MATH1211/1011(2)/Asol0/NKT
Department of Mathematics, The University of Hong Kong
MATH1211 (2010–11, 2nd semester) Suggested solution to Assignment 0
Note
: Some solutions given below are outline only. You may need to give more details in your solution. The
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are remarks for you to read; they are not a part of the solution.
1.
(Easy problem!)
(a)
f
0
(
x
) = 6
x
2

3
2
√
x

4
x

5
.
(b)
g
0
(
u
) =
(sin
u
+
u
cos
u
)(5 + cos
u
) +
u
sin
2
u
(5 + cos
u
)
2
=
u
+ 5 sin
u
+ 5
u
cos
u
+ sin
u
cos
u
(5 + cos
u
)
2
.
(c)
h
0
(
t
) = 2
e
2
t

2
t
t
2
+ 1
.
(d)
k
0
(
s
) =
2
s
1 +
s
4
.
2.
f
0
(
x
) = 6
x
2
+ 6
x

12 = 6(
x
+ 2)(
x

1). Critical points are
x
=

2
,
1.
At critical points,
f
(

2) = 21
, f
(1) =

6.
At end points of domain [

3
,
3],
f
(

3) = 10
, f
(3) = 46.
f
00
(
x
) = 12
x
+ 6 = 12
(
x
+
1
2
)
=
⇒
f
00
(
x
)
<
0
when
x <

1
2
,
= 0
when
x
=

1
2
,
>
0
when
x >

1
2
.
(a)
f
0
(

2) = 0 and
f
00
(

2)
<
0 =
⇒
f
attains relative maximum at
x
=

2 with value
f
(

2) = 21.
f
0
(1) = 0 and
f
00
(1)
>
0 =
⇒
f
attains relative minimum at
x
= 1 with value
f
(1) =

6.
At left end point
x
=

3,
f
0
(

3)
>
0 =
⇒
f
attains relative minimum with value
f
(

3) = 10.
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 Spring '11
 wang
 Math, Calculus, lim, Mathematical analysis, Convex function, Concave function

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