Download Individual Project

1)
a)
Find the domain of the following:
f (t )  log( t  5)
Answer: {t | t > 5}
Show your work or explain how you obtained your answer here:
The domain of log(t) is t > 0. We have to shift it to the right by 5 units.
b)
g ( x)  5e x
Answer: all real numbers
Explain how you obtained your answer here: There are no numbers that make the exponential
function invalid.
c)
g ( x)  ln( x  4)
Answer: {x | x > -4}
Show your work or explain how you obtained your answer here:
The domain of ln(x) is x > 0. We have to shift it to the left 4 units.
d)
g (t )  5t
Answer: all real numbers
Explain how you obtained your answer here: There are no numbers that make the exponential
function invalid.
2)
Describe the transformations on the following graph of f ( x)  e x . State the placement of the
horizontal asymptote and y-intercept after the transformation. For example, “left 1” or “reflected about
the y-axis” are descriptions.
10
9
8
7
6
5
4
3
2
1
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
Y
X
1 2 3 4 5 6 7 8 9 10
a)
g ( x)  e x  3
Description of transformation: shifted up by 3
Equation(s) for the Horizontal Asymptote(s): y = 3
y-intercept in (x, y) form: (0,4)
b)
h( x)  e  x
Description of transformation: reflected about the y axis
Equation(s) for the Horizontal Asymptote(s): y = 0
y-intercept in (x, y) form: (0,1)
3)
Describe the transformations on the following graph of f ( x)  log( x) . State the placement of
the vertical asymptote and x-intercept after the transformation. For example, “left 1” or “stretched
vertically by a factor of 2” are descriptions.
10
9
8
7
6
5
4
3
2
1
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
a)
Y
X
1 2 3 4 5 6 7 8 9 10
g ( x)  log( x  2)
Description of transformation: shifted left by 2 units
Equation(s) for the Vertical Asymptote(s): x = -2
x-intercept in (x, y) form: (-1,0)
b)
g ( x)   log( x)
Description of transformation: reflected about the x axis
Equation(s) for the Vertical Asymptote(s): x=0
x-intercept in (x, y) form: (1,0)
4)
The formula for calculating the amount of money returned for an initial deposit into a bank account or
CD (certificate of deposit) is given by
nt
r

A  P1  
 n
A is the amount of the return.
P is the principal amount initially deposited.
r is the annual interest rate (expressed as a decimal).
n is the number of compound periods in one year.
t is the number of years.
Carry all calculations to six decimals on each intermediate step, then round the final answer to the
nearest cent.
Suppose you deposit $4,000 for 8 years at a rate of 7%.
a)
Calculate the return (A) if the bank compounds annually (n = 1). Round your answer to the hundredth's
place.
Answer: 6872.74
Show work in this space. Use ^ to indicate the power or use the Equation Editor in MS Word.
A = P(1+r/n)^(nt)
A = 4000*(1+0.07/1)^(1*8)
A = 6872.744719
b)
Calculate the return (A) if the bank compounds monthly (n = 12). Round your answer to the hundredth's
place.
Answer: 6991.31
Show work in this space
A = P(1+r/n)^(nt)
A = 4000*(1+0.07/12)^(12*8)
A = 6991.305824
c)
Does compounding annually or monthly yield more interest? Explain why.
Answer: Monthly compounding
Explain: Monthly compounding allows you to earn additional interest on your interest. Since
you get interest each month, instead of each year, there is more time for those interest payments to
earn interest.
If a bank compounds continuously, then the formula used is A  Pe
where e is a constant and equals approximately 2.7183.
Calculate A with continuous compounding. Round your answer to the hundredth's place.
rt
d)
Answer: 7002.69
Show work in this space
A = 4000*e^(0.07*8)
A = 4000*e^(0.56)
A = 4000*1.7506725
A = 7002.690001
A commonly asked question is, “How long will it take to double my money?” At 7% interest rate
and continuous compounding, what is the answer? Round your answer to the hundredth's
place.
e)
Answer: 9.90 years
Show work in this space
Set A = 2P
2P = Pe^(0.07t)
Divide by P:
2 = e^(0.07t)
Take ln:
Ln(2) = 0.07t
Divide by 0.07:
T = ln(2)/0.07
T = 9.9021025
5)
Suppose that the function P  12  50 ln x represents the percentage of inbound e-mail in the
U.S. that is considered spam, where x is the number of years after 2001.
Carry all calculations to six decimals on each intermediate step, when necessary.
a)
Use this model to determine the percentage of spam in the year 2005. Round your answer to
two decimal places.
Answer: 81.31%
Show your work in this space:
2005 is 4 years after 2001, so x= 4
P = 12 + 50*ln(4)
P = 12 + 69.314718
P = 81.314718
b)
Use this model to determine how many years it will take (to two decimal places) for the
percent of spam to reach 90% provided that law enforcement regarding spammers does
not change?
Answer: 4.76 years
Show your work in this space:
Set P = 90:
90 = 12 + 50ln(x)
Subtract 12:
78 = 50ln(x)
Divide by 50:
Ln(x) = 78/50
E to each side:
X = e^(78/50)
X = 4.758821