Chapter ISG, Problem 9.9.1.7P

a.

To determine

To write a function for stock A and stock B.

Answer to Problem 9.9.1.7P

  $f(x)=45{\left(1-\frac{4}{100}\right)}^x$

  $g(x)=22{\left(1+\frac{22.7}{100}\right)}^x$

Explanation of Solution

Given information:

J comparing the price of two stocks

Stock A is $45

Decrease by 4%

Stock B is $22

Increase $2.25 per month

Formula used:

  $f(x)=P{\left(1+\frac{r}{100}\right)}^x$

P is initial amount r is rate of interest

x is time in years

Calculation:

Stock A initial amount in dollars $=45$

Decrease by rate $=-4$

Functions of amount of stock A in x years using formula

  $f(x)=45{\left(1-\frac{4}{100}\right)}^x$

Stock B initial amount in dollars $=22$

Increase by rate per month in dollars $=2.25$

Increase by rate per year in dollars $=27$

Using formula to calculate rate per year

  $27=22{\left(1+\frac{r}{100}\right)}^1$

On solving,

  $\frac{27}{22}=1+\frac{r}{100}$

On solving,

  $\frac{27}{22}-1=\frac{r}{100}$

On solving,

  $22.7\%=r$

Functions of amount of stock B in x years using formula

  $g(x)=22{\left(1+\frac{22.7}{100}\right)}^x$

b.

To determine

To sketch the graph and find how many month foramount in stock A and stock B will be same.

Answer to Problem 9.9.1.7P

Hence, 35 month after both stocks have same amount

Explanation of Solution

Given information:

Stock A

  $f(x)=45{\left(1-\frac{4}{100}\right)}^x$

Stock B

  $g(x)=22{\left(1+\frac{22.7}{100}\right)}^x$

Calculation:

Stock A

  $f(x)=45{\left(1-\frac{4}{100}\right)}^x$

Table for stock A

  

Graph

  

Stock B

  $g(x)=22{\left(1+\frac{22.7}{100}\right)}^x$

Table for stock B

  

Graph

  

Table for same amount

  

It is clear that at $x=2.916$ the amount of stocks are same

Then 2.916 years after amount of both stocks are same

In month

  $x=2.916\times 12$

So

  $x=2.916\times 12$

So

  $x\approx 34.99$

Approximately

  $x=35$

Hence, 35 month after both stocks have same amount