Chapter 1.3, Problem 41E

a.

To determine

To find: $d\left(p\right)$ .

Answer to Problem 41E

  $d\left(p\right)=0.88p$

Explanation of Solution

Given information:

The variables are as follows:

  $p=$ Original price of a computer

  $d\left(p\right)=$ Price of computer after discount

  $d=$ Discounted price

  $r\left(d\right)$ =Price after rebate

The given conditions are as follows:

  $d=12\%p$

Calculation:

The equation for price after discount can be written as follows:

  $d\left(p\right)=p-d$

Put the value of $d$ in the above equation and evaluate $d\left(p\right)$ as follows:

  $\begin{array}{c}d\left(p\right)=p-12\%p\\ =p-\frac{12}{100}p\\ =\frac{100p}{100}-\frac{12p}{100}\left[\text{Taking common denominator}\right]\\ =\frac{88}{100}p\\ =0.88p\end{array}$

Thus the function is, $d\left(p\right)=0.88p$ .

b.

To determine

To find: $r\left(d\right)$ .

Answer to Problem 41E

  $r\left(d\right)=d-100$

Explanation of Solution

Given information:

The variables are as follows:

  $p=$ Original price of a computer

  $d\left(p\right)=$ Price of computer after discount

  $d=$ Discounted price

  $r\left(d\right)$ =Price after rebate

The given conditions are as follows:

  $\text{rebate}\left(r\right)=\$100$

Calculation:

The equation for price after discount can be written as follows:

  $r\left(d\right)=d-r$

Put the value of $r$ in the above equation and evaluate $r\left(d\right)$ as follows:

  $r\left(d\right)=d-\$100$

Thus the function is, $r\left(d\right)=d-\$100$ .

c.

To determine

To find: the function related to selling price to original price of computer by using the composition of functions.

Answer to Problem 41E

  $r\left(p\right)=0.88p-\$100$

Explanation of Solution

Given information:

The variables are as follows:

  $p=$ Original price of a computer

  $d\left(p\right)=$ Price of computer after discount

  $d=$ Discounted price

  $r\left(d\right)$ =Price after rebate

Calculation:

The equations evaluated in part a. and b. are as follows:

  $d\left(p\right)=0.88p$

  $r\left(d\right)=d-\$100$

Since discounted price and price after discount is same. So, $d\left(p\right)=d$

Plug in the values of $d$ in the function of $r\left(d\right)$ as follows:

  $r\left(d\left(p\right)\right)=d\left(p\right)-\$100$

Now, put the value of $d\left(p\right)$ in the above equation as follows:

  $\begin{array}{l}r\left(d\left(p\right)\right)=0.88p-\$100\\ r\left(p\right)=0.88p-\$100\left[r\text{ in terms of }p\right]\end{array}$

Thus, the function related to the selling price to original price of a computer is as follows:

  $r\left(p\right)=0.88p-\$100$

d.

To determine

To find: the selling price of computer with the given original prices.

Answer to Problem 41E

  $\begin{array}{l}r\left(\$799.99\right)=\$603.99\\ r\left(\$999.99\right)=\$779.99\\ r\left(\$1499.99\right)=\$1219.99\end{array}$

Explanation of Solution

Given information:

The variable $p$ the original price of a computer

  $\begin{array}{l}{p}_1=\$799.99\\ p_2=\$999.99\\ p_3=\$1499.99\end{array}$

Calculation:

The equation for selling price of computer to original price evaluated in part c. is as follows:

  $r\left(p\right)=0.88p-\$100$ … (1)

Put $p_1=\$799.99$ in equation (1) and evaluate $r\left(p_1\right)$ as follows:

  $\begin{array}{c}r\left(p_1\right)=0.88p_1-\$100\\ r\left(\$799.99\right)=0.88\left(\$799.99\right)-\$100\\ =\$703.99-\$100\\ =\$603.99\end{array}$

Put $p_2=\$999.99$ in equation (1) and evaluate $r\left(p_2\right)$ as follows:

  $\begin{array}{c}r\left(p_2\right)=0.88p_2-\$100\\ r\left(\$999.99\right)=0.88\left(\$999.99\right)-\$100\\ =\$879.99-\$100\\ =\$779.99\end{array}$

Put $p_3=\$1499.99$ in equation (1) and evaluate $r\left(p_3\right)$ as follows:

  $\begin{array}{c}r\left(p_3\right)=0.88p_3-\$100\\ r\left(\$1499.99\right)=0.88\left(\$1499.99\right)-\$100\\ =\$1319.99-\$100\\ =\$1219.99\end{array}$

Thus, the selling price of computer to original price $p_1,p_2$ and $p_3$ is $\$603.99,\$779.99$ and $\$1219.99$ respectively.