Solutions to Test 3, Version A
Math 1501, Fall 02.
WG
*
November 8, 2002
Exercise 1.
(a) We have that
f
0
(
x
)=

2

2
x
if

2
<x<
0

1i
f
0
2
f
2
3
(
x

2)
2
if
3
4
.
Since
f
0

(0) =

2and
f
0
+
(0) =

1 we deduce that
f
0
(0) does not exist.
Similarly,
f
0

(2) =

1and
f
0
+
(2) = 1 and so
f
0
(2) does not exist. We have that
f
0

(3) = 1 =
f
0
+
(3)
and so,
f
0
(3) = 1
.
The only point where
f
0
vanishes is

1
.
Therefore, the critical points of
f
are

1
,
0
,
2
.
(b) and (c). Since
f
0
is positive on (

2
,

1), negative on (

1
,
0) we conclude that
f
has a local maximum at

1
.
Also,
f
is negative on (0
,
2) and positive on (2
,
3)
.
This
proves that
f
has a local minimum at 2
.
Exercise 2.
(a) We have that
f
0
(
x

4sin
x
cos
x

2
x
and so,
f
”(
x

4[cos
2
x

sin
2
x
]

2=

6+8sin
2
x
=8(s
in
x

√
3
2
)(sin
x
+
√
3
2
)
.
Note that sin
x
+
√
3
2
≥
√
3
2
on [0
,π
] and so, the sign of
f
” depends only on the sign of
sin
x

√
3
2
.
(b)
f
”(
x
) = 0 if and only if sin
x
=
√
3
2
.
This means
x
=
π
3
or
x
=
2
π
3
.
Also
f
”(
x
)
<
0
if and only if sin
x<
√
3
2
,
which means that
x
∈
(0
,
π
3
)
∪
(
2
π
3
)
.
Hence
f
” changes sign at
*
School of Mathematics, Georgia Institute of technology.
1
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=
π
3
and
x
=
2
π
3
.
These two points are points of inﬂection.
Exercise 3.
We have that
f
0
(
x
)=6
/
5(
x

2
5

x
1
5
)andso
,
lim
x
→
0
f
0
(
x
/
5 lim
x
→
0
x

2
5
=+
∞
.
This proves that the graph of
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 Fall '08
 N/A
 Math, Calculus, Sin, Mathematical analysis, Version A Math

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