MAT1613
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, 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
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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]
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JTS Maths Tutoring
Keeping you mathematically informed
