(b) [2 points] How fast was the ball thrown?
Solution:
h
0
(
t
) =

7
2
t
+ 14, so
h
(0) = 14. Therefore, the ball was thrown
upward at 14ft/sec.
Page 4
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View Full DocumentMath 171: Final Exam
(c) [2 points] What is the acceleration due to gravity on the alien planet?
Solution:
h
00
(
t
) =

7
/
2, so the acceleration due to gravity is
7
2
ft/sec
2
(in the
downward direction, of course).
9. Let
f
(
x
) =
√
3
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
p
3(
x
+
h
) + 1

√
3
x
+ 1
h
.
(b) [6 points]
Using the limit deﬁnition of the derivative,
ﬁnd
f
0
(
x
).
Solution:
f
0
(
x
) = lim
h
→
0
p
3(
x
+
h
) + 1

√
3
x
+ 1
h
= lim
h
→
0
p
3(
x
+
h
) + 1

√
3
x
+ 1
h
±
p
3(
x
+
h
) + 1 +
√
3
x
+ 1
²
±
p
3(
x
+
h
) + 1 +
√
3
x
+ 1
²
= lim
h
→
0
(3(
x
+
h
) + 1)

(3
x
+ 1)
h
±
p
3(
x
+
h
) + 1 +
√
3
x
+ 1
²
= lim
h
→
0
3
x
+ 3
h
+ 1

3
x

1
h
±
p
3(
x
+
h
) + 1 +
√
3
x
+ 1
²
= lim
h
→
0
3
h
h
±
p
3(
x
+
h
) + 1 +
√
3
x
+ 1
²
= lim
h
→
0
3
p
3(
x
+
h
) + 1 +
√
3
x
+ 1
=
3
p
3(
x
+ 0) + 1 +
√
3
x
+ 1
=
3
2
√
3
x
+ 1
.
10. [12 points] Sketch a graph of
y
=
(2
x

1)(
x

1)
(
x

2)(
x
+ 1)
. Find and label the horizontal and
vertical asymptotes and all intercepts.
Solution:
The numerator is zero when
x
= 1
/
2
,
1, while the denominator is zero
when
x
=

1
,
2. This tells us that the
x
intercepts (zeros) are
x
= 1
/
2
,
1 and the
vertical asymptotes are at
x
=

1
,
2.
Page 5
Math 171: Final Exam
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
/
2), (1
/
2
,
1), (1
,
2), and (2
,
∞
). 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

1
/
2.
For the horizontal asymptote,
lim
x
→∞
(2
x

1)(
x

1)
(
x

2)(
x
+ 1)
= lim
x
→∞
2
x
2
+ smaller order terms
x
2
+ smaller order terms
= lim
x
→∞
2
x
2
x
2
= lim
x
→∞
2 = 2
.
So the graph has a horizontal asymptote at
y
= 2. Similarly, lim
x
→∞
(2
x

1)(
x

1)
(
x

2)(
x
+1)
= 2.
(We also know that rational function can have at most one horizontal asymptote.)
We now have everything we need to sketch a graph:
Page 6
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 Fall '07
 GOMEZ,JONES
 Math, Calculus, Algebra, Trigonometry, Sin, Quadratic equation, Mathematics in medieval Islam

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