Question

1. (a) a long - distance company charges a monthly fee of $4.95 and $0.02 per minute. fill in the following chart with the correct values for f(x), the monthly long - distance bill, where x represents total minutes of long - distance time during the month.

x	f(x)
15
30
45
60

(b) what is the initial value of this function? what is the rate of change of this function? (c) write a linear function that gives a rule for the amount charged. (d) evaluate f(110) and explain what it means.

Explanation:

Step1: Find the linear - function formula

The monthly fee is a fixed cost of $\$4.95$ and the cost per minute is $\$0.02$. The linear - function formula for the monthly long - distance bill $f(x)$ in terms of the number of minutes $x$ is $f(x)=0.02x + 4.95$.

Step2: Calculate $f(x)$ for different values of $x$

When $x = 0$:
$f(0)=0.02\times0 + 4.95=4.95$
When $x = 15$:
$f(15)=0.02\times15 + 4.95=0.3+4.95 = 5.25$
When $x = 30$:
$f(30)=0.02\times30 + 4.95=0.6 + 4.95=5.55$
When $x = 45$:
$f(45)=0.02\times45+4.95 = 0.9+4.95=5.85$
When $x = 60$:
$f(60)=0.02\times60 + 4.95=1.2+4.95 = 6.15$

Step3: Find the initial value and rate of change

The initial value of the function is the value of $f(x)$ when $x = 0$. From the formula $f(x)=0.02x + 4.95$, when $x = 0$, $f(0)=4.95$. The rate of change of a linear function $y=mx + b$ is the slope $m$. For $f(x)=0.02x + 4.95$, the rate of change is $0.02$.

Step4: Evaluate $f(110)$

$f(110)=0.02\times110+4.95=2.2 + 4.95=7.15$. This means that if a customer uses 110 minutes of long - distance calls in a month, the monthly long - distance bill will be $\$7.15$.

Answer:

$x$	$f(x)$
15	$5.25$
30	$5.55$
45	$5.85$
60	$6.15$

(b) Initial value: $4.95$, Rate of change: $0.02$
(c) $f(x)=0.02x + 4.95$
(d) $f(110)=7.15$. It means the monthly long - distance bill is $\$7.15$ when 110 minutes of long - distance calls are made.