The Fermat’s Theorem
y
(c, f (c)

y = f (x)

(d, f (d)

c d x

f attains local maximum at c and local minimum at d

Duong T. PHAM 10 / 53
The Fermat’s Theorem
y
(c, f (c)

y = f (x)

(d, f (d)

c d x

f attains local maximum at c and local minimum at d =⇒ What is special

about f at x = c and x = d?

Theorem (The Fermat’s Theorem:).

If f attains local maximum or local minimum at c and if f 0 (c) exists, then
f 0 (c) = 0
Duong T. PHAM 10 / 53
Proof of Fermat’s Theorem

Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof:

Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

(c, f (c))

(c1 , f (c1 )) (c2 , f (c2 ))

Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

=⇒ ∀x ∈ (c1 , c2 ) : f (x) ≤ f (c) (c, f (c))

(c1 , f (c1 )) (c2 , f (c2 ))

(x, f (x))

Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

=⇒ ∀x ∈ (c1 , c2 ) : f (x) ≤ f (c) (c, f (c))

f (x) − f (c)

 ≥ 0 ∀c1 < x < c (c1 , f (c1 )) (c2 , f (c2 ))
 x −c


(x, f (x))
=⇒
 f (x) − f (c) ≤ 0 ∀c < x < c2



x −c

Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

=⇒ ∀x ∈ (c1 , c2 ) : f (x) ≤ f (c) (c, f (c))

f (x) − f (c)

 ≥ 0 ∀c1 < x < c (c1 , f (c1 )) (c2 , f (c2 ))
 x −c


(x, f (x))
=⇒
 f (x) − f (c) ≤ 0 ∀c < x < c2



x −c
f (x) − f (c)

 lim ≥0
x −c

x→c −

⇒ .
f (x) − f (c)


≤0

 lim
x→c + x −c

Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

=⇒ ∀x ∈ (c1 , c2 ) : f (x) ≤ f (c) (c, f (c))

Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

=⇒ ∀x ∈ (c1 , c2 ) : f (x) ≤ f (c) (c, f (c))

f (x) − f (c)

 ≥ 0 ∀c1 < x < c (c1 , f (c1 )) (c2 , f (c2 ))
 x −c


(x, f (x))
=⇒
 f (x) − f (c) ≤ 0 ∀c < x < c2



x −c
f (x) − f (c)

 lim ≥0
x −c

x→c −

f (x) − f (c)
⇒ . Since f 0 (c) exists, lim = M exists
x→c x −c
f (x) − f (c)


≤0

 lim
x→c + x −c
f (x) − f (c) f (x) − f (c)
=⇒ M = lim− = lim+
x→c x −c x→c x −c
Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

=⇒ ∀x ∈ (c1 , c2 ) : f (x) ≤ f (c) (c, f (c))

f (x) − f (c)

 ≥ 0 ∀c1 < x < c (c1 , f (c1 )) (c2 , f (c2 ))
 x −c


(x, f (x))
=⇒
 f (x) − f (c) ≤ 0 ∀c < x < c2



x −c
f (x) − f (c)

 lim ≥0
x −c

x→c −

f (x) − f (c)
⇒ . Since f 0 (c) exists, lim = M exists
x→c x −c
f (x) − f (c)


≤0

 lim
x→c + x −c
(
f (x) − f (c) f (x) − f (c) M≥0
=⇒ M = lim− = lim+ =⇒
x→c x −c x→c x −c M≤0
Duong T. PHAM 11 / 53
Proof of Fermat’s Theorem
The Fermat’s Theorem: If f attains local maximum or local minimum at c and
if f 0 (c) exists, then f 0 (c) = 0

Proof: Suppose f attains local maximum at c.

=⇒ ∀x ∈ (c1 , c2 ) : f (x) ≤ f (c) (c, f (c))

f (x) − f (c)

 ≥ 0 ∀c1 < x < c (c1 , f (c1 )) (c2 , f (c2 ))
 x −c


(x, f (x))
=⇒
 f (x) − f (c) ≤ 0 ∀c < x < c2



x −c
f (x) − f (c)

 lim ≥0
x −c

x→c −

f (x) − f (c)
⇒ . Since f 0 (c) exists, lim = M exists
x→c x −c
f (x) − f (c)


≤0

 lim
x→c + x −c
(
f (x) − f (c) f (x) − f (c) M≥0
=⇒ M = lim− = lim+ =⇒ =⇒ M = 0
x→c x −c x→c x −c M≤0
Duong T. PHAM 11 / 53
The Fermat’s Theorem

Duong T. PHAM 12 / 53
The Fermat’s Theorem
Remark: ”Local maximum and local minimum at c” + existence of f 0 (c)
=⇒ f 0 (c) = 0. But (⇐=) is NOT true.

Ex: Let f (x) = x 3 .

Duong T. PHAM 12 / 53
The Fermat’s Theorem
Remark: ”Local maximum and local minimum at c” + existence of f 0 (c)
=⇒ f 0 (c) = 0. But (⇐=) is NOT true.

Ex: Let f (x) = x 3 . We have f 0 (x) = 3x 2 and f 0 (0) = 0 but f does NOT
attain local max. or local min. at 0.

Duong T. PHAM 12 / 53
The Fermat’s Theorem
Remark: ”Local maximum and local minimum at c” + existence of f 0 (c)
=⇒ f 0 (c) = 0. But (⇐=) is NOT true.

Ex: Let f (x) = x 3 . We have f 0 (x) = 3x 2 and f 0 (0) = 0 but f does NOT
attain local max. or local min. at 0.
y

Duong T. PHAM 12 / 53
Local minimum and local maximum

Duong T. PHAM 13 / 53
Local minimum and local maximum

Ex: Function f (x) = |x| attains local minimum at x = 0 but it is NOT

differentiable at x = 0.

Duong T. PHAM 13 / 53
Local minimum and local maximum

Ex: Function f (x) = |x| attains local minimum at x = 0 but it is NOT

differentiable at x = 0.

0 x

Duong T. PHAM 13 / 53
Critical number of a function

Duong T. PHAM 14 / 53
Critical number of a function

Definition.
A number c is called a critical number of a function f if f 0 (c) = 0 or
f 0 (c) does not exists.

Duong T. PHAM 14 / 53
Critical number of a function

Definition.
A number c is called a critical number of a function f if f 0 (c) = 0 or
f 0 (c) does not exists.

f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist

Duong T. PHAM 14 / 53
Critical number of a function

Definition.
A number c is called a critical number of a function f if f 0 (c) = 0 or
f 0 (c) does not exists.

f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist
√
Ex: Find critical points of f (x) = x(1 − x).

Duong T. PHAM 14 / 53
Critical number of a function

Definition.
A number c is called a critical number of a function f if f 0 (c) = 0 or
f 0 (c) does not exists.

f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist
√
Ex: Find critical points of f (x) = x(1 − x).
1−x √
Ans: f 0 (x) = √ − x
2 x

Duong T. PHAM 14 / 53
Critical number of a function

Definition.
A number c is called a critical number of a function f if f 0 (c) = 0 or
f 0 (c) does not exists.

f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist
√
Ex: Find critical points of f (x) = x(1 − x).
1−x √ 1 − x − 2x
Ans: f 0 (x) = √ − x = √
2 x 2 x

Duong T. PHAM 14 / 53
Critical number of a function

Definition.
A number c is called a critical number of a function f if f 0 (c) = 0 or
f 0 (c) does not exists.

f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist
√
Ex: Find critical points of f (x) = x(1 − x).
1−x √ 1 − x − 2x 1 − 3x
Ans: f 0 (x) = √ − x = √ = √
2 x 2 x 2 x
1 − 3x 1
We have f 0 (x) = 0 ⇐⇒ √ = 0 ⇐⇒ x =
2 x 3

Duong T. PHAM 14 / 53
Critical number of a function

Definition.
A number c is called a critical number of a function f if f 0 (c) = 0 or
f 0 (c) does not exists.

Duong T. PHAM 14 / 53
Critical number of a function

Definition.
A number c is called a critical number of a function f if f 0 (c) = 0 or
f 0 (c) does not exists.

Duong T. PHAM 14 / 53
Critical numbers of a function

Duong T. PHAM 15 / 53
Critical numbers of a function

Corollary.
If f has a local minimum or local maximum at c then c is a critical
number of f

Duong T. PHAM 15 / 53
Critical numbers of a function

Corollary.
If f has a local minimum or local maximum at c then c is a critical
number of f

Proof:

Duong T. PHAM 15 / 53
Critical numbers of a function

Corollary.
If f has a local minimum or local maximum at c then c is a critical
number of f

Proof:
f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist

Duong T. PHAM 15 / 53
Critical numbers of a function

Corollary.
If f has a local minimum or local maximum at c then c is a critical
number of f

Proof:
f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist

f has local min. or local max. at c

Duong T. PHAM 15 / 53
Critical numbers of a function

Corollary.
If f has a local minimum or local maximum at c then c is a critical
number of f

Proof:
f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist

f has local min. or local max. at c =⇒ There are 2 cases:





Duong T. PHAM 15 / 53
Critical numbers of a function

Corollary.
If f has a local minimum or local maximum at c then c is a critical
number of f

Proof:
f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist

f has local min. or local max. at c =⇒ There are 2 cases:


f 0 (c) does not exist =⇒ c is a critical number



Duong T. PHAM 15 / 53
Critical numbers of a function

Corollary.
If f has a local minimum or local maximum at c then c is a critical
number of f

Proof:
f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist

f has local min. or local max. at c =⇒ There are 2 cases:


f 0 (c) does not exist =⇒ c is a critical number


f 0 (c) exists

Duong T. PHAM 15 / 53
Critical numbers of a function

Corollary.
If f has a local minimum or local maximum at c then c is a critical
number of f

Proof:
f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist

f has local min. or local max. at c =⇒ There are 2 cases:


f 0 (c) does not exist =⇒ c is a critical number


Fermat’s Th. 0
f 0 (c) exists =⇒ f (c) = 0

Duong T. PHAM 15 / 53
Critical numbers of a function

Corollary.
If f has a local minimum or local maximum at c then c is a critical
number of f

Proof:
f 0 (c) = 0

=⇒ c is a critical number of f
f 0 (c) does NOT exist

f has local min. or local max. at c =⇒ There are 2 cases:


f 0 (c) does not exist =⇒ c is a critical number


Fermat’s Th. 0
f 0 (c) exists =⇒ f (c) = 0 =⇒ c is a critical number

Duong T. PHAM 15 / 53
Finding absolute minimum and absolute maximum

Duong T. PHAM 16 / 53
Finding absolute minimum and absolute maximum
The closed interval method: Let f : [a, b] → R be a continuous function.

Duong T. PHAM 16 / 53
Finding absolute minimum and absolute maximum
The closed interval method: Let f : [a, b] → R be a continuous function. To
find the absolute max. and absolute min. of f , we follows the steps:
1 Find the values of f at critical numbers of f in (a, b);
2 Find the values f (a) and f (b);
3 The largest number in steps 1 and 2 is the absolute maximum value of f ,
and the smallest number in steps 1 and 2 is the absolute minimum of f .

Ex: Find abs. max. and abs. min. of f (x) = 2x 3 − 9x 2 + 12x + 2 in [0, 3]

Ex: Find abs. max. and abs. min. of f (x) = 2x 3 − 9x 2 + 12x + 2 in [0, 3]
Ans: f 0 (x) = 6x 2 − 18x + 12

Ex: Find abs. max. and abs. min. of f (x) = 2x 3 − 9x 2 + 12x + 2 in [0, 3]
Ans: f 0 (x) = 6x 2 − 18x + 12 = 6(x 2 − 3x + 2)

Ex: Find abs. max. and abs. min. of f (x) = 2x 3 − 9x 2 + 12x + 2 in [0, 3]
Ans: f 0 (x) = 6x 2 − 18x + 12 = 6(x 2 − 3x + 2)
1 f 0 (x) = 0 ⇐⇒ x = 1 or x = 2,

Ex: Find abs. max. and abs. min. of f (x) = 2x 3 − 9x 2 + 12x + 2 in [0, 3]
Ans: f 0 (x) = 6x 2 − 18x + 12 = 6(x 2 − 3x + 2)
1 f 0 (x) = 0 ⇐⇒ x = 1 or x = 2, and f (1) = 7, f (2) = 6
2 f (0) = 2 and f (3) = 11

Ex: Find abs. max. and abs. min. of f (x) = 2x 3 − 9x 2 + 12x + 2 in [0, 3]
Ans: f 0 (x) = 6x 2 − 18x + 12 = 6(x 2 − 3x + 2)
1 f 0 (x) = 0 ⇐⇒ x = 1 or x = 2, and f (1) = 7, f (2) = 6
2 f (0) = 2 and f (3) = 11
3 Comparing the 4 values in Steps 1 and 2, we conclude

max f = f (3) = 11 and min f = f (0) = 2

[0,3] [0,3]

Duong T. PHAM 16 / 53
The closed interval method

Duong T. PHAM 17 / 53
The closed interval method

y
y = 2x 3 − 9x 2 + 12x + 5

7
6

0 1 2 3 x

Duong T. PHAM 17 / 53
Rolle’s Theorem

Duong T. PHAM 18 / 53
Rolle’s Theorem
Theorem.
Rolle’s Theorem: Let f : [a, b] → R be a function satisfying

Duong T. PHAM 18 / 53
Rolle’s Theorem
Theorem.
Rolle’s Theorem: Let f : [a, b] → R be a function satisfying
1 f is continuous in [a, b]

Duong T. PHAM 18 / 53
Rolle’s Theorem
Theorem.
Rolle’s Theorem: Let f : [a, b] → R be a function satisfying
1 f is continuous in [a, b]
2 f is differentiable in (a, b)

Duong T. PHAM 18 / 53
Rolle’s Theorem
Theorem.
Rolle’s Theorem: Let f : [a, b] → R be a function satisfying
1 f is continuous in [a, b]
2 f is differentiable in (a, b)
3 f (a) = f (b).

Duong T. PHAM 18 / 53
Rolle’s Theorem
Theorem.
Rolle’s Theorem: Let f : [a, b] → R be a function satisfying
1 f is continuous in [a, b]
2 f is differentiable in (a, b)
3 f (a) = f (b). Then there exists a c ∈ (a, b) such that f 0 (c) = 0

Proof:

Proof: The case: f (x) = c ∀x ∈ [a, b].

Proof: The case: f (x) = c ∀x ∈ [a, b]. Then clearly

f 0 (c) = 0 ∀c ∈ (a, b).

Proof: The case: f (x) = c ∀x ∈ [a, b]. Then clearly

f 0 (c) = 0 ∀c ∈ (a, b).
The case: ∃x ∈ (a, b) s.t. f (x) > f (a) = f (b) .

Proof: The case: f (x) = c ∀x ∈ [a, b]. Then clearly

f 0 (c) = 0 ∀c ∈ (a, b).
The case: ∃x ∈ (a, b) s.t. f (x) > f (a) = f (b) . Since f is continuous in
ExtremeValueTh.
[a, b] =⇒ ∃c ∈ (a, b) such that
f (c) = max f
[a,b]

Proof: The case: f (x) = c ∀x ∈ [a, b]. Then clearly

f 0 (c) = 0 ∀c ∈ (a, b).
The case: ∃x ∈ (a, b) s.t. f (x) > f (a) = f (b) . Since f is continuous in
ExtremeValueTh.
[a, b] =⇒ ∃c ∈ (a, b) such that
f (c) = max f
[a,b]

=⇒ f has a local max. at c

Proof: The case: f (x) = c ∀x ∈ [a, b]. Then clearly

f 0 (c) = 0 ∀c ∈ (a, b).
The case: ∃x ∈ (a, b) s.t. f (x) > f (a) = f (b) . Since f is continuous in
ExtremeValueTh.
[a, b] =⇒ ∃c ∈ (a, b) such that
f (c) = max f
[a,b]

=⇒ f has a local max. at c + ”f 0 (c) exists”

Proof: The case: f (x) = c ∀x ∈ [a, b]. Then clearly

f 0 (c) = 0 ∀c ∈ (a, b).
The case: ∃x ∈ (a, b) s.t. f (x) > f (a) = f (b) . Since f is continuous in
ExtremeValueTh.
[a, b] =⇒ ∃c ∈ (a, b) such that
f (c) = max f
[a,b]

Fermat’s Th.
=⇒ f has a local max. at c + ”f 0 (c) exists” =⇒ f 0 (c) = 0

Proof: The case: f (x) = c ∀x ∈ [a, b]. Then clearly

f 0 (c) = 0 ∀c ∈ (a, b).
The case: ∃x ∈ (a, b) s.t. f (x) > f (a) = f (b) . Since f is continuous in
ExtremeValueTh.
[a, b] =⇒ ∃c ∈ (a, b) such that
f (c) = max f
[a,b]

Fermat’s Th.
=⇒ f has a local max. at c + ”f 0 (c) exists” =⇒ f 0 (c) = 0
The case: ∃x ∈ (a, b) s.t. f (x) < f (a) = f (b) .
Duong T. PHAM 18 / 53
Rolle’s Theorem
Theorem.
Rolle’s Theorem: Let f : [a, b] → R be a function satisfying
1 f is continuous in [a, b]
2 f is differentiable in (a, b)
3 f (a) = f (b). Then there exists a c ∈ (a, b) such that f 0 (c) = 0

Proof: The case: f (x) = c ∀x ∈ [a, b]. Then clearly

f 0 (c) = 0 ∀c ∈ (a, b).
The case: ∃x ∈ (a, b) s.t. f (x) > f (a) = f (b) . Since f is continuous in
ExtremeValueTh.
[a, b] =⇒ ∃c ∈ (a, b) such that
f (c) = max f
[a,b]

Fermat’s Th.
=⇒ f has a local max. at c + ”f 0 (c) exists” =⇒ f 0 (c) = 0
The case: ∃x ∈ (a, b) s.t. f (x) < f (a) = f (b) . (Similarly as above)
Duong T. PHAM 18 / 53
Rolle’s Theorem

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans:

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

f is a polynomial =⇒ f is differentiable (and thus continuous) in (−∞, ∞).

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

f is a polynomial =⇒ f is differentiable (and thus continuous) in (−∞, ∞).
f (−1) = −2 and f (0) = 1. Since f is continuous in [−1, 0] and since
f (−1) < 0 < f (0), the Intermediate value Theorem =⇒ ∃c ∈ (−1, 0) s.t.
f (c) = 0 =⇒ the equation has one root c ∈ (−1, 0)
Suppose that d 6= c is another root of the equation.
Rolle’s Th.
If c < d, then f is differentiable in [c, d] and f (c) = f (d) = 0 =⇒
exists e ∈ (c, d) s.t. f 0 (e) = 0.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Ex: Prove that equation x 3 + x 2 + 3x + 1 = 0 has exactly one real root.

Ans: Denote f (x) = x 3 + x 2 + 3x + 1.

Duong T. PHAM 19 / 53
Rolle’s Theorem

Duong T. PHAM 20 / 53
Rolle’s Theorem

y
y = x 3 + x 2 + 3x + 1

1
−1 c
0 x
−2

Duong T. PHAM 20 / 53
Recall the Rolle’s Theorem

Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

(a, f (a)) (b, f (b))

Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

(c, f (c))
t

(a, f (a)) (b, f (b))

Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

(c, f (c))
t

(a, f (a)) (b, f (b))

Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

(c, f (c))
t

(a, f (a)) (b, f (b))

f (b) − f (a)
The slope of secant connecting (a, f (a)) and (b, f (b)) is
b−a

Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

(c, f (c))
t

(a, f (a)) (b, f (b))

f (b) − f (a)
The slope of secant connecting (a, f (a)) and (b, f (b)) is =0
b−a

Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

(c, f (c))
t

(a, f (a)) (b, f (b))

f (b) − f (a)
The slope of secant connecting (a, f (a)) and (b, f (b)) is =0
b−a
Rolle’s Theorem: ∃c ∈ (a, b) s.t. f 0 (c) = 0

Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

(c, f (c))
t

(a, f (a)) (b, f (b))

f (b) − f (a)
The slope of secant connecting (a, f (a)) and (b, f (b)) is =0
b−a
Rolle’s Theorem: ∃c ∈ (a, b) s.t. f 0 (c) = 0
f (b) − f (a)
=⇒ f 0 (c) =
b−a
Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

(c, f (c))
t

(a, f (a)) (b, f (b))

f (b) − f (a)
The slope of secant connecting (a, f (a)) and (b, f (b)) is =0
b−a
Rolle’s Theorem: ∃c ∈ (a, b) s.t. f 0 (c) = 0
f (b) − f (a)
=⇒ f 0 (c) = ⇐⇒ t//L
b−a
Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

y
t
(c, f (c))

(b, f (b))
(c, f (c))
t

(a, f (a)) (b, f (b))

y
t
(c, f (c))

(b, f (b))

(a, f (a))

f (b) − f (a)
The slope of secant connecting (a, f (a)) and (b, f (b)) is
b−a

f (b) − f (a)
=⇒ f 0 (c) = ⇐⇒ t//L
b−a
Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

y
t
(c, f (c))

(b, f (b))

(a, f (a))

f (b) − f (a)
The slope of secant connecting (a, f (a)) and (b, f (b)) is
b−a

f (b) − f (a)
=⇒ f 0 (c) = ⇐⇒ t//L
b−a
Duong T. PHAM 21 / 53
Recall the Rolle’s Theorem

y
t
(c, f (c))

(b, f (b))

(a, f (a))

f (b) − f (a)
The slope of secant connecting (a, f (a)) and (b, f (b)) is
b−a

f (b) − f (a)
=⇒ f 0 (c) = ⇐⇒ t//L
b−a
Duong T. PHAM 21 / 53
The Mean Value Theorem

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b]
f is differentiable in (a, b)

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

Proof:

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

f (b) − f (a)
Proof: Denote h(x) = f (x) − (x − a) − f (a).
b−a

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

f (b) − f (a)
Proof: Denote h(x) = f (x) − (x − a) − f (a). Then h is
b−a
continuous in [a, b] and h is differentiable in (a, b).

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

f (b) − f (a)
Proof: Denote h(x) = f (x) − (x − a) − f (a). Then h is
b−a
continuous in [a, b] and h is differentiable in (a, b). Furthermore,

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

f (b) − f (a)
Proof: Denote h(x) = f (x) − (x − a) − f (a). Then h is
b−a
continuous in [a, b] and h is differentiable in (a, b). Furthermore,
h(a) = f (a)− f (b)−f
b−a
(a)
(a − a) − f (a) = 0

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

f (b)−f (a)
h(b) = f (b) − b−a (b − a) − f (a) = 0

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

f (b) − f (a)
Proof: Denote h(x) = f (x) − (x − a) − f (a). Then h is
b−a
continuous in [a, b] and h is differentiable in (a, b). Furthermore,

h(a) = f (a)− f (b)−f
b−a
(a)
(a − a) − f (a) = 0  

f (b)−f (a) 
h(b) = f (b) − b−a (b − a) − f (a) = 0 

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

f (b) − f (a)
Proof: Denote h(x) = f (x) − (x − a) − f (a). Then h is
b−a
continuous in [a, b] and h is differentiable in (a, b). Furthermore,

h(a) = f (a)− f (b)−f
b−a
(a)
(a − a) − f (a) = 0  
=⇒ h(a) = h(b).
f (b)−f (a) 
h(b) = f (b) − b−a (b − a) − f (a) = 0 

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

Rolle’s Theorem: there is a c ∈ (a, b) such that h0 (c) = 0

Duong T. PHAM 22 / 53
The Mean Value Theorem
Theorem.
Let f : [a, b] → R. Then
(
f is continuous in [a, b] f (b) − f (a)
=⇒ ∃c ∈ (a, b) s.t. f 0 (c) =
f is differentiable in (a, b) b−a

Rolle’s Theorem: there is a c ∈ (a, b) such that h0 (c) = 0 =⇒

f (b) − f (a)
f 0 (c) =
b−a
Duong T. PHAM 22 / 53
The Mean Value Theorem

Duong T. PHAM 23 / 53
The Mean Value Theorem

Theorem.
Let f : (a, b) → R be a function which is differentiable in (a, b) and
f 0 (x) = 0 for all x ∈ (a, b). Then f is a constant function.

Duong T. PHAM 23 / 53
The Mean Value Theorem

Theorem.
Let f : (a, b) → R be a function which is differentiable in (a, b) and
f 0 (x) = 0 for all x ∈ (a, b). Then f is a constant function.

Proof: Let x1 , x2 ∈ (a, b).

Duong T. PHAM 23 / 53
The Mean Value Theorem

Theorem.
Let f : (a, b) → R be a function which is differentiable in (a, b) and
f 0 (x) = 0 for all x ∈ (a, b). Then f is a constant function.

Proof: Let x1 , x2 ∈ (a, b). Then f is continuous in [x1 , x2 ] and

differentiable in (x1 , x2 ). By the Mean Value Theorem, there is a
c ∈ (x1 , x2 ) such that
f (x2 ) − f (x1 )
f 0 (c) = .
x2 − x1

Duong T. PHAM 23 / 53
The Mean Value Theorem

Theorem.
Let f : (a, b) → R be a function which is differentiable in (a, b) and
f 0 (x) = 0 for all x ∈ (a, b). Then f is a constant function.

Proof: Let x1 , x2 ∈ (a, b). Then f is continuous in [x1 , x2 ] and

differentiable in (x1 , x2 ). By the Mean Value Theorem, there is a
c ∈ (x1 , x2 ) such that
f (x2 ) − f (x1 )
f 0 (c) = .
x2 − x1
Since f 0 (x) = 0 for all x ∈ (x1 , x2 ), we have
f (x2 ) − f (x1 )
= 0.
x2 − x1

Duong T. PHAM 23 / 53
The Mean Value Theorem

Theorem.
Let f : (a, b) → R be a function which is differentiable in (a, b) and
f 0 (x) = 0 for all x ∈ (a, b). Then f is a constant function.

Proof: Let x1 , x2 ∈ (a, b). Then f is continuous in [x1 , x2 ] and

differentiable in (x1 , x2 ). By the Mean Value Theorem, there is a
c ∈ (x1 , x2 ) such that
f (x2 ) − f (x1 )
f 0 (c) = .
x2 − x1
Since f 0 (x) = 0 for all x ∈ (x1 , x2 ), we have
f (x2 ) − f (x1 )
= 0.
x2 − x1
It means that f (x1 ) = f (x2 ) and this is true for any x1 , x2 ∈ (a, b). Hence,
f is a constant function in (a, b).
Duong T. PHAM 23 / 53
Derivative and shape of a graph

Duong T. PHAM 24 / 53
Derivative and shape of a graph
Increasing and decreasing Test: Let f : (a, b) → R be a differentiable
function.

Duong T. PHAM 24 / 53
Derivative and shape of a graph
Increasing and decreasing Test: Let f : (a, b) → R be a differentiable
function.
1 If f 0 (x) > 0, ∀x ∈ (a, b), then f is increasing in (a, b).

y = f (x)

a b c d e x

y = f (x)

a b c d e x

Duong T. PHAM 24 / 53
Increasing and decreasing Test

Duong T. PHAM 25 / 53
Increasing and decreasing Test

Ex: Determine when the function f (x) = 3x 4 − 4x 3 − 12x 2 + 5 is increas-

ing and decreasing.

Duong T. PHAM 25 / 53
Increasing and decreasing Test

Ex: Determine when the function f (x) = 3x 4 − 4x 3 − 12x 2 + 5 is increas-

ing and decreasing.

Ans: f 0 (x) = 12x 3 − 12x 2 − 24x

Duong T. PHAM 25 / 53
Increasing and decreasing Test

Ex: Determine when the function f (x) = 3x 4 − 4x 3 − 12x 2 + 5 is increas-

ing and decreasing.

Ans: f 0 (x) = 12x 3 − 12x 2 − 24x = 12x(x − 2)(x + 1)

Duong T. PHAM 25 / 53
Increasing and decreasing Test

Ex: Determine when the function f (x) = 3x 4 − 4x 3 − 12x 2 + 5 is increas-

ing and decreasing.

Ans: f 0 (x) = 12x 3 − 12x 2 − 24x = 12x(x − 2)(x + 1)

=⇒ f 0 (x) = 0 ⇐⇒ x = −1 ∨ x = 0 ∨ x = 2

Duong T. PHAM 25 / 53
Increasing and decreasing Test

Ex: Determine when the function f (x) = 3x 4 − 4x 3 − 12x 2 + 5 is increas-

ing and decreasing.

Ans: f 0 (x) = 12x 3 − 12x 2 − 24x = 12x(x − 2)(x + 1)

=⇒ f 0 (x) = 0 ⇐⇒ x = −1 ∨ x = 0 ∨ x = 2

x −∞ −1 0 2 ∞
x −2 − | − | − 0 +
x − | − 0 + | +
x +1 − 0 + | + | +
f 0 (x) − 0 + 0 − 0 +
f (x) & 0 % 5 & −27 %

Duong T. PHAM 25 / 53
Increasing and decreasing Test

Duong T. PHAM 26 / 53
Increasing and decreasing Test

2
−1 0 x

y = 3x 4 − 4x 3 − 12x 2 + 5

Duong T. PHAM 26 / 53
Duong T. PHAM 27 / 53
The First derivative Test: Suppose that c is a critical number of a
continuous function f .
(i) If f 0 changes from positive to negative at c, then f has a local
maximum at c.
(ii) If f 0 changes from negative to positive at c, then f has a local
minimum at c.
(iii) If f 0 does not change sign at c (for example, if f 0 is positive on both
sides of c or negative on both sides), then f has no local maximum
or minimum at c.

Duong T. PHAM 27 / 53
The First derivative Test: Suppose that c is a critical number of a
continuous function f .
(i) If f 0 changes from positive to negative at c, then f has a local
maximum at c.
(ii) If f 0 changes from negative to positive at c, then f has a local
minimum at c.
(iii) If f 0 does not change sign at c (for example, if f 0 is positive on both
sides of c or negative on both sides), then f has no local maximum
or minimum at c.
y

f0 < 0 f0 > 0

c x

local minimum

f0 < 0 f0 > 0 f0 > 0 f0 < 0

c x c x

local minimum local maximum

f0 > 0

f0 < 0 f0 > 0 f0 > 0 f0 < 0 f0 > 0

c x c x c x
local minimum local maximum no local min. or max.

f0 < 0
f0 > 0

f0 < 0 f0 > 0 f0 > 0 f0 < 0 f0 > 0 f0 < 0

c x c x c x c x
local minimum local maximum no local min. or max.
no local min. or max

Duong T. PHAM 27 / 53
The first derivative Test

Duong T. PHAM 28 / 53
The first derivative Test

Ex: Find the local minimum and maximum values of the function f (x) =
3x 4 − 4x 3 − 12x 2 + 5

Duong T. PHAM 28 / 53
The first derivative Test

Ex: Find the local minimum and maximum values of the function f (x) =
3x 4 − 4x 3 − 12x 2 + 5

Ans: f 0 (x) = 12x 3 − 12x 2 − 24x

Duong T. PHAM 28 / 53
The first derivative Test

Ex: Find the local minimum and maximum values of the function f (x) =
3x 4 − 4x 3 − 12x 2 + 5

Ans: f 0 (x) = 12x 3 − 12x 2 − 24x = 12x(x − 2)(x + 1)

Duong T. PHAM 28 / 53
The first derivative Test

Ex: Find the local minimum and maximum values of the function f (x) =
3x 4 − 4x 3 − 12x 2 + 5

Ans: f 0 (x) = 12x 3 − 12x 2 − 24x = 12x(x − 2)(x + 1)

=⇒ f 0 (x) = 0 ⇐⇒ x = −1 ∨ x = 0 ∨ x = 2

Duong T. PHAM 28 / 53
The first derivative Test

Ex: Find the local minimum and maximum values of the function f (x) =
3x 4 − 4x 3 − 12x 2 + 5

Ans: f 0 (x) = 12x 3 − 12x 2 − 24x = 12x(x − 2)(x + 1)

=⇒ f 0 (x) = 0 ⇐⇒ x = −1 ∨ x = 0 ∨ x = 2

x −∞ −1 0 2 ∞
x −2 − | − | − 0 +
x − | − 0 + | +
x +1 − 0 + | + | +
f 0 (x) − 0 + 0 − 0 +
f (x) & 0 % 5 & −27 %

Duong T. PHAM 28 / 53
The first derivative Test

Ex: Find the local minimum and maximum values of the function f (x) =
3x 4 − 4x 3 − 12x 2 + 5

Ans: f 0 (x) = 12x 3 − 12x 2 − 24x = 12x(x − 2)(x + 1)

=⇒ f 0 (x) = 0 ⇐⇒ x = −1 ∨ x = 0 ∨ x = 2

x −∞ −1 0 2 ∞
x −2 − | − | − 0 +
x − | − 0 + | +
x +1 − 0 + | + | +
f 0 (x) − 0 + 0 − 0 +
f (x) & 0 % 5 & −27 %

f attains local minimum at −1 and 2; and attains local maximum at

Duong T. PHAM 28 / 53
The first derivative Test

Duong T. PHAM 29 / 53
The first derivative Test

2
−1 0 x

y = 3x 4 − 4x 3 − 12x 2 + 5

Duong T. PHAM 29 / 53
Duong T. PHAM 30 / 53
Definition.
Let f : I → R.

Duong T. PHAM 30 / 53
Definition.
Let f : I → R.
If the graph of f lies above all of its tangent lines, then f is said to be
upward concave on I

upwad concave
Duong T. PHAM 30 / 53
Definition.
Let f : I → R.
If the graph of f lies above all of its tangent lines, then f is said to be
upward concave on I
If the graph of f lies below all of its tangent lines, then f is said to be
downward concave on I
y

x x

upwad concave downward concave

Duong T. PHAM 30 / 53
Concavity Test

Duong T. PHAM 31 / 53
Concavity Test
Concavity Test: Let f : I → R. Then

Duong T. PHAM 31 / 53
Concavity Test
Concavity Test: Let f : I → R. Then
(a) If f 00 (x) > 0 ∀x ∈ I , then the graph of f is concave upward on I .

Duong T. PHAM 31 / 53
Concavity Test
Concavity Test: Let f : I → R. Then
(a) If f 00 (x) > 0 ∀x ∈ I , then the graph of f is concave upward on I .
(b) If f 00 (x) < 0 ∀x ∈ I , then the graph of f is concave downward on I .

Discussion:

Discussion:
f 00 (x) > 0 for all x ∈ I =⇒ f 0 (x) is increasing on I

x
f 0 increases → upwad concave
Duong T. PHAM 31 / 53
Concavity Test
Concavity Test: Let f : I → R. Then
(a) If f 00 (x) > 0 ∀x ∈ I , then the graph of f is concave upward on I .
(b) If f 00 (x) < 0 ∀x ∈ I , then the graph of f is concave downward on I .

Discussion:
f 00 (x) > 0 for all x ∈ I =⇒ f 0 (x) is increasing on I
f 00 (x < 0) for all x ∈ I =⇒ f 0 (x) is decreasing on I
y

Discussion:
f 00 (x) > 0 for all x ∈ I =⇒ f 0 (x) is increasing on I
f 00 (x < 0) for all x ∈ I =⇒ f 0 (x) is decreasing on I
y y

x x
f 0 increases → upwad concave f 0 decreases → downward concave
Duong T. PHAM 31 / 53
Inflection Point

Duong T. PHAM 32 / 53
Inflection Point

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 or from concave downward to concave upward at P

Duong T. PHAM 32 / 53
Inflection Point

(c, f (c)) (c, f (c))

x x

(c, f (c)) is inflection point (c, f (c)) is inflection point

Duong T. PHAM 32 / 53
The second derivative Test

Duong T. PHAM 33 / 53
The second derivative Test

The second derivative Test: Let f be a function such that f 00 is contin-

uous near c.

Duong T. PHAM 33 / 53
The second derivative Test

The second derivative Test: Let f be a function such that f 00 is contin-

uous near c. Then

Duong T. PHAM 33 / 53
The second derivative Test

The second derivative Test: Let f be a function such that f 00 is contin-

uous near c. Then
(a) If f 0 (c) = 0 and f 00 (c) > 0, then f has a local minimum at c,

Duong T. PHAM 33 / 53
The second derivative Test

The second derivative Test: Let f be a function such that f 00 is contin-

uous near c. Then
(a) If f 0 (c) = 0 and f 00 (c) > 0, then f has a local minimum at c,
(b) If f 0 (c) = 0 and f 00 (c) < 0, then f has a local maximum at c.

Duong T. PHAM 33 / 53
The second derivative Test

Duong T. PHAM 34 / 53
The second derivative Test
Ex: Discuss the concavity, inflection points, local maxima and local min-
ima of the curve y = x 4 − 4x 3 .

Ans:

Duong T. PHAM 34 / 53
The second derivative Test
Ex: Discuss the concavity, inflection points, local maxima and local min-
ima of the curve y = x 4 − 4x 3 .

Ans: Denote f (x) = x 4 − 4x 3 . Then

Duong T. PHAM 34 / 53
The second derivative Test
Ex: Discuss the concavity, inflection points, local maxima and local min-
ima of the curve y = x 4 − 4x 3 .

Ans: Denote f (x) = x 4 − 4x 3 . Then

f 0 (x) = 4x 3 − 12x 2 = 4x 2 (x − 3)

Duong T. PHAM 34 / 53
The second derivative Test
Ex: Discuss the concavity, inflection points, local maxima and local min-
ima of the curve y = x 4 − 4x 3 .

Ans: Denote f (x) = x 4 − 4x 3 . Then

f 0 (x) = 4x 3 − 12x 2 = 4x 2 (x − 3)
f 00 (x) = 12x 2 − 24x = 12x(x − 2)

Duong T. PHAM 34 / 53
The second derivative Test
Ex: Discuss the concavity, inflection points, local maxima and local min-
ima of the curve y = x 4 − 4x 3 .

Ans: Denote f (x) = x 4 − 4x 3 . Then

f 0 (x) = 4x 3 − 12x 2 = 4x 2 (x − 3)
f 00 (x) = 12x 2 − 24x = 12x(x − 2)

f 0 (x) = 0 ⇐⇒ x = 0 ∨ x = 3

Duong T. PHAM 34 / 53
The second derivative Test
Ex: Discuss the concavity, inflection points, local maxima and local min-
ima of the curve y = x 4 − 4x 3 .

Ans: Denote f (x) = x 4 − 4x 3 . Then

f 0 (x) = 4x 3 − 12x 2 = 4x 2 (x − 3)
f 00 (x) = 12x 2 − 24x = 12x(x − 2)

f 0 (x) = 0 ⇐⇒ x = 0 ∨ x = 3 and f 00 (x) = 0 ⇐⇒ x = 0 ∨ x = 2

Duong T. PHAM 34 / 53
The second derivative Test
Ex: Discuss the concavity, inflection points, local maxima and local min-
ima of the curve y = x 4 − 4x 3 .

Ans: Denote f (x) = x 4 − 4x 3 . Then

f 0 (x) = 4x 3 − 12x 2 = 4x 2 (x − 3)
f 00 (x) = 12x 2 − 24x = 12x(x − 2)

f 0 (x) = 0 ⇐⇒ x = 0 ∨ x = 3 and f 00 (x) = 0 ⇐⇒ x = 0 ∨ x = 2

x 0 2 3
x2 + 0 + | + | +
x −3 − | − | − 0 +
x −2 − | − 0 + | +
f 0 (x) − 0 − | − 0 +
f 00 (x) + 0 − 0 + | +
f (x) & 0 & −16 & −27 %
up.con. infl.p. down.con. infl.p. up.con. local min. up.con.

Duong T. PHAM 34 / 53
The second derivative Test

Duong T. PHAM 35 / 53
The second derivative Test
y

inflection point 2 3

0 x

−16 inflection point

−27
local min.

Duong T. PHAM 35 / 53
Indeterminate forms

Duong T. PHAM 36 / 53
Indeterminate forms
f (x)
Indeterminate forms: consider lim . Then
x→a g (x)

Duong T. PHAM 36 / 53
Indeterminate forms
f (x)
Indeterminate forms: consider lim . Then
x→a g (x)
(a) If f (x) → 0 and g (x) → 0 when x → a, then the limit is called an
indeterminate form of type 00
(b) If f (x) → ∞ (or −∞) and g (x) → ∞ (or −∞) when x → a, then
the limit is called an indeterminate form of type ∞
∞

x3 − x
Ex: Evaluate lim
x→1 x − 1

Ans:

x3 − x
Ex: Evaluate lim
x→1 x − 1

Ans: x3 − x x(x 2 − 1) x(x − 1)(x + 1)

lim = lim = lim
x→1 x − 1 x→1 x −1 x→1 x −1

Duong T. PHAM 36 / 53
Indeterminate forms
f (x)
Indeterminate forms: consider lim . Then
g (x)
x→a
(a) If f (x) → 0 and g (x) → 0 when x → a, then the limit is called an
indeterminate form of type 00
(b) If f (x) → ∞ (or −∞) and g (x) → ∞ (or −∞) when x → a, then
the limit is called an indeterminate form of type ∞
∞

x3 − x
Ex: Evaluate lim
x→1 x − 1

Ans: x3 − x x(x 2 − 1) x(x − 1)(x + 1)

lim = lim = lim
x→1 x − 1 x→1 x −1 x→1 x −1
= lim [x(x + 1)] = 1(1 + 1) = 2.
x→1

x3 − x
Ex: Evaluate lim
x→1 x − 1

Ans: x3 − x x(x 2 − 1) x(x − 1)(x + 1)

lim = lim = lim
x→1 x − 1 x→1 x −1 x→1 x −1
= lim [x(x + 1)] = 1(1 + 1) = 2.
x→1

ln x
Ex: lim
x→1 x − 1

x3 − x
Ex: Evaluate lim
x→1 x − 1

Ans: x3 − x x(x 2 − 1) x(x − 1)(x + 1)

lim = lim = lim
x→1 x − 1 x→1 x −1 x→1 x −1
= lim [x(x + 1)] = 1(1 + 1) = 2.
x→1

ln x
Ex: lim =⇒ we need a tool to evaluate this limit.
x→1 x − 1

Duong T. PHAM 36 / 53
L’Hopital’s Rule

Duong T. PHAM 37 / 53
L’Hopital’s Rule

L’Hopital’s Rule: Let f and g be differentiable functions on an interval

(b, c), except possibly at number x = a ∈ (b, c). Suppose further that
g 0 (x) 6= 0 for all x ∈ (b, c)\{a}. If
lim f (x) = lim g (x) = 0
x→a x→a
or lim f (x) = ±∞ and lim g (x) = ±∞,
x→a x→a
then
f (x) f 0 (x)
lim = lim 0
x→a g (x) x→a g (x)

provided the limit on the right hand side exists (or is ∞ or −∞)

Duong T. PHAM 37 / 53
L’Hopital’s Rule

L’Hopital’s Rule: Let f and g be differentiable functions on an interval

provided the limit on the right hand side exists (or is ∞ or −∞)

ln x
Ex: lim
x→1 x − 1

Duong T. PHAM 37 / 53
L’Hopital’s Rule

L’Hopital’s Rule: Let f and g be differentiable functions on an interval

provided the limit on the right hand side exists (or is ∞ or −∞)

ln x L’Hopital. (ln x)0

Ex: lim = lim
x→1 x − 1 x→1 (x − 1)0

Duong T. PHAM 37 / 53
L’Hopital’s Rule

L’Hopital’s Rule: Let f and g be differentiable functions on an interval

provided the limit on the right hand side exists (or is ∞ or −∞)

1
ln x L’Hopital. (ln x)0
Ex: lim = lim = lim x
x→1 x − 1 x→1 (x − 1)0 x→1 1

Duong T. PHAM 37 / 53
L’Hopital’s Rule

L’Hopital’s Rule: Let f and g be differentiable functions on an interval

provided the limit on the right hand side exists (or is ∞ or −∞)

1
ln x L’Hopital. (ln x)0
Ex: lim = lim = lim x = 1.
x→1 x − 1 x→1 (x − 1)0 x→1 1

Duong T. PHAM 37 / 53
L’Hopital’s Rule

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex
lim
x→∞ x 2

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0
lim = lim
x→∞ x 2 x→∞ (x 2 )0

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0 ex
lim = lim = lim
x→∞ x 2 x→∞ (x 2 )0 x→∞ 2x

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0 ex
lim = lim = lim
x→∞ x 2 x→∞ (x 2 )0 x→∞ 2x
x
(e ) 0
L’Hopital.
= lim
x→∞ (2x)0

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0 ex
lim = lim = lim
x→∞ x 2 x→∞ (x 2 )0 x→∞ 2x
x
(e ) 0 ex
L’Hopital.
= lim = lim
x→∞ (2x)0 x→∞ 2

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0 ex
lim = lim = lim
x→∞ x 2 x→∞ (x 2 )0 x→∞ 2x
x
(e ) 0 ex
L’Hopital.
= lim = lim = ∞.
x→∞ (2x)0 x→∞ 2

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0 ex
lim = lim = lim
x→∞ x 2 x→∞ (x 2 )0 x→∞ 2x
x
(e ) 0 ex
L’Hopital.
= lim = lim = ∞.
x→∞ (2x)0 x→∞ 2

Ex:

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0 ex
lim = lim = lim
x→∞ x 2 x→∞ (x 2 )0 x→∞ 2x
x
(e ) 0 ex
L’Hopital.
= lim = lim = ∞.
x→∞ (2x)0 x→∞ 2

Ex:
sin x
lim
x→0 x

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0 ex
lim = lim = lim
x→∞ x 2 x→∞ (x 2 )0 x→∞ 2x
x
(e ) 0 ex
L’Hopital.
= lim = lim = ∞.
x→∞ (2x)0 x→∞ 2

Ex:
sin x L’Hopital. (sin x)0
lim = lim
x→0 x x→0 (x)0

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0 ex
lim = lim = lim
x→∞ x 2 x→∞ (x 2 )0 x→∞ 2x
x
(e ) 0 ex
L’Hopital.
= lim = lim = ∞.
x→∞ (2x)0 x→∞ 2

Ex:
sin x L’Hopital. (sin x)0 cos x
lim = lim = lim
x→0 x x→0 (x)0 x→0 1

Duong T. PHAM 38 / 53
L’Hopital’s Rule

Ex:
ex L’Hopital. (e x )0 ex
lim = lim = lim
x→∞ x 2 x→∞ (x 2 )0 x→∞ 2x
x
(e ) 0 ex
L’Hopital.
= lim = lim = ∞.
x→∞ (2x)0 x→∞ 2

Ex:
sin x L’Hopital. (sin x)0 cos x
lim = lim = lim = 1.
x→0 x x→0 (x)0 x→0 1

Duong T. PHAM 38 / 53
Indeterminate products

Duong T. PHAM 39 / 53
Indeterminate products

Intermediate products: The limit lim [f (x)g (x)] in which f (x) → 0 and
x→a
g (x) → ∞ (or −∞) as x → a is called an indeterminate form of type
0·∞

Duong T. PHAM 39 / 53
Indeterminate products

Intermediate products: The limit lim [f (x)g (x)] in which f (x) → 0 and
x→a
g (x) → ∞ (or −∞) as x → a is called an indeterminate form of type
0·∞

Ex: Evaluate lim+ x ln x

x→0

Duong T. PHAM 39 / 53
Indeterminate products

Intermediate products: The limit lim [f (x)g (x)] in which f (x) → 0 and
x→a
g (x) → ∞ (or −∞) as x → a is called an indeterminate form of type
0·∞

Ex: Evaluate lim+ x ln x

x→0

Ans:

Duong T. PHAM 39 / 53
Indeterminate products

Intermediate products: The limit lim [f (x)g (x)] in which f (x) → 0 and
x→a
g (x) → ∞ (or −∞) as x → a is called an indeterminate form of type
0·∞

Ex: Evaluate lim+ x ln x

x→0

Ans: We have

Duong T. PHAM 39 / 53
Indeterminate products

Intermediate products: The limit lim [f (x)g (x)] in which f (x) → 0 and
x→a
g (x) → ∞ (or −∞) as x → a is called an indeterminate form of type
0·∞

Ex: Evaluate lim+ x ln x

x→0

Ans: We have

ln x
lim+ x ln x = lim+ 1
x→0 x→0
x

Duong T. PHAM 39 / 53
Indeterminate products

Intermediate products: The limit lim [f (x)g (x)] in which f (x) → 0 and
x→a
g (x) → ∞ (or −∞) as x → a is called an indeterminate form of type
0·∞

Ex: Evaluate lim+ x ln x

x→0

Ans: We have

ln x L’Hopital. (ln x)0

lim+ x ln x = lim+ 1
= lim+ 1 0

x→0 x→0 x→0
x x

Duong T. PHAM 39 / 53
Indeterminate products

Intermediate products: The limit lim [f (x)g (x)] in which f (x) → 0 and
x→a
g (x) → ∞ (or −∞) as x → a is called an indeterminate form of type
0·∞

Ex: Evaluate lim+ x ln x

x→0

Ans: We have
1
ln x L’Hopital. (ln x)0 x
lim+ x ln x = lim+ 1
= lim+ = lim+
1 0
x→0 x→0
x
x→0
x
x→0 − x12

Duong T. PHAM 39 / 53
Indeterminate products

Intermediate products: The limit lim [f (x)g (x)] in which f (x) → 0 and
x→a
g (x) → ∞ (or −∞) as x → a is called an indeterminate form of type
0·∞

Ex: Evaluate lim+ x ln x

x→0

Ans: We have
1
ln x L’Hopital. (ln x)0 x
lim+ x ln x = lim+ 1
= lim+ = lim+
1 0
x→0 x→0
x
x→0
x
x→0 − x12
= lim+ (−x)
x→0

Duong T. PHAM 39 / 53
Indeterminate products

Intermediate products: The limit lim [f (x)g (x)] in which f (x) → 0 and
x→a
g (x) → ∞ (or −∞) as x → a is called an indeterminate form of type
0·∞

Ex: Evaluate lim+ x ln x

x→0

Ans: We have
1
ln x L’Hopital. (ln x)0 x
lim+ x ln x = lim+ 1
= lim+ = lim+
1 0
x→0 x→0
x
x→0
x
x→0 − x12
= lim+ (−x) = 0.
x→0

Duong T. PHAM 39 / 53
Indeterminate differences

Duong T. PHAM 40 / 53
Indeterminate differences

Indeterminate difference: If lim f (x) = ∞ and lim g (x) = ∞, then the

x→a x→a
limit
lim [f (x) − g (x)]
x→a

is called an indeterminate form of type ∞ − ∞.

Duong T. PHAM 40 / 53
Indeterminate differences

Indeterminate difference: If lim f (x) = ∞ and lim g (x) = ∞, then the

x→a x→a
limit
lim [f (x) − g (x)]
x→a

is called an indeterminate form of type ∞ − ∞.

Ex: Compute lim (sec x − tan x)

x→0

Duong T. PHAM 40 / 53
Indeterminate differences

Indeterminate difference: If lim f (x) = ∞ and lim g (x) = ∞, then the

x→a x→a
limit
lim [f (x) − g (x)]
x→a

is called an indeterminate form of type ∞ − ∞.

Ex: Compute lim (sec x − tan x)

x→0

Ans: We have

Duong T. PHAM 40 / 53
Indeterminate differences

Indeterminate difference: If lim f (x) = ∞ and lim g (x) = ∞, then the

x→a x→a
limit
lim [f (x) − g (x)]
x→a

is called an indeterminate form of type ∞ − ∞.

Ex: Compute lim (sec x − tan x)

x→0

Ans: We have

1 sin x
lim (sec x − tan x) = lim −
x→(π/2)− x→(π/2)− cos x cos x

Duong T. PHAM 40 / 53
Indeterminate differences

Indeterminate difference: If lim f (x) = ∞ and lim g (x) = ∞, then the

x→a x→a
limit
lim [f (x) − g (x)]
x→a

is called an indeterminate form of type ∞ − ∞.

Ex: Compute lim (sec x − tan x)

x→0

Ans: We have

1 sin x 1 − sin x
lim (sec x − tan x) = lim − = lim
x→(π/2)− x→(π/2)− cos x cos x x→(π/2)− cos x

Duong T. PHAM 40 / 53
Indeterminate differences

Indeterminate difference: If lim f (x) = ∞ and lim g (x) = ∞, then the

x→a x→a
limit
lim [f (x) − g (x)]
x→a

is called an indeterminate form of type ∞ − ∞.

Ex: Compute lim (sec x − tan x)

x→0

Ans: We have

1 sin x 1 − sin x
lim (sec x − tan x) = lim − = lim
x→(π/2)− x→(π/2)− cos x cos x x→(π/2)− cos x

L’Hopital. (1 − sin x)0

= lim
x→(π/2)− (cos x)0

Duong T. PHAM 40 / 53
Indeterminate differences

Indeterminate difference: If lim f (x) = ∞ and lim g (x) = ∞, then the

x→a x→a
limit
lim [f (x) − g (x)]
x→a

is called an indeterminate form of type ∞ − ∞.

Ex: Compute lim (sec x − tan x)

x→0

Ans: We have

1 sin x 1 − sin x
lim (sec x − tan x) = lim − = lim
x→(π/2)− x→(π/2)− cos x cos x x→(π/2)− cos x
(1 − sin x) 0 − cos x
L’Hopital.
= lim 0
= lim
x→(π/2)− (cos x) x→(π/2)− − sin x

Duong T. PHAM 40 / 53
Indeterminate differences

Indeterminate difference: If lim f (x) = ∞ and lim g (x) = ∞, then the

x→a x→a
limit
lim [f (x) − g (x)]
x→a

is called an indeterminate form of type ∞ − ∞.

Ex: Compute lim (sec x − tan x)

x→0

= 0.

Duong T. PHAM 40 / 53
Indeterminate powers

Duong T. PHAM 41 / 53
Indeterminate powers

The limit lim [f (x)]g (x) when f and g satisfy

x→a

Duong T. PHAM 41 / 53
Indeterminate powers

The limit lim [f (x)]g (x) when f and g satisfy

x→a
(a) lim f (x) = lim g (x) = 0 type 00
x→a x→a

Duong T. PHAM 41 / 53
Indeterminate powers

The limit lim [f (x)]g (x) when f and g satisfy

x→a
(a) lim f (x) = lim g (x) = 0 type 00
x→a x→a
(b) lim f (x) = ∞ and lim g (x) = 0 type ∞0
x→a x→a

Duong T. PHAM 41 / 53
Indeterminate powers

The limit lim [f (x)]g (x) when f and g satisfy

x→a
(a) lim f (x) = lim g (x) = 0 type 00
x→a x→a
(b) lim f (x) = ∞ and lim g (x) = 0 type ∞0
x→a x→a
(c) lim f (x) = 1 and lim g (x) = ±∞ type 1∞ are indeterminate
x→a x→a
powers.

Duong T. PHAM 41 / 53
Indeterminate powers

The limit lim [f (x)]g (x) when f and g satisfy

Remark: To evaluate the above limits we can either

Duong T. PHAM 41 / 53
Indeterminate powers

The limit lim [f (x)]g (x) when f and g satisfy

Remark: To evaluate the above limits we can either

1 write y = [f (x)]g (x) and thus ln y = g (x) ln f (x), and then evaluate
lim ln y first; after that, we deduce lim y .
x→a x→a

Duong T. PHAM 41 / 53
Indeterminate powers

The limit lim [f (x)]g (x) when f and g satisfy

Remark: To evaluate the above limits we can either

1 write y = [f (x)]g (x) and thus ln y = g (x) ln f (x), and then evaluate
lim ln y first; after that, we deduce lim y .
x→a x→a
2 We can write [f (x)]g (x) = e g (x) ln f (x) .

Duong T. PHAM 41 / 53
Indeterminate powers

The limit lim [f (x)]g (x) when f and g satisfy

Remark: To evaluate the above limits we can either

1 write y = [f (x)]g (x) and thus ln y = g (x) ln f (x), and then evaluate
lim ln y first; after that, we deduce lim y .
x→a x→a
2 We can write [f (x)]g (x) = e g (x) ln f (x) .
3 By these ways, we transfer it to the problem of finding limit of the
type 0 · ∞.

Duong T. PHAM 41 / 53
Indeterminate powers

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Then

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Then
lim+ ln y = lim+ cot x ln(1 + 4 sin 4x)
x→0 x→0

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Then
ln(1 + 4 sin 4x)
lim+ ln y = lim+ cot x ln(1 + 4 sin 4x) = lim+
x→0 x→0 x→0 tan x

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Then
ln(1 + 4 sin 4x)
lim+ ln y = lim+ cot x ln(1 + 4 sin 4x) = lim+
x→0 x→0 x→0 tan x
16 cos 4x
lim 1 + 4 sin 4x
L’Hopital.
=
x→0+ 1
cos2 x

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Then
ln(1 + 4 sin 4x)
lim+ ln y = lim+ cot x ln(1 + 4 sin 4x) = lim+
x→0 x→0 x→0 tan x
16 cos 4x
16 cos 4x · cos2 x
lim+ 1 + 4 sin 4x = lim+
L’Hopital.
=
x→0 1 x→0 1 + 4 sin 4x
cos2 x

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Then
ln(1 + 4 sin 4x)
lim+ ln y = lim+ cot x ln(1 + 4 sin 4x) = lim+
x→0 x→0 x→0 tan x
16 cos 4x
16 cos 4x · cos2 x
lim+ 1 + 4 sin 4x = lim+
L’Hopital.
=
x→0 1 x→0 1 + 4 sin 4x
cos2 x
= 16

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Then
ln(1 + 4 sin 4x)
lim+ ln y = lim+ cot x ln(1 + 4 sin 4x) = lim+
x→0 x→0 x→0 tan x
16 cos 4x
16 cos 4x · cos2 x
lim+ 1 + 4 sin 4x = lim+
L’Hopital.
=
x→0 1 x→0 1 + 4 sin 4x
cos2 x
= 16

Hence, lim+ y = lim+ e ln y

x→0 x→0

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Then
ln(1 + 4 sin 4x)
lim+ ln y = lim+ cot x ln(1 + 4 sin 4x) = lim+
x→0 x→0 x→0 tan x
16 cos 4x
16 cos 4x · cos2 x
lim+ 1 + 4 sin 4x = lim+
L’Hopital.
=
x→0 1 x→0 1 + 4 sin 4x
cos2 x
= 16

lim ln y
Hence, lim+ y = lim+ e ln y = e x→0+
x→0 x→0

Duong T. PHAM 42 / 53
Indeterminate powers

Ex: Evaluate lim+ (1 + 4 sin 4x)cot x

x→0

Ans: Denote y = (1 + 4 sin 4x)cot x =⇒ ln y = cot x ln(1 + 4 sin 4x)

Then
ln(1 + 4 sin 4x)
lim+ ln y = lim+ cot x ln(1 + 4 sin 4x) = lim+
x→0 x→0 x→0 tan x
16 cos 4x
16 cos 4x · cos2 x
lim+ 1 + 4 sin 4x = lim+
L’Hopital.
=
x→0 1 x→0 1 + 4 sin 4x
cos2 x
= 16

lim ln y
Hence, lim+ y = lim+ e ln y = e x→0+ = e 16 .
x→0 x→0

Duong T. PHAM 42 / 53
Optimization problems

Duong T. PHAM 43 / 53
Optimization problems
Ex: A man wants to build a rectangular fish pond in his garden and he
wants to save money by using bricks left out from his house construction.
The amount of bricks is enough to build 50 m of the pond banks . One
side of the rectangular fish pond will be built from rocks which there
are in abundance around his house. Question: Determine the shape of
the fish pond which has the largest area and is built from the materials
the man has.

y
made of bricks

made of bricks

x x

y
made of rocks

y
made of bricks

made of bricks

x x

y
made of rocks

2x + y = 50
made of bricks

made of bricks

x x

y
made of rocks

2x + y = 50
made of bricks

made of bricks

x x Area = x · y

y
made of rocks

2x + y = 50
made of bricks

made of bricks

x x Area = x · y
max Area?
y
made of rocks

Duong T. PHAM 43 / 53
Optimization problems

Duong T. PHAM 44 / 53
Optimization problems

Find max (xy )?
Ex: Mathematical model:
2x + y = 50

Duong T. PHAM 44 / 53
Optimization problems

Find max (xy )?
Ex: Mathematical model:
2x + y = 50
Ans:

Duong T. PHAM 44 / 53
Optimization problems

Find max (xy )?
Ex: Mathematical model:
2x + y = 50
Ans: We have A = xy and since 2x + y = 50, we replace y = 50 − 2x in
A to obtain
A(x) = x(50 − 2x).

Duong T. PHAM 44 / 53
Optimization problems

Find max (xy )?
Ex: Mathematical model:
2x + y = 50
Ans: We have A = xy and since 2x + y = 50, we replace y = 50 − 2x in
A to obtain
A(x) = x(50 − 2x).
Our task is now to find: max A(x), where 0 ≤ x ≤ 25. We have
A0 = 50 − 4x, and
A0 = 0 ⇐⇒ 50 − 4x = 0 ⇐⇒ x = 12.5.

Duong T. PHAM 44 / 53
Optimization problems

Duong T. PHAM 45 / 53
Optimization problems

25m
12.5m

12.5m
Area = 312.5m2

25m

Duong T. PHAM 45 / 53
Applications in business and economics

Duong T. PHAM 46 / 53
Applications in business and economics
The cost function C (x) = the cost of producing x units of a certain
product.

Duong T. PHAM 46 / 53
Applications in business and economics
The cost function C (x) = the cost of producing x units of a certain
product.
The marginal cost (= C 0 (x)) is the change of C (x) w.r.t. x
The demand function (or price function), denoted by p(x), is the
price per unit that the company can charge if it sells x units

R(x) = xp(x)

The derivative R 0 is the marginal revenue function

R(x) = xp(x)

The derivative R 0 is the marginal revenue function

If x units are sold, the total profit

P(x) = R(x) − C (x)

and P is called the profit function.

R(x) = xp(x)

The derivative R 0 is the marginal revenue function

If x units are sold, the total profit

P(x) = R(x) − C (x)

and P is called the profit function.

P 0 is called the marginal profit function
Duong T. PHAM 46 / 53
Applications in business and economics

Duong T. PHAM 47 / 53
Applications in business and economics
Ex: A store sells 200 DVD burners a week at $350 each.

Duong T. PHAM 47 / 53
Applications in business and economics
Ex: A store sells 200 DVD burners a week at $350 each.
Survey: if each $10 rebate is offered, 20 more units will be sold
every week.

Duong T. PHAM 47 / 53
Applications in business and economics
Ex: A store sells 200 DVD burners a week at $350 each.
Survey: if each $10 rebate is offered, 20 more units will be sold
every week.
Q.: Find demand and revenue functions. How large a rebate should be
to maximize its revenue?
Ans: Denote by x the number of units sold every week =⇒ weekly increase
in sales is x − 200.
If the price per unit decreases by $10, more 20 units are sold.
=⇒ the price per unit so that the weekly sale is x units, is
x − 200 x
p(x) = 350 − · 10 = 450 − .
20 2

Duong T. PHAM 48 / 53
Applications in business and economics

x2
Our task: Find the absolute maximum of R(x) = 450x −
2

Duong T. PHAM 48 / 53
Applications in business and economics

x2
Our task: Find the absolute maximum of R(x) = 450x −
2
R 0 (x) = 450 − x,

Duong T. PHAM 48 / 53
Applications in business and economics

x2
Our task: Find the absolute maximum of R(x) = 450x −
2
R 0 (x) = 450 − x, and R 0 (x) = 0 ⇐⇒ x = 450.

Duong T. PHAM 48 / 53
Applications in business and economics

x2
Our task: Find the absolute maximum of R(x) = 450x −
2
R 0 (x) = 450 − x, and R 0 (x) = 0 ⇐⇒ x = 450.
consider the table

Duong T. PHAM 48 / 53
Applications in business and economics

x2
Our task: Find the absolute maximum of R(x) = 450x −
2
R 0 (x) = 450 − x, and R 0 (x) = 0 ⇐⇒ x = 450.
consider the table
x 350 450
R 0 (x) 100 + 0 −
R(x) 96250 % 101250 &

Duong T. PHAM 48 / 53
Applications in business and economics

x2
Our task: Find the absolute maximum of R(x) = 450x −
2
R 0 (x) = 450 − x, and R 0 (x) = 0 ⇐⇒ x = 450.
consider the table
x 350 450
R 0 (x) 100 + 0 −
R(x) 96250 % 101250 &
The revenue has an absolute maximum (101250) when the number of
weekly sold units is 450.

Duong T. PHAM 48 / 53
Applications in business and economics

Duong T. PHAM 48 / 53
Newton’s method

Duong T. PHAM 49 / 53
Newton’s method

Some examples:

Duong T. PHAM 49 / 53
Newton’s method

Some examples:
Solve 2x + 5 = 0

Duong T. PHAM 49 / 53
Newton’s method

Some examples:
Solve 2x + 5 = 0 ⇐⇒ x = − 25

Duong T. PHAM 49 / 53
Newton’s method

Some examples:
Solve 2x + 5 = 0 ⇐⇒ x = − 25

Solve x 2 − 5x + 4 = 0.

Duong T. PHAM 49 / 53
Newton’s method

Some examples:
Solve 2x + 5 = 0 ⇐⇒ x = − 25

Solve x 2 − 5x + 4 = 0.
∆ = (−5)2 − 4 · 1 · 4 = 9

Duong T. PHAM 49 / 53
Newton’s method

Some examples:
Solve 2x + 5 = 0 ⇐⇒ x = − 25

Solve x 2 − 5x + 4 = 0.
∆ = (−5)2 − 4 · 1 · 4 = 9
The solutions
√
5− ∆

 x1 = 2
=

 √
5+ ∆

x2 =
2

Duong T. PHAM 49 / 53
Newton’s method

Some examples:
Solve 2x + 5 = 0 ⇐⇒ x = − 25

Solve x 2 − 5x + 4 = 0.
∆ = (−5)2 − 4 · 1 · 4 = 9
The solutions
√ √
5− ∆ 5− 9
 
 x1 = 2
=  x1 = 2
⇐⇒
 
 √  √
5+ ∆ 5+ 9
 
x2 = x2 =
2 2

Duong T. PHAM 49 / 53
Newton’s method

Some examples:
Solve 2x + 5 = 0 ⇐⇒ x = − 25

Solve x 2 − 5x + 4 = 0.
∆ = (−5)2 − 4 · 1 · 4 = 9
The solutions
√ √
5− ∆ 5− 9
 
 x1 = =  x1 =

2 2 x1 = 1
⇐⇒ ⇐⇒ 
 
 √  √
5+ ∆ 5+ 9 x2 = 4
 
x2 = x2 =
2 2

Duong T. PHAM 49 / 53
Newton’s method

Some examples:
Solve 2x + 5 = 0 ⇐⇒ x = − 25

Solve cos x − x = 0?

Duong T. PHAM 49 / 53
Newton’s method

Duong T. PHAM 50 / 53
Newton’s method

r
x
y = f (x)

Duong T. PHAM 50 / 53
Newton’s method

f (x1 )

r
x1 x
y = f (x)

t1 : y = f 0 (x1 )(x − x1 ) + f (x1 )

Duong T. PHAM 50 / 53
Newton’s method

f (x1 )

r
x2 x1 x
y = f (x)

t1 : y = f 0 (x1 )(x − x1 ) + f (x1 )

Duong T. PHAM 50 / 53
Newton’s method

f (x1 )

f (x2 )

r
x2 x1 x
y = f (x)

t1 : y = f 0 (x1 )(x − x1 ) + f (x1 )

Duong T. PHAM 50 / 53
Newton’s method

f (x1 )

f (x2 )

r
x2 x1 x
y = f (x)

t2 t1

t1 : y = f 0 (x1 )(x − x1 ) + f (x1 )

t2 : y = f 0 (x2 )(x − x2 ) + f (x2 )

Duong T. PHAM 50 / 53
Newton’s method

f (x1 )

f (x2 )

r
x3 x2 x1 x
y = f (x)

t2 t1

t1 : y = f 0 (x1 )(x − x1 ) + f (x1 )

t2 : y = f 0 (x2 )(x − x2 ) + f (x2 )

Duong T. PHAM 50 / 53
Newton’s method

y Step 1: Choose some x1

f (x1 )

f (x2 )

r
x3 x2 x1 x
y = f (x)

t2 t1

t1 : y = f 0 (x1 )(x − x1 ) + f (x1 )

t2 : y = f 0 (x2 )(x − x2 ) + f (x2 )

Duong T. PHAM 50 / 53
Newton’s method

y Step 1: Choose some x1

x2 = x1 − ff0(x(x11))

f (x1 )

f (x2 )

r
x3 x2 x1 x
y = f (x)

t2 t1

t1 : y = f 0 (x1 )(x − x1 ) + f (x1 )

t2 : y = f 0 (x2 )(x − x2 ) + f (x2 )

Duong T. PHAM 50 / 53
Newton’s method

y Step 1: Choose some x1

x2 = x1 − ff0(x(x11))
f (x2 )
x3 = x2 − f 0 (x2 )
f (x1 )

f (x2 )

r
x3 x2 x1 x
y = f (x)

t2 t1

t1 : y = f 0 (x1 )(x − x1 ) + f (x1 )

t2 : y = f 0 (x2 )(x − x2 ) + f (x2 )

Duong T. PHAM 50 / 53
Newton’s method

y Step 1: Choose some x1

x2 = x1 − ff0(x(x11))
f (x2 )
x3 = x2 − f 0 (x2 )
f (x1 ) f (xn−1 )
Step 2: xn = xn−1 − f 0 (xn−1 )

f (x2 )

r
x3 x2 x1 x
y = f (x)

t2 t1