Question

mike is a veterinarian. he is placing a 9-line job ad for an assistant. the following piecewise function gives the price of an x-line ad.
$a(x)= \

$$\begin{cases} 45 & \\text{when } x \\leq 3 \\\\ 45 + 9(x - 3) & \\text{when } x > 3 \\end{cases}$$

$
a. find the difference between the cost of a two-line ad and the cost of a three-line ad.
b. find the cost of a 10-line ad.
c. find the cost of an 11-line ad.

Explanation:

Part a

Step1: Cost of 2 - line ad

For \(x = 2\), since \(2\leq3\), use \(a(x)=45\). So cost of 2 - line ad is \(45\).

Step2: Cost of 3 - line ad

For \(x = 3\), since \(3\leq3\), use \(a(x)=45\). So cost of 3 - line ad is \(45\).

Step3: Find the difference

Difference \(=45 - 45=0\).

Part b

Step1: Determine the function to use

For \(x = 10\), since \(10>3\), use \(a(x)=45 + 9(x - 3)\).

Step2: Substitute \(x = 10\)

\(a(10)=45+9\times(10 - 3)\)
\(=45 + 9\times7\)
\(=45+63\)
\(=108\).

Part c

Step1: Determine the function to use

For \(x = 11\), since \(11>3\), use \(a(x)=45 + 9(x - 3)\).

Step2: Substitute \(x = 11\)

\(a(11)=45+9\times(11 - 3)\)
\(=45 + 9\times8\)
\(=45+72\)
\(=117\).

Answer:

s:
a. \(\boldsymbol{0}\)

b. \(\boldsymbol{108}\)

c. \(\boldsymbol{117}\)