Math 108
Name:
Spring 2016
Score:
/41
Show all your work
Dr. Lily Yen
No Calculator allowed in this part.
Problem 1: Use the rules of differentiation to find the derivative of each of these functions.
Perform any obvious simplifications—coefficients, exponents, etc.
d √
3
x4 − π ln(3x) .
a. Find
dx
√
√
3
d
4 − π ln(3x) = d x4/3 − π ln(3x) = 4 x1/3 − π 3 = 4 3 x − π
x
dx
dx
3
3x
3
x
Test 2
b. Differentiate f (x) = 7xπ −
√
3
Score:
/2
Score:
/2
Score:
/2
2
π + e−4x .
2
f 0 (x) = 7πxπ−1 − 0 + e−4x (−8x) = 7πxπ−1 − 8xe−4x
2
√
3
t3 (9t − 5 t2 )
c. Find the derivative of g(t) =
3t5
2/3
2/3
Note that g(t) = 9t−5t
= 3t9t2 − 5t3t2 = 3t−1 − 35 t−4/3 , so
3t2
−4 −7/3
5
0
−2
= −3t−2 + 20
t−7/3 .
g (t) = −3t − 3 × 3 t
9
Problem 2: Determine the following derivatives using differentiation rules. Do not simplify.
√
4
4
a. f (x) = 3x2 − 11x5 3−x + log7 (5x) − tan(x)
3
√
5
4 3−x + log7 (5x) − tan(x) −3−x ln(3) + 5x ln(7)
− sec2 (x) 4 3x2 − 11x5 + 3−x +
4
log7 (5x) − tan(x) 41 (3x2 − 11x5 )−3/4 (6x − 55x4 )
Score:
sin(x2 − 5) + cot(2x)
b. y =
sec2 (3x) − cos(x4 )
2x cos(x2 −5)−2 csc2 (2x)
sec2 (3x)−cos(x4 ) − sin(x2 −5)+cot(2x)
/4
6 sec2 (3x) tan(3x)+4x3 sin(x4 )
2
sec2 (3x)−cos(x4 )
Score:
/4
Problem 3: Using the given graph of g 00 (x), answer the following questions for g 0 and g.
y
1
−1
−1
1
2
3
4
5
6
7
8
9
x
g 00
−2
−3
a. List open intervals of decrease for g 0 .
g 0 is decreasing where g 00 (x) < 0, so (−1, −3) ∪ (9, 10).
b. List open intervals where g is concave up.
g is concave up where g 00 (x) > 0 so (3, 5) ∪ (5, 9).
c. List all x-coordinates of inflection points of g.
g has an inflection point where the sign of g 00 (x) changes, so at x = 3 and at x = 9.
d. List open intervals where g may have a relative maximum.
Where g has a relative maximum, g 0 is either zero or undefined. Since g 00 (x) is defined
for all x in (−1, 10), both g 0 (x) and g(x) are defined for all x in (−1, 10). Thus, where
g has a relative maximum, g 0 is zero and g 00 < 0. Hence g may have a relative
maximum in (−1, 3) ∪ (9, 10).
Score:
/6
Problem 4: Sketch the graph of a function satisfying all of the following conditions.
• lim + f (x) = −∞; • lim f (x) = 0; • f 0 (x) < 0 if x > 4; • f (0) = 5; • f 0 (x) > 0 if
x→∞
x→−1
−5 < x < −1 or −1 < x < 4; • lim − f (x) = ∞, and • f (−5) = 0.
x→−1
y
5
x
−5
−1
4
Score:
Page 2
/4
Math 108
Math 108
Name:
Spring 2016
Show all your work
Dr. Lily Yen
Calculators allowed in this part.
Problem 5: Suppose f (3) = 4, g(3) = 2, f 0 (3) = −6, and g 0 (3) = 5, f 0 (2) = 1/2,
g 0 (4) = −2/5, find the following.
a. Dx (f − 2g)x=3
Dx (f − 2g)x=3 = (f 0 − 2g 0 )x=3 = f 0 (3) − 2g 0 (3) = −6 − 2 × 5 = −16.
Test 2
d
(f (g(x))
b.
dx
x=3
d
0
0
(f
(g(x))
=
f
(g(x))g
(x)
= f 0 (g(3))g 0 (3) = f 0 (2) × 5 =
dx
x=3
x=3
1
2
×5=
5
2
Score:
/4
2
Problem 6: Let f : f (x) = 5e−0.2(x−6) describe the amount (in tons) of a particular kind of
algae growing in a region in the Strait of Vancouver as a function of temperature in degrees
Celsius.
a. Draw the graph and discuss the range of temperature for which this function is relevant.
Give reasons.
y/ton
I never saw the strait freeze over, and in a Vancouver summer the water temperature is unlikely to reach 20 ◦C. Thus 0 ≤ x ≤ 20 seems
reasonable.
4
3
2
1
x/◦C
2
4
6
8
10
12
14
16
18
Score:
/1
b. Find and interpret f (5). Give 4-decimal place accuracy.
If the water temperature is 5 ◦C, the water contains f (5) = 4.0937 ton of the algæ.
Score:
/1
c. Find and interpret f 0 (5). Give 4-decimal place accuracy.
2
2
Since f 0 (x) = 5e−0.2(x−6) (−0.2)2(x − 6) = −2(x − 6)e−0.2(x−6) , if the water
temperature were to rise a small amount, ∆x, from 5 ◦C, then the amount of algæ
would rise approximately f 0 (5) × ∆x = 1.6375 ton/◦C × ∆x.
Score:
Page 3
/2
Math 108
Problem 7: Suppose the cost in dollars to make x oboe reeds is given by
C(x) = 5 log2 (x) + 10
a. Find the marginal cost when 31 reeds are made. Interpret your answer.
5
C 0 (x) = x ln(2)
, so C 0 (31) = $0.23 /reed. Thus the cost of making one more reed is
approximately $0.23.
Score:
/2
b. Find the marginal average cost when 16 reeds are sold. Interpret your answer.
C(x)
, and the marginal average cost
x
5
x−(5
log2 (x)+10)
5−5 ln(x)−10 ln(2)
x ln(2)
The average cost is C̄(x) =
C 0 (x)x−C(x)
is the derivative that,
so C̄ 0 (x) =
=
=
. Thus
x2
x2
x2 ln(2)
C̄ 0 (16) = $−0.09 /reed/reed.
Hence, if you were to make one more reed, the average cost would drop about
$0.09 /reed.
Score:
/2
Problem 8: At Mauna Loa, Hawaii, atmospheric carbon dioxide levels in parts per million
(ppm) have been measured regularly since 1958 (Source: Greenhouse Earth). The function
defined by
L(t) = 0.022t2 + 0.55t + 316 + 3.5 sin(2πt)
can be used to model these levels, where t is in years and t = 0 corresponds to 1960.
a. Graph L(t) for t in [0, 30].
L
350
340
330
320
t
310
5
10
15
20
25
30
Score:
/2
Score:
/1
b. Find the carbon dioxide level in 2016.
L(2016 − 1960) = L(56) = 415.8 ppm
c. Find L0 (60.5) and interpret what your answer means.
L0 (t) = 0.044t + 0.55 + 7π cos(2πt), so L0 (60.5) = −18.78 ppm/year. Thus, in the
summer of 2020, the carbon dioxide levels are expected to fall at a rate of
18.78 ppm/year.
Score:
Page 4
/2
Math 108