Solving - Math Problem Example

Math Problem

1.

1. a) Using synthetic division:

-3

3

5

-6

18

-9

12

-18

3

-4

6

0

2

3

-4

6

6

4

3

2

10

As zero remainder is received when dividing by -3, x = -3 is a zero Division by 2 produced the non-zero remainder, and, therefore, x = 2 is not a zero.

b)

-4

3

11

-2

8

-12

4

-8

3

-1

2

0

2

3

-1

2

6

10

3

5

12

As zero remainder is received for the synthetic division by -4, x = -4 is zero, and (x+4) is the factor of the given polynomial. However, the subsequent division by 2 produced the non-zero remainder. x=2 is not a zero.

2.  
3. a) For f(x) = 2x 3 -5x2-4x+3 from the Rational roots test we can suggest that +/-3 (multiple of the factor 3/2) can be the possible solutions. Using synthetic division we find that x=3 is a zero of the polynomial:

3

2

-5

-4

3

6

3

-3

2

1

-1

0

-1

2

1

-1

-2

1

2

-1

0

Thus, f(x) = 2x 3 -5x2-4x+3 can be rewritten as f(x) = (x-3)* (2x2+x-1), and further as

f(x) =(x-3)*(x+1)*(2x-1) with zeroes at x=3, x=-1 and x = ½.

1. x = 3, x = -1 and x = ½.
2. x =1, x = -3 and x = -1 ½.
3. x = 3/2, x = -1/2 and x = ¼.

3.

1. a) The horizontal asymptote of the graph of the function N(t) = (0.8t + 1000 ) / (5t+4) would be at 5t + 4 = 0, thus, at t = -4/5.

N(t) approaches 0.16 ( equals 0.8/5)  as t approaches infinity.

1. b) As a sufficient amount of time passes by, the influence of medication diminishes and the body concentration approaches the natural level of 0.16 parts per million.

4.  
5. a) For x = 6, g(x) = 2(x+12) = 2(6+12) = 36.

        For x = 10, g(x) = 2(x+12) = 2(10+12) = 44.

        For x = 14, g(x) = 2(x+12) = 2(14+12) = 52.

1. b) To find the formula for the inverse of the function, we solve g(x) = 2(x+12) for x:

g(x) = 2x + 24;

2x = g(x) – 24;

x = g(x)/2 – 12.

Or, rewritten f(x) = 1/2x – 12, where x – is the size in Italy, and f(x) – is the size of clothing in the US.

6. c) For x = 36, f(x) = 1/2x – 12 = 1/2*36 – 12 = 18 – 12 = 6.

    For x = 44, f(x) = 1/2x – 12 = 1/2*44 – 12 = 22 – 12 = 10.

    For x = 60, f(x) = 1/2x – 12 = 1/2*60 – 12 = 30 – 12 = 20.

5.

1. a) The formula for continuously compounded interest for 5 periods and a $2,000 starting amount:

2000*e5k = 2983.65

Solving the equation, we find:

e5k = 1.492

5k = ln 1.492

5k = 0.400

k = 0.08

Thus, interest rate equals 8.0%

1. b) The exponential growth function, in general terms, can be written as F = P* en*k,

where P – starting amount,

n – quantity of years,

k – annual interest rate,

F – future amount.

1. c) To find F after 10 years, let n = 10:

F = 2000*e 10*0.08 = 2000 * 2.226 = $4, 451.08

1. d) Let's say the $2,000 starting amount will double after z years:

2,000*ez*0.08 = 4,000

ez*0.08 = 2

0.08z = ln 2

0.08z = 0.693

Thus, z = 8.66

$2,000 will double in 8.66 years.