MA 115 Homework Solutions for Week 3
2.7
#4
(a) Since
g
(5) =
−
3, the point (5
,
−
3) is on the graph of
g
. Since
g
′
(5) = 4, the slope of
the tangent line at
x
= 5 is 4. Using the pointslope form of a line gives us
y
−
(
−
3) =
4(
x
−
5), or
y
= 4
x
−
23.
(b) Since (4
,
3) is on
y
=
f
(
x
),
f
(4) = 3. The slope of the tangent line between (0
,
2) and
(4
,
3) is
1
4
, so
f
′
(4) =
1
4
.
2.7
#26
v
(5) =
f
′
(5)
=
lim
h
→
0
f
(5 +
h
)
−
f
(5)
h
= lim
h
→
0
[(5 +
h
)
−
1
−
(5 +
h
)]
−
(5
−
1
−
5)
h
=
lim
h
→
0
1
5+
h
−
5
−
h
−
1
5
+ 5
h
= lim
h
→
0
1
5+
h
−
h
−
1
5
h
= lim
h
→
0
5
−
5
h
(5+
h
)
−
(5+
h
)
5(5+
h
)
h
=
lim
h
→
0
5
−
25
h
−
5
h
2
−
5
−
h
5
h
(5 +
h
)
= lim
h
→
0
−
5
h
2
−
26
h
5
h
(5 +
h
)
= lim
h
→
0
h
(
−
5
h
−
26)
5
h
(5 +
h
)
=
lim
h
→
0
−
5
h
−
26
5(5 +
h
)
=
−
26
25
m/s
The speed when
t
= 5 is
vextendsingle
vextendsingle
−
26
25
vextendsingle
vextendsingle
=
26
25
= 1
.
04 m/s.
2.7
#27
(a)
f
′
(
x
) is the rate of change of the production cost with respect to the number of ounces
of gold produced. Its units are dollars per ounce.
(b) After 800 ounces of gold have been produced, the rate at which the production cost
is increasing is $17 per ounce. So the cost of producing the 800th (or 801st) ounce is
about $17.
(c) In the short term, the values of
f
′
(
x
) will decrease because more efficient use is made
of startup costs as
x
increases. But eventually
f
′
(
x
) might increase due to largescale
operations.
2.8
#3
(a) II. Since from left to right, the slopes of the tangents to graph (a) start out negative,
become 0, then positive, then 0, then negative again. The actual function values in
graph II follow the same pattern.
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 Fall '08
 Mahalanobis
 Derivative, Slope, lim, $17, 1.04 m/s, 800 ounces, lim h0

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