MAT1613
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1. Educational Aid: These study notes are designed to serve as educational aids and should not be considered as a
substitute for individual research, critical thinking, or professional guidance. Students are encouraged to
conduct their own extensive research and consult with their instructors or academic advisors for specific
assignment requirements.
2. Personal Responsibility: While every effort has been made to ensure the accuracy and reliability of the
information provided in these study notes, the seller cannot guarantee the completeness or correctness of all
the content. It is the responsibility of the buyer to verify the accuracy of the information and use their own
judgment when applying it to their assignments.
3. Academic Integrity: It is crucial for students to uphold academic integrity and adhere to their institution's
policies and guidelines regarding plagiarism, citation, and referencing. These study notes should be used as a
tool for learning and inspiration, but any direct reproduction of the content without proper acknowledgment and
citation may constitute academic misconduct.
4. Limited Liability: The seller of these study notes shall not be held liable for any direct or indirect damages,
losses, or consequences arising from the use of the notes. This includes, but is not limited to, poor grades,
academic penalties, or any other negative outcomes resulting from the application or misuse of the information
provided.
, MAT1613
Oct/Nov 2021
Memorandum
QUESTION 1
If the derivative f ′ (x) of a function f is given by
x2
f ′ (x) = 1 −
(3x − 2)2
(a) Write f ′ (x) in quotient form and use the sign pattern to determine the interval(s) (6)
over which f increases and over which it decreases.
(b) Determine f ′′ (x) and use the sign pattern of f ′′ (x) to determine:
(i) the intervals where the graph of f is concave up and where it is (7)
concave down
(3)
(ii) the x − coordinate(s) of the local extreme point(s).
[16]
Solution
(a) ′ (x)
x2 (6)
f =1−
(3x − 2)2
(3x − 2)2 − x 2
f ′ (x) =
(3x − 2)2
′ (x)
9x 2 − 12x + 4 − x 2
f =
(3x − 2)2
8x 2 − 12x + 4
f ′ (x) =
(3x − 2)2
′ (x)
4(2x 2 − 3x + 1)
f =
(3x − 2)2
4[(2x − 1)(x − 1)]
f ′ (x) =
(3x − 2)2
(b) 1 2 (5)
Critical points for f ′ (x) are: , and 1. These critical points split the number
2 3
line into four distinct intervals:
1 1 2 2
x< , <x< , < x < 1, x>1
2 2 3 3
Note:
Critical points (stationary points) are points on a graph where f ′ (x) = 0
or f ′ (x) does not exist.
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, -2- MAT1613
Oct/Nov 2021
Memorandum
Sign pattern:
undefined
1 1 2 2 x>1
x< <x< <x<1
2 2 3 3
(2x − 1) − + + +
(x − 1) − − − +
f ′ (x) + − − +
1 2
1
2 3
Note:
2
f ′ (x) is undefined at and (3x − 2)2 is always positive, i. e. , it
3
doesn′ t affect the sign pattern, hence we do not include it on the sign
table.
f rises where they are positive signs on the sign table.
f falls where they are negative signs on the sign table.
1 1 2 2
f rises over (−∞, ) ∪ (1, ∞) and f falls over ( , ) ∪ ( , 1)
2 2 3 3
(b) 2x(3x − 2)2 − 6x 2 (3x − 2) (7)
f ′′ (x) = 0 − [ ]
(3x − 2)4
′′ (x)
(3x − 2)[2x(3x − 2) − 6x 2 ]
f = −[ ]
(3x − 2)4
6x 2 − 4x − 6x 2
f ′′ (x) = − [ ]
(3x − 2)3
−4x
f ′′ (x) = − [ ]
(3x − 2)3
4x
f ′′ (x) =
(3x − 2)3
© 2021
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, -3- MAT1613
Oct/Nov 2021
Memorandum
(i) 2
Critical points for f ′′ (x) are: 0 and
. These critical points split the
3
number line into three distinct intervals:
2 2
x < 0, 0<x< , x>
3 3
Sign pattern:
undefined
x<0 2 2
0<x< x>
3 3
4x − + +
(3x − 2) − − +
f ′ ′(x) + − +
2
0
3
Note:
2
f ′ (x) is undefined at and (3x − 2)3 is not always positive,
3
i. e. , it affects the sign pattern, hence it must be investigated.
f is concave up where they are positive signs on the sign
table.
f is concave down where they are negative signs on the sign
table.
2
f is concave up on (−∞, 0) ∪ ( , ∞) and f is concave down
3
2
down on (0 < x < ).
3
Local extreme point(s) occur whenf ′ (x) = 0
(ii)
4[(2x − 1)(x − 1)] 1
2
= 0 ⇒ 4[(2x − 1)(x − 1)] = 0 ⇒ x = or x = 1
(3x − 2) 2
1
∴ the x − coordinate(s) of the local extreme poit(s) are x = and x = 1.
2
[16]
© 2021
JTS Maths Tutoring
Keeping you mathematically informed
EXAM PACK
Recent exam questions and answers
Summarised study notes
Exam tips and guidelines
DISCLAIMER & TERMS OF USE
1. Educational Aid: These study notes are designed to serve as educational aids and should not be considered as a
substitute for individual research, critical thinking, or professional guidance. Students are encouraged to
conduct their own extensive research and consult with their instructors or academic advisors for specific
assignment requirements.
2. Personal Responsibility: While every effort has been made to ensure the accuracy and reliability of the
information provided in these study notes, the seller cannot guarantee the completeness or correctness of all
the content. It is the responsibility of the buyer to verify the accuracy of the information and use their own
judgment when applying it to their assignments.
3. Academic Integrity: It is crucial for students to uphold academic integrity and adhere to their institution's
policies and guidelines regarding plagiarism, citation, and referencing. These study notes should be used as a
tool for learning and inspiration, but any direct reproduction of the content without proper acknowledgment and
citation may constitute academic misconduct.
4. Limited Liability: The seller of these study notes shall not be held liable for any direct or indirect damages,
losses, or consequences arising from the use of the notes. This includes, but is not limited to, poor grades,
academic penalties, or any other negative outcomes resulting from the application or misuse of the information
provided.
, MAT1613
Oct/Nov 2021
Memorandum
QUESTION 1
If the derivative f ′ (x) of a function f is given by
x2
f ′ (x) = 1 −
(3x − 2)2
(a) Write f ′ (x) in quotient form and use the sign pattern to determine the interval(s) (6)
over which f increases and over which it decreases.
(b) Determine f ′′ (x) and use the sign pattern of f ′′ (x) to determine:
(i) the intervals where the graph of f is concave up and where it is (7)
concave down
(3)
(ii) the x − coordinate(s) of the local extreme point(s).
[16]
Solution
(a) ′ (x)
x2 (6)
f =1−
(3x − 2)2
(3x − 2)2 − x 2
f ′ (x) =
(3x − 2)2
′ (x)
9x 2 − 12x + 4 − x 2
f =
(3x − 2)2
8x 2 − 12x + 4
f ′ (x) =
(3x − 2)2
′ (x)
4(2x 2 − 3x + 1)
f =
(3x − 2)2
4[(2x − 1)(x − 1)]
f ′ (x) =
(3x − 2)2
(b) 1 2 (5)
Critical points for f ′ (x) are: , and 1. These critical points split the number
2 3
line into four distinct intervals:
1 1 2 2
x< , <x< , < x < 1, x>1
2 2 3 3
Note:
Critical points (stationary points) are points on a graph where f ′ (x) = 0
or f ′ (x) does not exist.
© 2021
JTS Maths Tutoring
Keeping you mathematically informed
, -2- MAT1613
Oct/Nov 2021
Memorandum
Sign pattern:
undefined
1 1 2 2 x>1
x< <x< <x<1
2 2 3 3
(2x − 1) − + + +
(x − 1) − − − +
f ′ (x) + − − +
1 2
1
2 3
Note:
2
f ′ (x) is undefined at and (3x − 2)2 is always positive, i. e. , it
3
doesn′ t affect the sign pattern, hence we do not include it on the sign
table.
f rises where they are positive signs on the sign table.
f falls where they are negative signs on the sign table.
1 1 2 2
f rises over (−∞, ) ∪ (1, ∞) and f falls over ( , ) ∪ ( , 1)
2 2 3 3
(b) 2x(3x − 2)2 − 6x 2 (3x − 2) (7)
f ′′ (x) = 0 − [ ]
(3x − 2)4
′′ (x)
(3x − 2)[2x(3x − 2) − 6x 2 ]
f = −[ ]
(3x − 2)4
6x 2 − 4x − 6x 2
f ′′ (x) = − [ ]
(3x − 2)3
−4x
f ′′ (x) = − [ ]
(3x − 2)3
4x
f ′′ (x) =
(3x − 2)3
© 2021
JTS Maths Tutoring
Keeping you mathematically informed
, -3- MAT1613
Oct/Nov 2021
Memorandum
(i) 2
Critical points for f ′′ (x) are: 0 and
. These critical points split the
3
number line into three distinct intervals:
2 2
x < 0, 0<x< , x>
3 3
Sign pattern:
undefined
x<0 2 2
0<x< x>
3 3
4x − + +
(3x − 2) − − +
f ′ ′(x) + − +
2
0
3
Note:
2
f ′ (x) is undefined at and (3x − 2)3 is not always positive,
3
i. e. , it affects the sign pattern, hence it must be investigated.
f is concave up where they are positive signs on the sign
table.
f is concave down where they are negative signs on the sign
table.
2
f is concave up on (−∞, 0) ∪ ( , ∞) and f is concave down
3
2
down on (0 < x < ).
3
Local extreme point(s) occur whenf ′ (x) = 0
(ii)
4[(2x − 1)(x − 1)] 1
2
= 0 ⇒ 4[(2x − 1)(x − 1)] = 0 ⇒ x = or x = 1
(3x − 2) 2
1
∴ the x − coordinate(s) of the local extreme poit(s) are x = and x = 1.
2
[16]
© 2021
JTS Maths Tutoring
Keeping you mathematically informed
