MAT 137Y 200809 Winter Session, Solutions to Supplementary Problem Set #2
1
(a) By IVT, it is clear that the equation
f
(
x
) =
x
3

x
+
1 has a root in the interval
(

2
,

1
)
, since
f
(

2
) =

5
<
0 and
f
(

1
) =
1
>
0. We now show
f
(
x
) =
0 has only one unique solution.
Differentiating
f
, we have
f
0
(
x
) =
3
x
2

1. The function is increasing on
(

∞
,

1
√
3
)
, which is
where one of the roots is located. There can’t be another root at that interval since the function is
strictly increasing. The function is decreasing on
(

1
√
3
,
1
√
3
)
, but
f
(
1
√
3
) =
9

2
√
3
9
>
0, so there
can’t be any roots at that interval. Finally,
f
is increasing for
x
>
1
√
3
, so there obviously can’t
be any roots there.
f
(
x
)
has only one root.
(b) We use Newton’s Method with the initial guess of
x
0
=

3
2
. Then,
x
n
+
1
=
x
n

f
(
x
n
)
f
0
(
x
n
)
=
⇒
x
n
+
1
=
x
n

x
2
n

x
n
+
1
3
x
2
n

1
.
Applying a few iterations, we get
x
1
=
x
0

x
2
0

x
0
+
1
3
x
2
0

1
=
⇒
x
1
=

31
23
≈ 
1
.
347826
x
2
=
x
1

x
2
1

x
1
+
1
3
x
2
1

1
=
⇒
x
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 Spring '08
 UPPAL
 Calculus, lim, Order theory, Monotonic function

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