course Mth 272
Was thinking of quiting, but looking into a tutor to nock of the rust.
1) Evaluate each of the following expressions. (1/4)^3, (1/27)^(1/3), 8^(2/3), (3/7)^2, 100^(5/2), 25^(3/2)(1/4)^3 = 0.015
(1/27)^(1/3) = 0.333
8^(2/3) = 4
(3/7)^2 = 0.1836
100^(5/2) = 100,000
25^(3/2) = 125
2) Using the properties of exponents simplify the following expressions. (3^2)/(3^5), (1/6)^(-2), (32^(1/2))*(2^(1/2)), (8^(5/2)) * (1/2)^(5/2)
(3^2)/(3^5) = 3^2-5 = 3^-3 = 1/3^3 = 0.037
(1/6)^(-2) = 1/0.166^2 = .25
(32^(1/2))*(2^(1/2)) = (32 * 2)^1/2 = 64^1/2 = 8
(8^(5/2)) * (1/2)^(5/2) = (8 * ½)^5/2 = 4^5/2 = 32
3) Using the properties of exponents simplify the following expressions. (3^3)*(3^2), (1/5)^3*(5^3), (2^8)^(1/4), [(8^-2)(8^(5/3))]^3
(3^3)*(3^2) = (3 * 3)^3 + 2 = 9^5 = 59,049
(1/5)^3*(5^3) = (1/5 * 5)^3 = (0.20 * 5)^3 = 1^3 = 1
(2^8)^(1/4) = 2^8*1/4 = 2^8*.25 = 2^2 = 4
[(8^-2)(8^(5/3))]^3 = 8^(3 * 2) + (3 * 5/3) = 8^6 + 5 = 8^11 = 8,589,934,592
4) Let f(x) = 4^(x+1). Evaluate f(-4), f(-1/2), f(2), f(-5/2).
f(-4) = 4^(-4 +1) = 4^-3 = 1/(4^3) = 0.0156
f(-1/2) = 4^(-1/2+1) = 4^0.5 = 2
f(2) = 4^(2+1) = 4^3 = 64
f(-5/2) = 4^(-5/2+1) = 4^-1.5 = 1/(4^1.5) = 0.125
5) Let g(x) = 1.175^x. Evaluate g(1.3), g(170), g(40), g(12.5).
g(1.3) = 1.175^1.3 = 1.23
g(170) = 1.175^170 = 806,189,844,936
g(40) = 1.75 ^40 = 5,266,498,289.31
g(12.5) = 1.75^12.5 = 7.507
6) Solve the equation 4^(x+1) = 64 for x.
4^(x+1) = 64
4^(x+1) = 4^3 therefore
x+1 = 3 subtract 1 from both sides
x = 2
7) Solve the equation 4^2 = (x+2)^2 for x.
16 = x^2 + 4x + 4
0 = x^2 + 4x – 12 (subtracted 16 from both sides)
Used the quadratic formula to solve for x
X ={-4 + squrt 4^2 – (4*1*-12)} / 2*1
= (-4 + squrt 64) / 2
X = 2 and X = -6
X = 2 is the correct solution
8) Sketch and describe the graph of 2^(-x/2).
9) Sketch and describe the graph of f(x) = (1/3)^x
10) Sketch and describe the graph of f(x) = 4^(-x).
11) Sketch and describe the graph of f(x) = 4^|x| (|x| denotes the absolute value of x.)
12) Suppose that the annual rate of inflation averages over 6% per year over the next 10 years. Given this, the approximate cost of goods is given by C(t) = P(1.06)^t, 0 <= t <=10 where t it time in years and P is the present cost. If the price of a movie ticket is currently $7.95 estimate the cost of the ticket in 10 years.
C(t) = P(1.06)^t t = 10; P = $7.95
C(10) =7.95(1.06)^10
C(10) = 7.95(1.79)
C(10) = 14.23
C = 14.23 / 10
C = 1.43
The cost of the ticket in ten year is $7.95 + $1.43 = $9.37
13) After t years, the initial mass of 20 grams of a radioactive element whose half-life is 35 years is given by y = 20((1/2)^(t/35)), t >= 0. How much of the initial mass remains after 60 years? Using a graphing calculator or something similar, find the amount of time required for the mass to decay to only 1 gram
y = 20 [(1/2)^(t/35)] t = 60 years
y = 20*[(1/2)^(60/35)]
= 20 *[(1/2)^1.714]
= 20 * [0.3047]
y = 6.09
According to my graphing calculator the amount of time required for the mass to decay of 1 gram is 1.49 years.
Section 4.2
1) 4.2.2 Using the properties of exponents simplify the following expressions. (1/e)^-3, (e^2/e^5)^-1, e^2/e^4, 1/(e^-2)
(1/e)^-3 = (1^-3) / (e^-3) = 1 / (1 / e^3) = 1 / 0.0497 = 20.086
(e^2/e^5)^-1 = (e^2-5)^-1 = e^-2 + 5 = e^3 = 20.085
e^2/e^4 = e^2-4 = e^-2 = 1/e^2 = 0.1353
1/(e^-2) = 1/(1 / e^2) = 1 / 0.135 = 7.389
2) 4.2.4 Using the properties of exponents simplify the following expressions. (e^-5)^(3/5), (e^3)/(e^(-1/3)), (e^-1)^-3, (e^4)(e^-3/4)
(e^-5)^(3/5) = e^-5*0.60 = e^-3 = 1 / e^3 = 0.0497
(e^3)/(e^(-1/3)) = e^3-(-0.333) = e^3.3333 = 28.032
(e^-1)^-3 = e^-1*-3 = e^3 = 20.086
(e^4)(e^-3/4) = e^4+(-.75) = e^3.25 = 25.79
3) 4.2.6 Solve the equation e^x = 1 for x.
e^x = 1
x*LN e = LN 1
x*2.718281828 = 0
x = 0 / 2.718281828
x = 0
4) 4.2.8 Solve the equation e^(-1/2x) = sqrt(e) for x. (sqrt(e) is the square root of e or e^1/2)
5) *4.2.14 Sketch and describe the graph of the function f(x) = e^(-x/2).
6) 4.2.20 Sketch and describe the graph of the function f(x) = e^(2x)
7) 4.2.28 Use a graphing calculator or something similar to graph the function f(x) = (1/2)(e^x - e^(-x)). Determine whether the function has any horizontal asymptotes and discuss the continuity of the function. Explain how you could sketch the graph 'by hand', without the use of the calculator.
8) 4.2.32 Find the balance A of $2500 invested at 5% for 40 years when the interest is compounded 1, 2, 4, 12, and 365 times a year, and also when it is compounded continuously.
A = P (1 + (r /n))^nt
A1 = 2500 [(1 +(0.05/1)^1(40)] = 2500(1.05) = $17,599.97
A2 = 2500 [(1 + (0.05/2))^2(40)] = 2500(1.0506) = $18,023.92
A4 = 2500 [(1 + (0.05/4))^4(40)] = 2500(1.05094) = $18,245.05
A12 = 2500 [(1 + (0.05/12))^12(40)] = 2500(1.05116) = $18,396.04
A365 = 2500 [(1 + (0.05/365))^365(40)] = 2500(1.05127) = $18,470.11
Compounded Continuously
A = Pe^rt
A = 2500*e^0.05 *1 = $2,628.18
A = 2500*e^0.05*2 = $2762.93
A = 2500*e^0.05*4 = $3053.51
A = 2500*e^0.05*12 = $4,555.30
9) 4.2.42 How much should be deposited into an account paying 8.1% interest compounded monthly in order to have a balance of $17,201.66 three years from now?
P = A / (1 + (r/n))^nt
Given A = $17,201.66, r= 0.08, n=12, t=3
P = 17,201.66 / (1 + (0.08/12)^12*3
= 17201.66 / (1.0066)^36
= 17201.66 / 1.27
P = $13,542.09
10) If your typing rate at clock time t is N = 95 / (1 + 8.5 e^(-.12 t)) words / minute, where t is clock time in weeks, then
• How many words per minute are you typing at t = 10 weeks?
1hr = 60 min, 60min * 24hrs = 1,440 mins in one day * 7 = 10,080 min in one week.
N = 95 / (1+8.5*e^(-.12*t), t = 10
= 95 / 1+8.5*e^(-1.2)
= 95 / 1+8.5*(1/e^1.2)
= 95 / (1+2.5606)
= 95 /3.5606
N = 26.68 = 27 words / min in week ten
• What is the limiting value does your typing rate approach?
• At what clock time t is your typing rate 70 words / minute?
70 = 95 / 1 + 8.5 * e^(-.12*t)
Ln70 = 95 / 1 + 8.5(-0.12t) * 1
Ln70 = 95 / 1 + (-0.02t)
4.2485 (1+(-0.02t)) = 95
4.2485 + (-0.08497t) = 95
-0.08497t = 95 – 4.2485
-0.08497t = 90.7515
t = 90.7515 /-0.08497
t = 20.944 = 21 weeks
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Section 4.3
1) 4.3.4 Find the slope of the tangent line of the function y = e^(-3x) at the point (0,1)
y = e^-3x
2) 4.3.8 Find the derivative of f(x) = e^(1/x).
1/x = -x
Let u = -x then du / dx = 1
f (x) = e^u *(du / dx) = e^-x
3) 4.3.10 Find the derivative of g(x) = e^sqrt(x+3).
4) 4.3.12 Find the derivative of y = 5(x^2)(e^-x).
5) 4.3.16 Find the derivative of y = (2x^2)(e^x) - (4x)(e^x) + 4e^x.
6) 4.3.16 Find the derivative of y = (e^-x + e^x)^3
7) 4.3.24 Find dy/dx implicitly for the equation (x^3)y + xe^x - 12 = 3
8) 4.3.26 Find dy/dx implicitly for the equation e^(xy) + x^2- y^2 = 0.
9) 4.3.34 Graph the function f(x) = xe^(-x) and give its extrema, points of inflection, and asymptotes.
10) Give the equation of tangent line to the graph of e^(4x-2)^2 at the point (0, 1).
11) Find the extrema of the function y = x e^(-x).
12) 4.3.42 The Ebbinghaus Model for human memory is p = (100-a)e^(-bt) + a where p is percent retained after t weeks. If a = 20 and b = 0.5 at what rate is information being retained after 1 week and after 3 weeks? At what rate is memory being lost at t = 3 weeks?
11) *4.3.44 From 1995 to 2002, the numbers y (in millions) of employed people in the United States can be modeled by y = 115.46 + 1.592t + 0.0552t^2 - 0.00004e^t where t = 5 corresponds to the year 1995. Find the rates of change of the number of employed people in 1996, 1998, 2001.
12) *4.4.46 A survey of certain group of college students has determined that the mean height of females is 65 inches with a standard deviation of 2.9 inches. Assuming that the data can be modeled with the normal distribution, find a model for the data. Find the derivative of the model and show that f ' > 0 when x < mu and f ' < 0 when x > mu.
13) 4.4.48 Using a graphing calculator or something similar, graph the normal probability density function with sigma = 1 and mu = -3, 1, 2 in the same window.
• What effect does mu have on the function? Explain your reasoning.
• Explain how you could have answered this question analytically, without the use of your calculator.
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