# Sections 9A,9B

### exercises

...

Identifying a function. Exercise 12, p524

You make a list of your friends and their e-mail addresses. Are there two variables here related in a way that may be described as a function?

Recall generalities on functions.

A function: two variables, an independent variable and dependent variable such that a value of the independent variable determines the value of the dependent variable uniquely

Answer. The situation can be described as a function; the independent variable is the e-mail address, while the dependent variable is the name of a friend. Indeed, a e-mail address identifies uniquely the name of a friend.

Identifying a function. Exercise 12, p524 Some remarks.

1. Actually, several people may share the same e-mail address although that does not happen frequently.

In such a case, there is no function which may adequately model this situation.

2. No quantities are involved. Although indeed one considers functions operating with non-quantitative data, we typically do not do that in this class.

Functions and their graphs. Exercise 24, p524

Study the graph which represents hours of daylight at $$40^o$$N latitude.

a. Use the graph to estimate the number of hours of daylight on April 1 (91st day of the year) and October 1 (the 304th day of the year).

Exercise 24, p524 ... continued Study the graph which represents hours of daylight at $$40^o$$N latitude.

b. Use the graph to estimate the dates on which there are 13 hours of daylight.

Answer: about 110 and 230 days of the year. That corresponds to April, 20 and August, 18.

Linear functions. Exercise 12, p537

Consider the graph.

a. In words, describe the function shown on the graph.

Answer. The population increases linearly with time.

b. Find the slope of the graph, and express it as a rate of change

Answer. The slope is $$\frac{20}{2}=10$$. The rate of change of population is 10 thousand people per year.

Linear functions. Exercise 16, p537

Consider the graph.

a. In words, describe the function shown on the graph.

Answer. The record pace decreases linearly with the length of race.

b. Find the slope of the graph, and express it as a rate of change,

Answer. The slope is approximately $$\frac{-20}{10}=-2$$. The rate of change is -2 (km/hr)/km.

Rate of change. Exercise 20, p537

A gas station owner finds that for every penny increase in the price of gasoline, she sells 80 gallons fewer of gas per week. How much more or less gas will she sell if she raises the price by 8 cents per gallon?

Solution. At this gas station, the amount of gas sold depends linearly on the price. The rate of change is
$$-80$$ (gallons per week)/cent.
If the price increases by 8 cents, the change is
$$-80 \times 8 = -640$$
That is a 640 gallons less sold per week.

Rate of change. Exercise 18, p537

You run along a path at a constant speed of 5.5 miles per hour. How far do you travel in 1.5 hours?

Solution. The distance traveled is a linear function of time. It increases at a constant rate of $$5.5$$ miles per hour. The distance traveled in $$1.5$$ hours is
$$5.5 \times 1.5 = 8.25$$ miles.

Linear equations. Exercise 26, p537

The cost of leasing a car is $$\1,000$$ for a downpayment and processing fee plus $$\360$$ per month. For how many months can you lease a car with $$\3680$$?

Solution. The total amount spent $$y$$ is a linear function of time $$x$$ with a rate of change of 360, and initial value ($$y$$-intercept) of 1000. This function can be written as

$$y=360\times x + 1000$$

Solve the equation for $$x$$: $$x=\frac{y-1000}{360}$$, and find for y=3680:

$$x=\frac{3680-1000}{360} \approx 7.44$$. That means 7 full months of lease.

Exercise 26, p537 ... continued ... The theory behind this solution.

Recall that linear function is described by the formula

$$y=mx + b$$

This equation can always be solved for $$x$$:

$$x = \frac{y-b}{m}$$

and we always can find the value of independent variable $$x$$

for a given value of the dependent variable $$y$$

Exercise 26, p537 ... continued

The cost of leasing a car is $$\1,000$$ for a downpayment and processing fee plus $$\360$$ per month. For how many months can you lease a car with $$\3680$$?

Alternative solution. After paying $$\1,000$$ of downpayment and processing fee you are left with
$$3680-1000= \ 2680$$.
With a monthly payment of $$\360$$, that suffices for $$\frac{2680}{360} \approx 7.44$$,
or $$7$$ full months.

Linear equations. Exercise 30, p538

You can purchase a motorcycle for $$\6,500$$ or lease it for a downpayment of $$\200$$ and $$\150$$ per month. Find the function which describes how the cost of the lease depends on time. How long can you lease the motorcycle before you've paid more than its purchase price?

Solution. This is a linear function with a slope of 150, and an initial value ($$y$$-intercept) of 200.

The equation for this function thus is

$$y=150x+ 200$$.

Exercise 30, p538 ... continued

You can purchase a motorcycle for $$\6,500$$ or lease it for a downpayment of $$\200$$ and $$\150$$ per month. Find the function which describes how the cost of the lease depends on time. How long can you lease the motorcycle before you've paid more than its purchase price?

Solution. We found the linear function $$y=150x+ 200$$.

Solve the equation for $$x$$:

$$x=\frac{y-200}{150}$$,

and find for $$y=6500$$

$$x=\frac{6500-200}{150} =42$$ months (3.5 years) of lease.

Exercise 30, p538 ... continued

You can purchase a motorcycle for $$\6,500$$ or lease it for a downpayment of $$\200$$ and $$\150$$ per month. How long can you lease the motorcycle before you've paid more than its purchase price?

Alternative solution (no function involved).

After $$\200$$ of downpayment is paid, you are left with

$$6500-200=\6300$$.

With a monthly lease of $$\150$$, this amount suffices for

$$\frac{6300}{150}=42$$ months (equivalently, 3.5 years) of lease.

Linear equations. Exercise 34, p538

A mining company can extract $$2000$$ tons of gold ore per day with a purity of $$3$$ ounces of gold per ton. The cost of extraction is $$\1000$$ per ton. If $$p$$ is the price of gold in dollars per ounce, find the function that gives the daily profit/loss of the mine as it varies with the price of gold.

Solution. While the total cost of extraction is $$1000 \times 2000 = 2,000,000$$ dollars, the price of all gold extracted during a day is $$2000 \times 3 \times p = 6000\times p$$ dollars.

The daily profit $$P$$ is the difference between them:

$$P=6000p-2000000$$

Exercise 34, p538 ... continued

What is the minimum price of the gold which makes the mine profitable?

Solution. We found the profit $$P$$ as a function of the price $$p$$:

$$P=6000p-2000000$$

This becomes a loss, and the mine stops to be profitable when $$P=0$$:

$$0=6000p-2000000$$.

Solve for $$p$$ to find $$p=\frac{2000000}{6000} \approx \ 333.33$$