FUNCTIONS, ALGEBRA AND MODELS
LRT
01/30/2017
1. Functions
In this course a function is a rule which assigns to each number
in some subset of the real numbers another real number. The set of
numbers that you can input is called the domain and the set of numbers
that you get out is called the range.
There are several ways that functions will be given to us.
1. The rule will be described in words.
2. The rule will be given by a formula (and possibly 1 as well).
• Piecewise definition.
3. The rule will be given by a table (and almost certainly 1 as
well).
4. The rule will be given by a graph (and possibly 1 as well).
1.1. Graphs. A curve is a 1-dimensional subset of the plane. We will
not make this precise but you’ll know it when you see it. A typical
source of curves is the solution set to an equation f (x, y) = 0.
y 2 = x3 − 1.883x2 + 1
is the curve from the home page on
the web site.
A more familiar set of examples are the circles (x−a)2 +(y−b)2 = r2 .
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Here is an example of a verbally defined function with a piecewise
linear verbal formula, namely the current IRS tax table for filing singly.
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1.2. Graphs of functions. The most important set of examples for
us will be graphs of the form y = f (x) where f is some function. Such
a graph consists of points of the form x, f (x) .
A class of
examples that we talked about earlier are the lines,
x, mx + b .
Graphs of functions have a property not shared by all curves:
A curve is the graph of a function if and only if each
vertical line intersects the curve in at most one point.
The domain of the function is the set of all points a so that the
vertical line at a intersects the curve in precisely one point. The value
of the function at a is b where (a, b) is the point on the curve.
2. Kinds of problems
Given a verbal description of a function, find a formula. Area of a
circle given the radius.
Find the domain of a function. The domain of a formula is the largest
subset of real numbers which can be plugged into the formula (unless
told otherwise).
Find the range of a function.
• Hard from a formula.
• Easier from a graph.
Graph functions by plotting points. (We will get better methods for
this problem later.)
Define function from a graph.
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3. Algebra of functions
•
•
•
•
•
Sum
Difference
Product
Quotient
Composition
3.1. Examples.
• Constant functions.
• The identity function.
• Polynomial functions.
• Rational functions.
• Power functions.
– Exponents
– Radicals
• Katie-bar-the-door.
4. Supply and demand
One application of the material thus far is to supply and demand.
In a market, there are two forces at work. Suppliers supply a good or
service and consumers purchase them.
The demand function expresses the relationship between the unit
price of the good or service and the amount per unit of time that the
consumers will demand at that price.
The supply function expresses the relationship between the amount
of the good or service per unit of time that the suppliers will offer at
each unit price.
There will be two versions of each of these functions. One version,
the one discussed in this section, takes as input the amount of goods
or service and yields as output the unit price.
If x represents the amount of goods or services per unit time and if p
represents the unit price then the demand function is written p = d(x).
The supply function is written p = s(x).
The main problem to be solved is that given a supply and a demand
function for a market to find the market equilibrium. The price where
the equilibrium occurs is called the equilibrium price and the quantity
is called the equilibrium quantity. By definition, the market equilibrium
FUNCTIONS, ALGEBRA AND MODELS
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is a point where the demand function and the supply function agree.
In other words:
A market equilibrium point is a point where d(x) = s(x) .
4.1. Demand functions. Typically, a demand function p = d(x) has
a finite interval as domain as does x = d(p). As p increases, the quantity
consumers are willing to purchase so x = d(p) is a decreasing function
of p. It follows that p = d(x) is also decreasing.
p
x
4.2. Supply functions. Typically, a supply function p = s(x) has
a finite interval as domain as does x = s(p). As the price goes up,
the amount suppliers will produce tends to go up, so x = s(p) is an
increasing function. It follows that x = s(p) is also increasing.
p
x
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4.3. Market equilibrium point. Here is a graphic illustration of a
market equilibrium.
p
•
x
5. Supply functions in business
If you own a business with many suppliers, you have little control
over the consumer demand function. But you do have your own supply
function: you will produce x units per unit time if the unit price is p
where the unit price is determined by the market over which you have
no control because you are such a small part of the overall market.
This function comes up when you go to compute revenue, R, for your
business. If your supply function is p = s(x), then R(x) = x · s(x). If
x = s(p) then R(p) = p · s(p).
You often want to know x because your cost function is typically a
function of x, C(x) and hence your profit function P (x) = x · s(x) −
C(x).
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6. Modeling problems
I would add: read the problem in several steps.
Pass 0. What are we being asked to do? If it is to construct a mathematical model, then proceed as below.
Pass 1. What is the quantity for which we trying to find a function?
Pass 2. What does this quantity depend on?
Pass 3. If your answer in 2 contains several variables, eliminate all but
one of them using algebra and further relations present in the
problem. The problem itself may specify which variable to use.
Pass 4. Check for any domain restrictions on your chosen variable.
Pass 5. Be sure you answer the question that was actually asked.
6.1. Example. Find the perimeter of a rectangle given that the area
is 7ft2 .
Let L be the perimeter. What does the perimeter depend on? Well,
the whole rectangle is determined when you give the width w and the
height h.
The perimeter is given by the formula L = 2w + 2h.
We have two variables, w and h, but we only want one. But we have
one more bit of information: the area A = 7. But the area is also given
by A = w · h so 7 = w · h.
There are two solutions to this problem depending on whether you
solve 7 = w · h for w or for h. The two answers are
14
L =2w +
(1)
w
14
(2)
L =2h +
h
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14
w
14
(2)
L =2h +
h
To get a well-defined problem we would say something like
Find the perimeter of a rectangle in terms of the width
given that the area is 7ft2 .
Then the correct answer is (1).
(1)
L =2w +
Jeopardy question: What question has (2) as the correct answer?
Further issues: What are the units for w, h, L and 14?
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7. More problems
Drug Dosages
This is not a modeling problem at all, despite coming from the section
in the book on modeling problems.
In the last sentence we are asked to compute that dosage for a 0.4 m2
child, given that the adult dosage is 500 mg.
Reading the second sentence we see a = 500 and S = 0.4. Then
(0.4)(500)
D(S) =
≈ 117.6470588235 mg
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Revenue Cost Profit
Here we are building three models, one for cost, one for revenue and
one for profit. The answer to d is found by plugging numbers into the
profit function.
Cost: From sentence 1, C(x) = 100, 000 + 14x
Revenue: From sentence 2, T (x) = 20x
Profit: P (x) = R(x)−C(x) = 20x− 100, 000+14x = 20x−100, 000−
14x = 6x − 100, 000.
For part d proceed as follows. When x = 12, 000, P (12, 000) =
6 · 12, 000 − 100, 000 = −28, 000. You are losing $28, 000 dollars each
for each unit of time.
When x = 20, 000, P (20, 000) = 6 · 20, 000 − 100, 000 = 20, 000. You
are making $20, 000 dollars each for each unit of time.
As for comparing these two answers, try real hard to make 20, 000
units.
You might also want to solve P (x) = 0 since this is the point
100, 000
where you break even. Hence 6x − 100, 000 = 0 so x =
≈
6
16, 666.6666666667 so you start to make money when you produce
16, 667 units and your profit increases steadily after that.
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Linear Depreciation
We want to derive a formula for linear depreciation. Let D(t) denote
the depreciated value at time t, 0 6 t 6 n. Since the formula is suppose
to be linear, the equation we get should be the equation of a line. Hence
we need a slope and a point.
By sentence 3, the line goes through the point (0, C) and the point
(n, S). At this point you should realize you are done.
C −S
S−C
=
0−n
n
There are two way to continue depending on if we want the line to
go through the first point or the second.
S−C
(1)
t
D − C =m(t − 0) = m · t =
n
S−C
D − S =m(t − n) =
(2)
(t − n)
n
The slope-intercept form for (1) is
S−C
D=
t+C
n
The slope-intercept form for (2) is
S−C
S−C
S−C
S−C
D=
(t − n) + S =
t−
n+S =
t−S +C +S
n
n
n
n
so either (1) or (2) gives the same slope-intercept form.
m=
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LRT 01/30/2017
Market Equilibrium
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1
You need to solve 144−x2 = p = 48+ x2 so x2 + x2 = 144−48 = 96
2
2
3 2
1 2
or x = 96. Divide by 3 so x = 32 and then multiply by 2 so x2 = 64
2
2
and x = ±8. The solution x = −8 makes no sense in the context of
this problem so the equilibrium quantity is x = 8.
The equilibrium price is
p = 144 − 82 = 144 − 64 = 80
or
1
p = 48 + 82 = 48 + 32 = 80
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