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Math 171: Exam 2
A rational function can only change sign at zeros and vertical asymptotes, so we
can determine when the function is positive or negative by considering the inter
vals (
1
,

1), (

1
,
0), (0
,
1
/
2), (1
/
2
,
1), and (1
,
1
). By checking points in those
intervals we find that it is +,

,

,

, and +, respectively.
To find the
y
intercept we just evaluate the function at
x
= 0, at which point it’s
zero (as we have already seen).
For the horizontal asymptote,
lim
x
!1
(
x

1)
x
2
(2
x

1)
2
(
x
+ 1)
= lim
x
!1
x
3
+ smaller order terms
4
x
3
+ smaller order terms
= lim
x
!1
x
3
4
x
3
= lim
x
!1
1
4
=
1
4
.
So the graph has a horizontal asymptote at
y
= 1
/
4. Similarly, lim
x
!1
(
x

1)
x
2
(2
x

1)
2
(
x
+1)
=
1
4
. (We also know that rational function can have at most one horizontal asymptote.)
We now have everything we need to sketch a graph (see the last page).
5. Let
f
(
x
) =
1
2
x

1
.
(a) [6 points] Express
f
0
(
x
) as a limit.
Solution:
By the definition of the derivative:
f
0
(
x
) = lim
h
!
0
1
2(
x
+
h
)

1

1
2
x

1
h
.
(b) [6 points]
Using the limit definition of the derivative,
find
f
0
(
x
).
Solution:
f
0
(
x
) = lim
h
!
0
1
2(
x
+
h
)

1

1
2
x

1
h
= lim
h
!
0
1
2(
x
+
h
)

1

1
2
x

1
h
(2(
x
+
h
)

1)(2
x

1)
(2(
x
+
h
)

1)(2
x

1)
= lim
h
!
0
(2
x

1)

(2(
x
+
h
)

1)
h
(2(
x
+
h
)

1)(2
x

1)
= lim
h
!
0
2
x

1

2
x

2
h
+ 1
h
(2(
x
+
h
)

1)(2
x

1)
= lim
h
!
0

2
h
h
(2(
x
+
h
)

1)(2
x

1)
= lim
h
!
0

2
(2(
x
+
h
)

1)(2
x

1)
=

2
(2(
x
+ 0)

1)(2
x

1)
=

2
(2
x

1)
2
.
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 Fall '07
 GOMEZ,JONES
 Math, Calculus, Algebra, Trigonometry, Derivative, Product Rule, lim

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