Question 3: What is function notation?
Problems in business and finance are often mathematical in nature. These problems
come from real-world situations that can be extremely complex. A mathematical model
is a mathematical representation of the situation. Often these representations take the
form of functions.
For instance, businesses operate by obtaining money by the sales of goods and
services. The amount of money obtained through the sales of goods and services is
called revenue. Revenue is modeled by multiplying the quantity a good or service by the
price of each unit of the good or service. We can write this model in mathematical form
by writing
revenue  price per unit  quantity This representation allows the revenue to be calculated if the price and quantity are
known. If we use letters to represent the quantities in the problem, we might write
R  PQ where R is the revenue obtained from selling Q units of a good or service at a price of P
per unit. As long as we know what each letter represents in the model, we can use it to
calculate any one of the three quantities as long as we have the other two quantities.
In many problems involving revenue, the price is fixed at some value and we are
interested in knowing how the revenue R changes as the quantity Q changes. In this
case, we know that Q is the independent variable and P is a constant. Function notation
helps us to indicate that Q is the independent variable. In function notation, a name is
given to the function such as R. The name of the function is followed by a set of
parentheses enclosing the independent variable.
By writing
R(Q)  PQ we are indicating that Q is a variable and P is a constant.
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The name of the function usually reflects the output or dependent variable in the
problem. We could have just as easily written this function as
Revenue(Q)  PQ This name emphasizes what the output from the function is.
Suppose we know that the price per unit for memory cards is fixed at $10 per unit. With
this information, we can form the revenue function based on this price as
R(Q)  10Q Using the function notation, we can specify the revenue at a specific quantity of memory
cards by substituting a number in place of Q. If we want to know the revenue when a
quantity of 200 memory cards is sold, we would write R(200) . The expression R(300)
indicates the revenue when 300 memory cards is sold. These values are calculated
from the function by substituting the appropriate value of Q in the formula on the right
side of the function’s definition:
R (200)  10  200   2000
R (300)  10  300   3000
This means that the revenue from selling 200 memory cards at $10 each is $2000 and
the revenue from selling 300 memory cards at $10 each is $3000.
We can also use function notation to indicate operation with functions. For instance, the
expression R(300)  R(200) represents the difference in revenue between selling 300
memory cards and 200 memory cards.
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Example 7
Revenue Function
Suppose the quantity of memory cards is fixed at 300 and we are
interested in varying the price of each memory card to see its impact on
revenue.
a. Use function notation to define revenue for this application as a
function of the price P.
Solution We need to write a revenue function we’ll call R as a function
of P, R( P) . Using the model for revenue, we can set the quantity equal
to 300 and define
R( P)  300 P
To find the revenue from selling 300 memory cards at a price P,
substitute the price into the function.
b. Use the function from part a to determine the increase in revenue
from increasing the price of memory cards from $10 per card to $12
per card.
Solution To find the change in revenue, we need to calculate the
difference between the revenue at a price of $12 per card and the
revenue at a price of $10 per card. To find this difference, substitute 10
and 12 into the revenue function:
R (10)  300(10)  3000
R (12)  300(12)  3600
The difference is
R(12)  R(10)  3600  3000  600
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This means that increasing the price from $10 to $12 yields $600 in
additional revenue.
In Example 7, the quantity Q of memory cards was fixed and the price P was variable.
For this reason, the revenue function was denoted by R( P) . In Example 8, we’ll apply
function notation to a cost function.
Example 8
Cost Function
Suppose that the cost of memory cards is given by the function
C (Q)  5Q  500,000 dollars where Q is the number of memory cards
produced.
a. Find and interpret C (0) .
Solution To find C (0) , substitute Q  0 into the right side of the
function,
C (0)  5  0   500, 000  500, 000
This means that the cost of producing no memory cards is $500,000.
These costs are called fixed costs and are incurred even though no
cards are produced.
b. Find and interpret C (100)  C (99) .
Solution We’ll start by evaluating the cost function at 99 memory cards
and 100 memory cards:
C (99)  5  99   500, 000  500, 495
C (100)  5 100   500, 000  500,500
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The difference in these costs is found by subtracting the costs,
C (100)  C (99)  500,500  500, 495  5
The cost of producing 100 memory cards is $500,500 and the cost of
producing 99 memory cards is $500,495. Thus the 100th memory card
costs C (100)  C (99) or $5.
c.
Find and interpret C (Q  1)  C (Q) .
Solution If C (Q)  5Q  500,000 , we can find C (Q  1) by replacing Q
with Q + 1,
C (Q  1)  5  Q  1  500, 000
 5Q  5  500, 000
Remove the parentheses
 5Q  500, 005
Simplify
To find the difference in the costs at the two production levels, we’ll
subtract the two formulas:
C (Q  1)  C (Q )  5  Q  1  500, 000   5Q  500, 000
 5Q  5  500, 000  5Q  500, 000
5
This means that the difference in costs between any two consecutive
production levels is $5.
The name of a function is arbitrary, but care needs to be taken so that names are not
confusing. Quantities beginning with the letter p are especially problematic. Two
different economic quantities begin with the letter p, price and profit. To distinguish
between them, we’ll need to name them carefully.
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Profit is the difference between revenue and cost. We can write this mathematically as
profit  revenue  cost
If the amount received from sales is greater than the cost, the profit is positive since the
revenue is greater than the cost. On the other hand, if the costs are greater than the
revenue, the profit is negative.
To name a profit function with an independent variable Q, we might want to write P(Q) .
Although this is perfectly acceptable, the name P might be confused with the variable P
representing price. To avoid this confusion, it would be wise to use the name Profit(Q) .
This function takes the quantity Q of some good or service and outputs the profit at that
production level.
If a word is used to name a function instead of simply a letter, we should probably
continue this pattern with other related function. Instead of R(Q) for the revenue
function, we could use the name Revenue(Q) . Instead of C (Q) for the cost function, we
could use the name Cost(Q) . The names are very descriptive of exactly what the
function does and allow us to write the relationship between these functions as
Profit(Q)  Revenue(Q)  Cost(Q) The name of a function is up to the user. Some textbooks might choose to use an entire
world while others might use a single letter. We’ll use both naming conventions so you
get used to them.
Example 9
Profit Function
The cost of producing robotic hamsters at an Asian manufacturing plant
is Cost(Q)  5Q  750,000 dollars where Q is the number of robotic
hamsters. The revenue from selling the toys is Revenue(Q)  10Q .
a. Find the profit function.
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Solution The profit function is formed by subtracting Cost(Q) from
Revenue(Q) ,
Profit(Q)  10Q   5Q  750, 000 
 5Q  750, 000
b. Find the profit at a production level of 100,000 robotic hamsters.
Solution Substitute Q equal to 100,000 into the profit function,
Profit(Q)  5Q  750,000 , to yield
Profit(100000)  5 100000   750, 000  250000
Since the amount is negative, the manufacturing plant loses $250,000 at
a production level of 100,000 robotic hamsters.
c.
The manufacturing plant breaks even when production is increased
to a level where the profit is equal to $0. Find the production level
where the plant breaks even.
Solution In this part we know the profit and want to find the
corresponding production level. Instead of substituting a value for Q in
the function and computing the profit from Profit(Q) (like in part b), we’ll
set Profit(Q)  0 and solve for Q:
5Q  750, 000  0
5Q  750, 000
Q
750, 000
5
Add 750,000 to both sides
Divide both sides by 5
To break even, the manufacturing plant must produce
750,000
5
or 150,000
robotic hamsters. At higher production levels, the plant makes money.
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