1. 12 1
Functions 1

2. 1.4 Functions Given by Words

3. Functions Given by Words
Many times mathematical functions are described verbally. In order for us to work effectively with such functions, it is essential that the verbal description be clearly understood.
Sometimes this is enough, but more often it is necessary to supplement the verbal description with a formula, graph, or table of values.

4. Getting Formulas from Words

5. Getting Formulas from Words
Many times verbal descriptions can be directly translated into formulas for functions.
Suppose, for example, that an engineering firm invests $78,000 in the design and development of a more efficient computer hard disk drive. For each disk drive sold, the engineering firm makes a profit of $98.
We want to get a formula that shows the company’s net profit as a function of the number of disk drives sold, taking into account the initial investment.
This is not a difficult problem, and you may find that you can do it without further help, but we want to show a general method that may be useful for more complicated descriptions.

6. Getting Formulas from Words
Step 1:
Identify the function and the things on which it depends, and write the relationships you know in a formula using words.
The function we are interested in is net profit. That is the profit on disk drives sold minus money spent on initial development.
The profit on drives sold depends in turn on the number of sales:
Net profit = Profit from sales − Initial investment (1.1)
= Profit per item  Number sold − Initial investment. (1.2)

7. Getting Formulas from Words
Step 2:
Select and record letter names for the function and for each of the variables involved, and state their units.
We can use any letters we want as long as we identify them and state clearly what they mean.
To say that we want to express net profit as a function of the number of disk drives sold means that net profit is the function and the number of disk drives is the variable.
We will use N to denote our function, the net profit in dollars, and d to represent the variable, the number of disk drives sold:
N = Net profit in dollars
d = Number of drives sold.

8. Getting Formulas from Words
Step 3:
Replace the words in step 1 by the letters identified in step 2 and appropriate information from the verbal description.
Here we simply replace the words Net profit in
Equation (1.2) by N, Profit per item by 98, Number sold by d, and Initial investment by 78,000:

9. Getting Formulas from Words
Thus we get the formula N = 98d − 78,000,
where d is the number of disk drives sold and N is the net profit in dollars.
We can use this formula to provide information about the engineering firm’s disk drive project.
For example, in functional notation, N(1200) represents the net profit in dollars if 1200 disk drives are sold.
We can calculate its value by replacing d by 1200:
N(1200) = 98  1200 − 78,000 = 39,600 dollars.

10. Getting Formulas from Words

11. Example 1.16 – Cutting a Diamond
The total investment a jeweler has in a gem-quality diamond is the price paid for the rough stone plus the amount paid to work the stone. Suppose the gem cutter earns $40 per hour.
Part 1: Choose variable and function names, and give a
formula for a function that shows the total investment in
terms of the cost of the stone and the number of hours
required to work the stone.
Part 2: Use functional notation to express the jeweler’s
investment in a stone that costs $320 and requires 5
hours and 15 minutes of labor by the gem cutter.

12. Example 1.16 – Cutting a Diamond
cont’d
Part 3: Use the formula you made in part 1 to calculate the
value from part 2.
Solution:
Part 1:
To say that we want the total investment as a function of the cost of the stone and the hours of labor means that total investment or total cost is our function and that cost of stone and hours of labor are our variables.
The total investment is the cost of the stone plus the cost
of labor, which in turn depends on the number of hours
needed to work the stone.

13. Example 1.16 – Solution
cont’d
We write out the relationship using a formula with words.
We will accomplish this in two steps:
Investment = Cost of stone + Cost of labor (1.3)
= Cost of stone + Hourly wage  Hours of labor (1.4)
The next step is to choose letters to represent the function and the variables.
We let c be the cost in dollars of the rough stone and h the number of hours of labor required.
We let I = I (c, h) be the total investment in dollars in the diamond.

14. Example 1.16 – Solution
cont’d
To get the formula we want, we use Equation (1.4), replacing Investment by I, Cost of stone by c, Hourly wage by 40, and Hours of labor by h:
The result is I = c + 40h, where I is the total investment in dollars, c is the cost in dollars of the rough stone, and h is the number of hours required to work the stone.
Note that, in our final presentation of the formula, we were careful to identify the meaning of each letter, noting the appropriate units.
Such descriptions are as important as the formula.

15. Example 1.16 – Solution
cont’d
Also, in this case you may find it easy enough to bypass much of what we presented here and go directly to the final answer.
You are very much encouraged, however, to put in the intermediate step using words, just as it appears here.
You will encounter many word problems, and this practice will make things much easier for you as the course progresses.

16. Example 1.16 – Solution Part 2:
cont’d
Part 2:
The cost of the rough stone is $320, so we use c = 320.
To get the value of h, we need to remember that h is measured in hours. Because 5 hours and 15 minutes is hours, h = 5.25 hours, and that is what we will put in for h.
Thus in functional notation, the total investment is I(320, 5.25).

17. Example 1.16 – Solution Part 3:
cont’d
Part 3:
In the formula we put 320 in place of c and 5.25 in place of h:
I(320, 5.25) =  5.25 = 530 dollars .

18. Cost, Revenue, and Profit Functions

19. C(x) = “variable costs” + “fixed costs”
Cost Function
The total cost C for a manufacturer during a given time period is a function of the number x of items produced during that period.
To determine a formula for the total cost, we need to know the manufacturer's fixed -costs (covering things such as plant maintenance and insurance), as well as the cost for each unit produced, which is called the variable cost.
To find the total cost, we multiply the variable cost by the number of items produced during that period and then add the fixed costs.
C(x) = “variable costs” + “fixed costs”

20. Revenue Function R = “unit price”  “number of units” R = px
The total revenue R for a manufacturer during a given time period is a function that specifies the total payment received from selling x items produced during that period.
To determine a formula for the total revenue, we need to know the selling price p per unit of the item.
To find the total revenue, we multiply this selling price p by the number x of items produced.
R = “unit price”  “number of units”
R = px

21. Profit Function Profit = Revenue – Cost
The profit P for a manufacturer is the total revenue minus the total cost.
If this number is positive, then the manufacturer turns a profit, whereas if this number is negative, then the manufacturer has a loss.
If the profit is zero, then the manufacturer is at a break-even point.
Profit = Revenue – Cost

22. Break-even Analysis
The break-even point is the level of production that results in no profit and no loss.
To find the break-even point we set the profit function equal to zero and solve for x.

23. Example P(x) = R(x) – C(x) = 12x – (3x + 3600) = 9x – 3600
A shirt producer has a fixed monthly cost of $3600. If each shirt has a cost of $3 and sells for $12 find:
The cost function
C(x) = 3x where x is the number of shirts produced.
The revenue function
R(x) = 12x where x is the number of shirts sold.
The profit function
P(x) = R(x) – C(x) = 12x – (3x ) = 9x – 3600

24. Example
A shirt producer has a fixed monthly cost of $3600. If each shirt has a cost of $3 and sells for $12 find:
The profit from selling 900 shirts
P(900) = 9(900) – 3600 = $4500

25. Example
A shirt producer has a fixed monthly cost of $3600. If each shirt has a cost of $3 and sells for $12 find:
The break even point is the solution of the equation C (x) = R (x), which in this case is
Therefore, at 400 units the break-even revenue is $4800.

26. Example: Geometric Representation

27. Demand and Supply Functions

28. Demand Function p (the price per item).
A demand function or demand equation expresses the
number q of items demanded as a function of the unit price
p (the price per item).
Thus, q(p) is the number of items demanded when the
price of each item is p.
q = items
q = items
Price p
Price p

29. Supply Function A supply function or supply equation expresses the
number q of items, a supplier is willing to make available, as
a function of the unit price p (the price per item).
Thus, q(p) is the number of items supplied when the price
of each item is p.
q = items
q = items
Price p
Price p

30. Market Equilibrium
Market Equilibrium occurs when the quantity produced is
equal to the quantity demanded.
supply curve q
demand curve
surplus
shortage
Equilibrium Point p

31. Market Equilibrium Equilibrium demand
Market Equilibrium occurs when the quantity produced is
equal to the quantity demanded.
supply curve q
demand curve
Equilibrium demand p
Equilibrium price

32. Market Equilibrium
Market Equilibrium occurs when the quantity produced is
equal to the quantity demanded.
To find the Equilibrium price set the demand equation equal to the supply equation and solve for the price p.
To find the Equilibrium demand evaluate the demand (or supply) function at the equilibrium price found in the previous step.

33. Example: Linear Demand Function
At a local store, the demand and supply curve for certain
computer model are:
where p is the price (in hundreds of dollars) of the computer.
What is the equilibrium price in dollars for this computer and
how many are bought and sold at this price?

34. Example: Linear Demand Function

35. Example: Nonlinear Revenue
As the operator of Workout Fever Health Club, you calculate
your demand equation to be q 0.06p + 84 where q is the
number of members in the club and p is the annual
membership fee you charge.
Part 1.
Your annual operating costs are a fixed cost of $20,000 per
year plus a variable cost of $20 per member. Find the annual
revenue and profit as functions of the membership price p.

36. Example: Nonlinear Revenue
Part 2.
At what price should you set the membership fee to obtain
the maximum revenue? What is the maximum possible
revenue?
Part 3.
the maximum profit? What is the maximum possible profit?
What is the corresponding revenue?

37. Example: Nonlinear Revenue
The annual revenue is given by
The annual cost as function of q is given by
The annual cost as function of p is given by

38. Example: Nonlinear Revenue
Thus the annual profit function is given by

39. The graph of the revenue function is

40. The graph of the revenue function is

41. The profit function is

42. The profit function is

43. Exponential Functions
A Quick Review

44. Exponential Functions
Here, we study a new class of functions called exponential functions. For example,
f (x) = 2x
is an exponential function (with base 2). Notice how quickly the values of this function increase:
f (3) = 23 = 8
f (10) = 210 = 1024

45. Exponential Functions
Compare this with the function g (x) = x2, where g(30) = 302 = 900.
The point is that when the variable is in the exponent, even a small change in the variable can cause a dramatic change in the value of the function.

46. Exponential Functions
We assume that a  1 because the function f (x) = 1x = 1 is just a constant function.
Notice that ax is defined for all values of x and that ax is always positive, ax > 0. Therefore, we have

47. Exponential Functions
The following Law of Exponents may be needed later.

48. Exponential Functions
Here are some examples of exponential functions:
f (x) = 2x g (x) = 3x h (x) = 10x
f (x) = ex,
where the base e = …..is an irrational number. This function is called, the natural exponential function. It is the only function with the property that its instantaneous rate of change is the function itself.

49. Example 1 – Evaluating Exponential Functions
Let f (x) = 3x, and evaluate the following:
(a) f (2) (b) f
(c) f () (d) f ( )
Solution:
We use a calculator to obtain the values of f.

50. Example 1 – Solution
cont’d

51. Exponential Functions
Graphs of
Exponential Functions

52. Graphs of Exponential Functions
We first graph exponential functions by plotting points.
We will see that the graphs of such functions have an easily recognizable shape.
Example: Draw the graph of each function. a) b)

53. Example 2 – Graphing Exponential Functions by Plotting Points-4
1/16 -3
1/8 -2
1/4 -1
1/2 1 2 4 3 8 x y -4
1/16 -3
1/8 -2
1/4 -1
1/2 1 2 4 3 8

54. Example 2 – Graphing Exponential Functions by Plotting Points-3 8 -2 4 -1 2 1
1/2
1/4 3
1/8
1/16 x y -3 8 -2 4 -1 2 1
1/2
1/4 3
1/8
1/16

55. Graphs of Exponential Functions
The figure shows the graphs of the family of exponential functions f (x) = ax for various values of the base a.
A family of exponential functions

56. Applications Involving Exponential Functions

57. Example: Exponential Growth
For example, there are initially 2000 bacteria in a Petri dish. The bacteria reproduce by cell division, and every three hours the number of bacteria doubles.
This is a verbal description of a function N = N(t), where N is the number of bacteria present at time t. It is common in situations like this to begin at time t = 0. Thus N(0) is the number of bacteria we started with, 2000.
Three hours later the population doubles, so N(3) = 4000 bacteria. In three more hours the population doubles again, giving N(6) = 8000 bacteria. Continuing, we see that N(9) = 16,000 bacteria.

58. Example: Exponential Growth
In order to find the pattern that models this situation we do not compute the function values but rather indicate what must be done to compute them, namely,
N(0) = 2000  20 (after waiting 0 three-hour periods)
N(3) = 2000  21 (after waiting 1 three-hour periods)
N(6) = 2000  22 (after waiting 2 three-hour periods)
N(9) = 2000  23 (after waiting 3 three-hour periods)
Thus, after waiting k three-hour periods the number of bacteria is given by
N(3k) = 2000  2k

59. Example: Exponential Growth
Now, knowing that after waiting k three-hour periods the number of bacteria is given by
N(3k) = 2000  2k
to get the number of bacteria after waiting t hours, all we have to do is count the number k of three-hour periods in t hours. This is simply k = t/3.
Therefore, the number of bacteria after waiting t hours is given by
N(t) = 2000  2t/3
This situation is an example of exponential growth.

60. Example: Exponential Growth
The graph of N(t) = 2000  2t/3 is shown below.

61. Example: Exponential Growth
Observe that the exponent in the formula N(t) = 2000  2t/3 is t/3 and carries no units. This because the exponent is just counting the number k of three-hour periods in t hours.
The units of t are [hours] and the units of the denominator, 3, are also [hours]. Therefore t/3 is dimensionless.
On the other hand, the factor 2000 has the units [bacteria]. Thus N(t) has the correct units, [bacteria].

62. Example: Exponential Growth
Now we can calculate N(t) at any time t.
For instance, the number of bacteria present after 11 hours is indicated by N(11) and its value is given by
N(11) = 2000  211/3 =
which is round it up to N(11) = bacteria

63. Example 1.14 – Purifying Water
Water that is initially contaminated with a concentration of 9 milligrams of pollutant per liter of water is subjected to a cleaning process. The cleaning process is able to reduce the pollutant concentration by 25% every 2 hours. Let C = C(t) denote the concentration, in milligrams per liter, of pollutant in the water t hours after the purification process begins.
Part 1: Find a formula for the concentration C = C(t).
Part 2: What is the concentration of pollutant in the water after 3 hours?
Part 3: Find the concentration of pollutant after hours 4 hours and 15 minutes of cleaning.

64. Example 1.14 – Purifying Water
cont’d
Part 1: The statement “the cleaning process is able to reduce the pollutant concentration by 25% every 2 hours” means that, after waiting 2 hours, the remaining concentration is 3/4 of the previous one (two hours ago). Thus,
C(0) = 9  (3/4)0 (after waiting 0 two-hour periods)
C(2) = 9  (3/4)1 (after waiting 1 two-hour periods)
C(4) = 9  (3/4)2 (after waiting 2 two-hour periods)
C(6) = 9  (3/4)3 (after waiting 3 two-hour periods)
Thus, after waiting k two-hour periods the concentration is given by
C(2k) = = 9  (3/4)k

65. Example 1.14 – Purifying Water
cont’d
Now, knowing that after waiting k two-hour periods the concentration is given by
C(2k) = 9  (3/4)k
to get the concentration after waiting t hours, all we have to do is count the number of k two-hour periods in t hours. This is simply k = t/2.
Therefore, the concentration after waiting t hours is given by
C(t) = 9  (3/4)t/2 = 9  0.75 t/2

66. Example: Exponential Growth
The graph of C(t) = 9  (3/4)t/2 = 9  0.75 t/2 is shown below.

67. Example 1.14 – Solution
Part 2: What is the concentration of pollutant in the water after 3 hours?
C(3) = 9  /2 = 5.85 milligrams/liter
Part 3: Find the concentration of pollutant after hours 4 hours and 15 minutes of cleaning.
C(4.25) = 9  /2 = 4.88 milligrams/liter

68. Example: Exponential Growth
A certain population has a yearly growth rate of 6.7% and the
initial population value is 2 million. Express using functional
notation the population after 6 years, and then calculate that
value in millions (round to the nearest hundredth).
a) N(6) = 5.22 million
b) N(6) = 2.32 million
c) N(6) = 2.95 million
d) N(2) = 5.22 million
e) N(2) = 2.95 million

69. Compound Interest
Exponential functions also occur in calculating compound interest.
If an amount of money P, called the principal, is invested at an interest rate i per time period, then after one time period the interest is Pi, and the amount A of money is
A = P + Pi = P(1 + i)
If the interest is reinvested, then the new principal is P(1 + i), and the amount after another time period is
A = P(1 + i)(1 + i) = P(1 + i)2.

70. Compound Interest Similarly, after a third time period the amount is
A = P(1 + i)3
In general, after k periods the amount is
A = P(1 + i)k
Notice that this is an exponential function with base 1 + i.
If the annual interest rate is r and if interest is compounded n times per year, then in each time period the interest rate is i = r /n, and there are nt time periods in t years.

71. Compound Interest
This leads to the following formula for the amount after t years.

72. Example – Calculating Compound Interest
A sum of $1000 is invested at an interest rate of 12% per year. Find the amounts in the account after 3 years if interest is compounded annually, semiannually, quarterly, monthly, and daily.
Solution:
We use the compound interest formula with P = $1000, r = 0.12, and t = 3.

73. Example – Calculating Compound Interest
We see from the previous example that the interest paid increases as the number of compounding periods n increases. Let’s see what happens as n increases indefinitely. If we let m = n/r, then

74. Example – Calculating Compound Interest
It is now easy to see that as m increases, the expression
approaches the number e.
For m = 365,
For m = 10000,

75. Continuously Compounded Interest
This leads to the following formula for the amount after t years.

76. Continuously Compounded Interest
Example: Find the accumulated amount of money after 25 years if $7500 is invested at 12% per year compounded continuously.