MATH 137
Fall 2010
Solutions to Assignment #5, Part II
1. For the function
y
=
x

1
x
+ 1
,
dy
dx
=
(
x
+ 1)

(
x

1)
(
x
+ 1)
2
=
2
(
x
+ 1)
2
.
Since the slope of the line
x

2
y
= 2 is
1
2
, we want to find the equation of all tangent lines to
the curve for which
dy
dx
=
1
2
.
Thus we solve the equation
2
(
x
+ 1)
2
=
1
2
, hence (
x
+ 1)
2
= 4 or
x
+ 1 =
±
2, so
x
= 1 or
x
=

3.
At the point (1
,
0), the equation of the tangent line to the curve is
y
=
1
2
(
x

1).
At the point (

3
,
2), the equation of the tangent line to the curve is
y

2 =
1
2
(
x
+ 3).
2. We first find the points where the graphs meet.
That is, we solve
x
3

3
x
+ 4 = 3
x
2

3
x
, or
x
3

3
x
2
+ 4 = 0.
Let
f
(
x
) =
x
3

3
x
2
+ 4. Then
f
(

1) = 0, so
x
+ 1 is a factor of
f
(
x
). Using synthetic (or
long) division we find the complete factorization of
f
(
x
) to be (
x
+ 1)(
x

2)
2
.
Therefore, the curves meet where
x
=

1 and where
x
= 2, or more specifically, the curves
meet at the points (

1
,
6) and (2
,
6).
It is also required that the curves have a common tangent at these points.
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 Fall '08
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