Math 108
Name:
Spring 2016
Score:
/40
Show all your work
Dr. Lily Yen
No Calculator allowed in this part.
Problem 1: Determine the following limits analytically showing all steps. Use the symbols
dne, ∞, and −∞ where appropriate.
Test 1
2x + 6
=
x→−3 9 − x2
limx→−3 2x+6
= limx→−3
9−x2
1/3
a. lim
2(x+3)
(3−x)(3+x)
= limx→−3
2
3−x
=
2
6
=
1
3
Score:
4x3 − 3x + x2
=
x→∞ 6x2 − 7x3 + 1
−4/7
b. lim
limx→∞
4x3 −3x+ x2
6x2 −7x3 +1
/2
= limx→∞
3
+ 24 )
x2
x
6
3
x ( x −7+ 13 )
x
x3 (4−
= limx→∞
3
+ 24
x2
x
6
−7+ 13
x
x
4−
=
4
−7
Score:
∞
(a2 − 2)(7 − 3a2 )
=
a→−∞
3 − a2
c. lim
lima→−∞
∞
(a2 −2)(7−3a2 )
3−a2
/2
= lima→−∞
2
)(7−3a2 )
a2
3
2
a ( 2 −1)
a
a2 (1−
= lima→−∞
(1−
2
)(7−3a2 )
a2
3
−1
a2
=
(1)(−∞)
−1
=
Score:
15 + 3x
=
x→−5 |x + 5|
= lim−5−
Since limx→−5− 15+3x
|x+5|
dne
d. lim
limx→−5+
exist.
15+3x
|x+5|
= lim−5+
3(5+x)
x+5
/2
3(5+x)
−(x+5)
= limx→−5− −3 = −3 and
= limx→−5− 3 = 3 are different, limx→−5
15+3x
|x+5|
does not
Score:
/2
Problem
2: Use the limit definition of the derivative to find the derivative of f (x) =
√
1 + 3x − 2. Use correct notation, show all steps and leave your answer in simplified form.
√
√
√
√
1+ 3(x+h)−2−(1+ 3x−2)
3(x+h)−2− 3x−2
f (x+h)−f (x)
0
f (x) = limh→0
= limh→0
= limh→0
=
h
h
h
√
√
√
2 √
√
√
2
3(x+h)−2− 3x−2
3(x+h)−2+ 3x−2
3(x+h)−2 − 3x−2
limh→0
×√
= limh→0 √
=
√
√
h
3(x+h)−2+ 3x−2
h( 3(x+h)−2+ 3x−2)
3(x+h)−2−(3x−2)
3x+3h−3x
3h
limh→0 √
= limh→0 √
= limh→0 √
√
√
√
h( 3(x+h)−2+ 3x−2)
h( 3(x+h)−2+ 3x−2)
h( 3(x+h)−2+ 3x−2)
3
3
limh→0 √
= 2√3x−2
√
3(x+h)−2+ 3x−2
=
Score:
/4
Math 108
Name:
Spring 2016
Show all your work
Dr. Lily Yen
Calculators allowed from here on.
Problem 3: The graph of y = f (x) is shown. Use the graph to answer the questions. Use
the symbols dne, ∞, and −∞ where appropriate.
Test 1
y
4
2
f0
x
−8
−6
−4
−2
2
4
−2
Parabolic arch
−4
−6
a. List the intervals for which f is continuous in (−∞, 0). Express in as few
intervals as possible.
(−∞, −4) ∪ (−4, −1) ∪ (−1, 0)
b. f (0) =
5
f.
0
lim f (x) =
x→−1−
−1
f (x) − f (e)
=
x→e
x−e
g. lim
dne
h. lim f (x) =
x→−1
c. lim− f (x) =
x→3
d. lim+ f (x) =
x→3
e. lim− f 0 (x) =
x→3
2
−1
−∞
i. Use the tangent line to estimate the
value of f 0 (−6).
0
j. In the same grid above, graph y = f 0 (x)
for the interval (−∞, ∞).
Score:
/12
Problem 4: State the definition of continuity at a point for a function f . Determine the
value of k so that f (x) is continuous on any interval.
( 2
3x +2x−8
, x 6= −2
x+2
f (x) =
3x + k,
x = −2
A function f is continuous at a point a if limx→a f (x) = f (a).
The function is obviously continuous (since it is a rational function) everywhere except
possibly at x = −2. Since
2 +2x−8
= limx→−2 (x+2)(3x−4)
= limx→−2 3x − 4 = −10 and
limx→−2 f (x) = limx→−2 3x x+2
x+2
f (−2) = −6 + k, continuity at x = −2 requires that −6 + k = −10, so k = −4.
Score:
Page 2
/4
Math 108
Problem 5: The value of one Canadian dollar in US currency from January 1st, 2015 to
January 1st, 2016 is given in the following table.
Jan. 1, 2015 Feb. 1
$0.8615
$0.7954
Mar. 1
$0.7995
Apr. 1
$0.7926
Aug. 1
$0.7639
May 1
$0.8224
Sep. 1
$0.7549
June 1
$0.7983
Oct. 1
$0.7551
July 1
$0.7942
Nov. 1
$0.7653
Dec. 1 Jan. 1, 2016
$0.7485
$0.7208
a. Examine and draw a scatter plot and determine the best linear model to fit the data.
State your model correct to 3 decimal places. Be sure to define your variables and
include units.
price in US$
0.85
0.83
0.8
0.78
0.75
0.73
2
4
6
8
10
12
If time is measured in months since December 1st, 2014 (so that time is 1 on
January 1st, 2015), then the price is
y{1,...,13} = −0.008227x + 0.8401.
If measured since January 1st, 2015, then
y{0,...,12} = −0.008227x + 0.8319.
time
Score:
/2
b. Use your model to predict the price of a Canadian dollar in the beginning of March,
2016, and comment on the accuracy of this prediction. Give 4 decimal place accuracy.
If x = 15, then y{1,...,13} = $0.7167.
Since the actual price of the dollar in mid-January is actually below $0.70, the model
may not be aggressive enough.
Score:
/1
c. According to your model, what was the value of one Canadian dollar in US currency
on January 1st, 2015? Find the error of the estimate. Give 2 decimal place accuracy.
If x = 1, then y{1,...,13} = 0.8319, which is 0.0296 less than the value in the table. This
looks like rather a lot in the figure but is actually only 3.44% off.
Score:
/2
d. According to your model, when will the value of one Canadian dollar fall below 60 cent
US? Comment on the accuracy of this prediction.
If y{1,...,13} = 0.60, then x = 29.18, so early May 2017.
Score:
/1
e. According to the scatter plot, can you propose another model more suitable than the
linear model you used? Give reasons.
price in US$
0.85
0.83
0.8
0.78
0.75
0.73
Page 3
2
4
6
8
time
10
12
The scatter plot could suggest a slight
downward bend, so maybe a quadratic
model. However, the best quadratic
/1
model, y1,...,13 = −0.000007892x2 Score:
−
0.008117x+0.8398, graphed here is hardly
any better than the linear model.
Math 108
Problem 6: The table below gives the diameter (in cm) of a melting snowball at various
times one afternoon in January in Whistler.
Time
12:00
Minutes
0
Size
12.0
12:30 1:00 1:45
30
60
105
11.2 10.4 9.0
2:10
130
8.1
2:40
160
6.9
3:00
180
6.0
3:15
195
5.2
3:30
210
4.2
3:45
225
3.0
3:50
230
2.4
3:55
235
1.7
a. Find the average rate of change of the diameter between 1:00 pm and 3:00 pm.
6.00 cm−10.4 cm
120 min
= −0.0367 cm/min
Score:
/1
b. Plot the data points and draw a curve through them.
d/cm
12
10
8
6
4
2
Time/min
30
60
90
120
150
180
210
240
Score:
/2
c. Draw a tangent line to your curve at 3:30 pm and use it to estimate the instantaneous
rate of change of the diameter at 3:30 pm.
The instantaneous rate of change (he slope of the red line above) is (around)
−0.0707 cm/min.
Score:
Page 4
/2
Math 108