MATH 1210 - Section 002

Fall 2010

Nicolas Fourrier

Due on November 4, 2010

Homework 8 Name: 1. Starting with the graph of y = ex , write the equation of the graph that results from: (a) shifting 2 units downward (b) shifting 2 units to the right (c) reflecting about the x-axis (d) reflecting about the y-axis (e) reflecting about the x-axis and y-axis.

(a) shifting 2 units downward: y = ex − 2

(b) shifting 2 units to the right: y = ex−2 (c) reflecting about the x-axis: y = −ex

(d) reflecting about the y-axis: y = e−x

(e) reflecting about the x-axis and y-axis: y = −e−x 2. Plot the graphs of f (x) = (1/2)x , g(x) = (1/3)x and h(x) = (1/4)x

3. Under ideal conditions a certain bacteria population is known to double every three hours. Suppose that there are initially 100 bacteria. (a) What is the size of the poulation after 16 hours ? (b) What is the size of the population after t hours ? (c) Estimate the size of the population after 20 hours ?

(a) Fifteen hours srepresents 5 doubling periods (one doubling period is three hours). 100.25 = 3200 (b) In t hours there will be t/3 doubling periods. The initial population is 100, so the population y at time t is y = 100.2t/3 (c) t = 20 ⇒ y = 100.220/3 = 10159

4. If $1000 is borrowed at interest 8%, find the amounts due at the end of 3 years if the interest is compounded annually, quaterly, monthly, weekly, hourly and continuously. Using A = P (1 + (a) $1000(1 + (b) $1000(1 + (c) $1000(1 + (d) $1000(1 + (e) $1000(1 +

r nt ) n

with A0 = 1000, r = 0.08 and t = 3

0.08 1.3 ) = $1259.71 with annual compounding 1 0.08 4.3 ) = $1268.24 with quaterly compounding 4 0.08 12.3 ) = $1270.24 with monthly compounding 12 0.08 52.3 ) = $1271.01 with weekly compounding 52 0.08 365.24.3 ) = $1258.25 with hourly compounding 365.24

(f) A = 1000e(0.08).3 = $1271.25 with continously compounding

5. Solve each equation for x. (a) 2ln(x) = 1 (b) e−x = 5 (c) ln(ln(x)) = 1 (d) ln(x) + ln(x − 1) = 1

(e) eax = Cebx where a 6= b (a) x =

√

e

(b) x = −ln(5) (c) x = ee

(d) x2 − x − e = 0 ⇒ x = 12 (1 + (e) x =

√

1 + 4e)

ln(C) a−b

6. The inflation rates in the U.S. economy in 2000 through 2003 were 3.4%, 2.8%, 1.6% and 2.3%. What was the purchasing power of a dollar at the beginning of 2004 compared to that at the beginning of 2000 ? 0.966 × 0.972 × 0.984 × 0.977 = 0.91$ 7. The inflation rates in the U.S. economy in 2000 through 2003 were 3.4%, 2.8%, 1.6% and 2.3%. What was the purchasing power of a dollar at the beginning of 2004 compared to that at the beginning of 2000 ? Use Compound interest formula: 32100 = 25350(1 + r)2 ⇒ r = 12.75% 8. How long will it take an investment to double in value if the interest rate is 6% compounded continuouly ? Solve P e0.06t = 2P , and get 0.06t = ln(2) ⇒ t = 11.55. So the investment will double in 11.55 years