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MATH 1001 R03 FINAL SOLUTION
FALL 2014 - MOON
(1) (a) (3 pts) Consider the line ` in the xy-plane, whose equation is y = 3x − 5.
Find the equation of the line m which is parallel to ` and passes through
(1, 2).
Because m is parallel to `, its slope is same with that of `, so it is 3. Also
from the assumption, it passes through (1, 2). Therefore the equation of the
line m is:
y − 2 = 3(x − 1) ⇒ y − 2 = 3x − 3 ⇒ y = 3x − 1.
• Finding the slope correctly: 1 pt.
• Getting the correct answer y = 3x − 1: 3 pts.
(b) (3 pts) Find the equation of the circle C whose center is (1, 2) and radius is
3.
(x − 1)2 + (y − 2)2 = 32 ⇒ (x − 1)2 + (y − 2)2 = 9
(c) (4 pts) Find the first coordinates of two intersection points of C and x-axis.
The x-axis has the equation y = 0. So we need to solve (x−1)2 +(y −2)2 = 9
and y = 0.
(x − 1)2 + (0 − 2)2 = 9 ⇒ (x − 1)2 + 4 = 9 ⇒ (x − 1)2 = 5
√
√
⇒x−1=± 5⇒x=1± 5
• Substituting y = 0 to the equation of C and getting (x−1)2 +(0−2)2 =
9: 2 pts.
√
• Sovling the equation and obtaining the answer 1 ± 5: 4 pts.
Date: December 16, 2014.
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MATH 1001 Final
Fall 2014 - Moon
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(2) (5 pts) Suppose that f (x) = 8x5 , g(x) = 2x 3 + 7. Find the formula of g ◦ f .
1
1
1
(g ◦ f )(x) = g(f (x)) = g(8x5 ) = 2(8x5 ) 3 + 7 = 2 · 8 3 (x5 ) 3 + 7
5
5
= 2 · 2x 3 + 7 = 4x 3 + 7
1
• Getting 2(8x5 ) 3 + 7: 3 pts.
5
• By using exponent laws, obtaining 4x 3 + 7: 5 pts.
(3) The following figure is the graph of f .
(a) (2 pts) Find the domain of f . Write the answer as an interval.
[−3, 2]
(b) (2 pts) Find the range of f . Write the answer as an interval.
[−2, 3]
(c) (2 pts) Find the interval on which f is decreasing.
[−1, 1]
(d) (2 pts) How many values of x satisfy the equation f (x) = 1?
3
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MATH 1001 Final
Fall 2014 - Moon
(4) (a) (6 pts) The function f such that f (x) is the temperature on the Fahrenheit
scale (◦ F) corresponding to temperature x on the Celsius scale (◦ C) is known
to be a linear function. The freezing temperature of water equals 0◦ C and
32◦ F. The boiling point of water equals 100◦ C and 212◦ F. Find f (x).
Because f is linear, f (x) = mx + b for two numbers m and b.
f (0) = 32 ⇒ m · 0 + b = 32 ⇒ b = 32, f (x) = mx + 32.
f (100) = 212 ⇒ m · 100 + 32 = 212 ⇒ 100m = 212 − 32 = 180
180
⇒m=
= 1.8 ⇒ f (x) = 1.8x + 32.
100
• Setting f (x) = mx + b: 2 pts.
• Finding b = 32: 4 pts.
• Getting f (x) = 1.8x + 32: 6 pts.
(b) (4 pts) The function g such that g(x) is the temperature on the Celsius scale
corresponding to temperature x on the Kelvin scale (◦ K) is given by
g(x) = x − 273.15.
Find the formula of a function h such that h(x) is the temperature on the
Fahrenheit scale corresponding to temperature x on the Kelvin scale.
h(x) = (f ◦ g)(x) = f (g(x)) = f (x − 273.15) = 1.8(x − 273.15) + 32
= 1.8x − 459.67
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MATH 1001 Final
Fall 2014 - Moon
(5) Suppose that g(x) = −x2 + 4x + 5.
(a) (4 pts) Find the vertex of g.
2 2 !
4
4
−x2 + 4x + 5 = − (x2 − 4x) + 5 = −
x−
− −
+5
2
2
= − (x − 2)2 + 4 + 5 = −(x − 2)2 + 9
Vertex: (2, 9)
2 2 !
4
4
− −
+
• Applying completing square and getting −
x−
2
2
5: 2 pts.
• Making given quadratic function into a standard form −(x − 2)2 + 9:
3 pts.
• Getting the vertext (2, 9): 4 pts.
(b) (3 pts) Sketch the graph of g.
(c) (2 pts) Find the maximum of g.
9
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MATH 1001 Final
Fall 2014 - Moon
(6) (7 pts) Write
in the form of f (x) +
c
2x − 1
8x2
2x − 1
, where f (x) is a polynomial and c is a number.
8x2 = 4x(2x − 1) + 4x
⇒
4x(2x − 1) + 4x
4x(2x − 1)
4x
4x
8x2
=
=
+
= 4x +
2x − 1
2x − 1
2x − 1
2x − 1
2x − 1
4x = 2(2x − 1) + 2
⇒
4x
2(2x − 1) + 2
2(2x − 1)
2
2
=
=
+
=2+
2x − 1
2x − 1
2x − 1
2x − 1
2x − 1
2
8x
4x
2
⇒
= 4x +
= 4x + 2 +
2x − 1
2x − 1
2x − 1
• Finding the expression 8x2 = 4x(2x − 1) + 4x: 2 pts.
4x
: 4 pts.
• Getting the first step 4x +
2x − 1
• Obtaining the expression 4x = 2(2x − 1) + 2: 6 pts.
2
: 7 pts.
• Finding the answer 4x + 2 +
2x − 1
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MATH 1001 Final
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(7) Suppose that
f (x) =
x2
5x + 1
.
+ 3x + 2
(a) (3 pts) Find the domain of f (x). Write it by using intervals.
The domain is the set of all numbers making nonzero denominators.
x2 + 3x + 2 = 0 ⇒ (x + 1)(x + 2) = 0 ⇒ x + 1 = 0 or x + 2 = 0
⇒ x = −1, −2
Domain: (−∞, −2) ∪ (−2, −1) ∪ (−1, ∞)
• Finding zeros x = −1, −2 of the denominator: 2 pts.
• Writing the answer (−∞, −2) ∪ (−2, −1) ∪ (−1, ∞) using appropriate
intervals: 3 pts.
(b) (3 pts) Is 1 in the range? Explain your answer.
5x + 1
= 1 ⇒ 5x + 1 = x2 + 3x + 2 ⇒ x2 + 3x + 2 − (5x + 1) = 0
2
x + 3x + 2
x2 − 2x + 1 = 0 ⇒ (x − 1)2 = 0 ⇒ x = 1
⇒ f (1) = 1
Therefore 1 is in the range.
• Writing the answer without any justification: 1 pt.
(c) (3 pts) Find all asymptotes of f .
If x → ∞,
5x + 1
5x
5
≈ 2 = ≈0
+ 3x + 2
x
x
So y = 0 is a horizontal asymptote.
There are two vertical asymptotes x = −1, x = −2.
f (x) =
x2
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MATH 1001 Final
Fall 2014 - Moon
(8) Suppose that f (x) = 2x − 3 and its domain is [0, 3]. The graph of f (x) is below.
(a) (3 pts) Suppose that g(x) = 2f (x + 4). Find its formula.
g(x) = 2f (x + 4) = 2(2x+4 − 3) = 2 · 2x · 24 − 2 · 3 = 32 · 2x − 6
• Finding 2(2x+4 − 3): 2 pts.
• Getting 32 · 2x − 6: 3 pts.
(b) (3 pts) Sketch the graph of g on the plane above.
(c) (2 pts) Find the domain and the range of g.
Domain: [−4. − 1]
Range: [−4, 10]
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MATH 1001 Final
Fall 2014 - Moon
(d) (3 pts) Find the formula of the inverse function f −1 of f .
y = 2x − 3 ⇒ 2x = y + 3 ⇒ log2 (y + 3) = x
⇒ f −1 (y) = log2 (y + 3)
• Setting up the equation y = 2x − 3: 1 pt.
• Getting the formula f −1 (y) = log2 (y + 3): 3 pts.
(e) (3 pts) Sketch the graph of f −1 below.
It is sufficient to take a flip across the diagonal line y = x.
(f) (2 pts) Find the domain and the range of f −1 .
The domain of f −1 is same with the range of f , and the range of f −1 is same
with the domain of f .
Domain: [−2, 5]
Range: [0, 3]
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MATH 1001 Final
Fall 2014 - Moon
(9) Suppose that x and y are two positive numbers such that log x = 2.41 and log y =
7.1.
(a) (3 pts) Evaluate log(x3 y −5 ).
log(x3 y −5 ) = log(x3 ) + log(y −5 ) = 3 log x + (−5) log y
= 3 · 2.41 − 5 · 7.1 = −28.27
(b) (4 pts) Find the whole number k such that xk has 80 digits.
log(x33 ) = 33 log x = 79.53
So x33 has 80 digits. k = 33.
(c) (3 pts) Which number is larger between x3000 and y 1000 ? Explain your answer.
log(x3000 ) = 3000 log x = 3000 · 2.41 = 7230,
log(y 1000 ) = 1000 log y = 1000 · 7.1 = 7100.
Because 7230 > 7100, x3000 > y 1000 .
• Unless there is a reasonable explanation, one cannot get the points.
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MATH 1001 Final
Fall 2014 - Moon
(10) (6 pts) A fossilized animal bone is unearthed on the ruins of Pompeii. It contains 80% of the carbon-14 found in living matter. About how old is the bone?
Round the answer to a whole number. (The function modeling the number of
the carbon-14 in the sample is
t
a(t) = a0 · 2− h
where a0 is the number of atoms at time 0, h is the half-life, and t is the number
of years. It takes 5730 years to be half of the initial amount of carbon-14.)
Because the half-life of the carbon-14 is 5730 years,
t
a(t)
= 2− 5730 .
a0
a(t)
From the given condition, we know that
= 0.8. So we have an equation
a0
t
a(t) = a0 · 2− 5730 , or
t
0.8 = 2− 5730 .
t
log 0.8 = log(2− 5730 ) = −
t
log 2
5730
−5730 log 0.8
≈ 1844.647.
log 2
Therefore the bone is approximately 1845 years old.
t=
t
• By using given conditions, setting up the equation 0.8 = 2− 5730 : 4 pts.
• Solving the equation and getting the answer 1845 years: 6 pts.
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MATH 1001 Final
Fall 2014 - Moon
(11) Sheldon deposits $3,000 in a saving account paying 6% interest per year, compounded monthly. The total amount in an account can be computed by using
the formula
r
P (1 + )nt
n
where P is the principal, r is the annual interest rates, n is the number of compounding periods per year, and t is the number of years.
(a) (4 pts) How much will be in the account at the end of 5 years? Round the
answer to the nearest cents.
0.06 12·5
3000(1 +
)
≈ 4046.5504
12
It becomes approximately $4046.55.
0.06 12·5
• Setting up the formula 3000(1 +
) : 2 pts.
12
• Getting the correct answer $4046.55: 4 pts.
(b) (6 pts) How long will it take for the saving account to reach $20,000? Round
the answer to one decimal place.
We want to find t such that
20000 = 3000(1 +
0.06 12t
) .
12
0.06 12t
20000
= (1 +
)
3000
12
20000
0.06 12t
0.06
log(
) = log((1 +
) ) = 12t log(1 +
)
3000
12
12
log( 20000
)
3000
t=
0.06 ≈ 31.6976
12 log(1 + 12 )
Approximately it takes 31.7 years.
0.06 12t
• Setting up the equation 20000 = 3000(1 +
) : 4 pts.
12
• Getting the answer 31.7 years: 6 pts.
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