Basic Calculus for Economists

1
Lecture 5

• Basic Calculus for Economists

2
Analyzing a Limit
We can examine what occurs at a particular point
by the limit. Using the function f (x) 2x 1,
lets examine what happens near x 2 through
the following chart
x 1.5 1.9 1.99 1.999 2 2.001 2.01 2.1 2.5
f (x) 2 2.8 2.98 2.998 ? 3.002 3.02 3.2 4
We see that as x approaches 2, f (x) approaches
3.
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Limits
In limit notation we have
3
Definition We write
2
or as x ? c, then f (x) ? L, if the
functional value of f (x) is close to the single
real number L whenever x is close to, but not
equal to, c (on either side of c).
4
One-Sided Limits

• We write
• and call K the limit from the left (or
  left-hand limit) if f (x) is close to K whenever
  x is close to c, but to the left of c on the
  real number line.
• We write
• and call L the limit from the right (or
  right-hand limit) if f (x) is close to L whenever
  x is close to c, but to the right of c on the
  real number line.
• In order for a limit to exist, the limit from the
  left and the limit from the right must exist and
  be equal.

5
Limit Properties

• Let f and g be two functions, and assume that
  the following two limits exist and are finite
• Then
• the limit of the sum of the functions is equal
  to the sum of the limits.
• the limit of the difference of the functions is
  equal to the difference of the limits.

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Limit Properties(continued)

• the limit of a constant times a function is equal
  to the constant times the limit of the function.
• the limit of the product of the functions is the
  product of the limits of the functions.
• the limit of the quotient of the functions is the
  quotient of the limits of the functions, provided
  M ? 0.
• the limit of the nth root of a function is the
  nth root of the limit of that function.

7
Examples
From these examples we conclude that
f any polynomial function r any rational
function with a nonzero denominator at x c
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Indeterminate Forms
It is important to note that there are
restrictions on some of the limit properties. In
particular if
then finding may present difficulties, since
the denominator is 0.
If and
, then is said to be indeterminate. The term
indeterminate is used because the limit may or
may not exist.
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Example
This example illustrates some techniques that can
be useful for indeterminate forms.
Algebraic simplification is often useful when the
numerator and denominator are both approaching 0.
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Difference Quotients
Let f (x) 3x - 1. Find Solution
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Definition of Continuity
A function f is continuous at a point x c if
1. 2. f (c) exists 3. A function f is
continuous on the open interval (a,b) if it is
continuous at each point on the interval. If a
function is not continuous, it is discontinuous.
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Example 1
f (x) x 1 at x 2. 1.
The limit exists! 2. f(2)
1 3. Therefore this function is continuous at x
2.
1
2
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Example 2
f (x) (x2 9)/(x 3) at x
-31.
The limit exists (reduce the fraction). 2. f
(-3) 0/0 is undefined! 3. The function is
not continuous at x -3. (Graph should have an
open circle there.)
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Example 3
f (x) x/x at x 0 and at x
1. 1. Does not
exist! 2. f (0) 0/0 Undefined! 3. The
function is not continuous at x 0. This
function is continuous at x 1.
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Limits in Mathematica
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Continuity Properties
If two functions are continuous on the same
interval, then their sum, difference, product,
and quotient are continuous on the same interval,
except for values of x that make the denominator
0.
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Examples of Continuous Functions

• A constant function is continuous for all x.
• For integer n gt 0, f (x) xn is continuous for
  all x.
• A polynomial function is continuous for all x.
• A rational function is continuous for all x,
  except those values that make the denominator 0.
• For n an odd positive integer, is
  continuous wherever f (x) is continuous.
• For n an even positive integer, is
  continuous wherever f (x) is continuous and
  nonnegative.

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Infinite Limits
There are various possibilities under which does
not exist. For example, if the one-sided limits
are different at x a, then the limit does not
exist. Another situation where a limit may fail
to exist involves functions whose values become
very large as x approaches a. The special symbol
? (infinity) is used to describe this type of
behavior.
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To illustrate this case, consider the function f
(x) 1/(x-1), which is discontinuous at x 1.
As x approaches 1 from the right, the values of f
(x) are positive and become larger and larger.
That is, f (x) increases without bound. We write
this symbolically as Since ? is not a real
number, the limit above does not actually exist.
We are using the symbol ? (infinity) to describe
the manner in which the limit fails to exist, and
we call this an infinite limit.
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Example(continued)
As x approaches 1 from the left, the values of f
(x) are negative and become larger and larger in
absolute value. That is, f (x) decreases through
negative values without bound. We write this
symbolically as
The graph of this function is as shown Note that
does not exist.
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Infinite Limits and Vertical Asymptotes
Definition The vertical line x a is a vertical
asymptote for the graph of y f (x) if f (x)
? ? or f (x) ? -? as x ? a or x ? a. That
is, f (x) either increases or decreases without
bound as x approaches a from the right or from
the left. Note If any one of the four
possibilities is satisfied, this makes x a a
vertical asymptote. Most of the time, the limit
will be infinite ( or -) on both sides, but it
does not have to be.
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Vertical Asymptotesof Polynomials
How do we locate vertical asymptotes? If a
function f is continuous at x a, then
Since all of the above limits exist and are
finite, f cannot have a vertical asymptote at x
a. In order for f to have a vertical
asymptote at x a, at least one of the limits
above must be an infinite limit, and f must be
discontinuous atx a. We know that polynomial
functions are continuous for all real numbers, so
a polynomial has no vertical asymptotes.
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Vertical Asymptotes of Rational Functions
Since a rational function is discontinuous only
at the zeros of its denominator, a vertical
asymptote of a rational function can occur only
at a zero of its denominator. The following is a
simple procedure for locating the vertical
asymptotes of a rational function If f (x)
n(x)/d(x) is a rational function, d(c) 0 and
n(c) ? 0, then the line x c is a vertical
asymptote of the graph of f. However, if both
d(c) 0 and n(c) 0, there may or may not be
a vertical asymptote at x c.
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Example
Let Describe the behavior of f at each point
of discontinuity. Use ? and -? when appropriate.
Identify all vertical asymptotes.
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Example(continued)
Let Describe the behavior of f at each point
of discontinuity. Use ? and -? when appropriate.
Identify all vertical asymptotes. Solution Let
n(x) x2 x - 2 and d(x) x2 - 1. Factoring
the denominator, we see that d(x) x2 - 1
(x1)(x-1) has two zeros, x -1 and x 1.
These are the points of discontinuity of f.
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Example(continued)
Since d(-1) 0 and n(-1) -2 ? 0, the theorem
tells us that the line x -1 is a vertical
asymptote. Now we consider the other zero of
d(x), x 1. This time n(1) 0 and the theorem
does not apply. We use algebraic simplification
to investigate the behavior of the function at x
1
Since the limit exists as x approaches 1, f
does not have a vertical asymptote at x 1. The
graph of f is shown on the next slide.
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Example(continued)
Vertical Asymptote
Point of discontinuity
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Limits at Infinity ofPower Functions
We begin our consideration of limits at infinity
by considering power functions of the form x p
and 1/x p, where p is a positive real number. If
p is a positive real number, then x p increases
as x increases, and it can be shown that there is
no upper bound on the values of x p. We indicate
this by writing or
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Power Functions (continued)
Since the reciprocals of very large numbers are
very small numbers, it follows that 1/x p
approaches 0 as x increases without bound. We
indicate this behavior by writing
or
This figure illustrates this behavior for f (x)
x2 and g(x) 1/x2.
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Power Functions (continued)
In general, if p is a positive real number and k
is a nonzero real number, then
Note k and p determine whether the limit at ? is
? or -?. The last limit is only defined if the
pth power of a negative number is defined. This
means that p has to be an integer, or a rational
number with odd denominator.
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Limits at Infinity of Polynomial Functions
What about limits at infinity for polynomial
functions? As x increases without bound in
either the positive or the negative direction,
the behavior of the polynomial graph will be
determined by the behavior of the leading term
(the highest degree term). The leading term will
either become very large in the positive sense or
in the negative sense (assuming that the
polynomial has degree at least 1). In the first
case the function will approach ? and in the
second case the function will approach -?. In
mathematical shorthand, we write this asThis
covers all possibilities.
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Limits at Infinity andHorizontal Asymptotes
A line y b is a horizontal asymptote for the
graph of y f (x) if f (x) approaches b as
either x increases without bound or decreases
without bound. Symbolically, y b is a
horizontal asymptote if In the first case, the
graph of f will be close to the horizontal line
y b for large (in absolute value) negative x.
In the second case, the graph will be close to
the horizontal line y b for large positive
x. Note It is enough if one of these conditions
is satisfied, but frequently they both are.
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Example
This figure shows the graph of a function with
two horizontal asymptotes, y 1 and y -1.
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Horizontal Asymptotes of Rational Functions
If then

• There are three possible cases for these limits.
• If m lt n, then The line y 0 (x axis) is a
  horizontal asymptote for f (x).
• 2. If m n, then The line y am/bn is a
  horizontal asymptote for f (x) .
• 3. If m gt n, f (x) does not have a horizontal
  asymptote.

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Horizontal Asymptotes of Rational Functions
(continued)
Notice that in cases 1 and 2 on the previous
slide that the limit is the same if x approaches
? or -?. Thus a rational function can have at
most one horizontal asymptote. (See figure).
Notice that the numerator and denominator have
the same degree in this example, so the
horizontal asymptote is the ratio of the leading
coefficients of the numerator and denominator.
y 1.5
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Example
Find the horizontal asymptotes of each function.
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Example Solution
Find the horizontal asymptotes of each function.
Since the degree of the numerator is less than
the degree of the denominator in this example,
the horizontal asymptote is y 0 (the x axis).
Since the degree of the numerator is greater than
the degree of the denominator in this example,
there is no horizontal asymptote.
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Summary

• An infinite limit is a limit of the form(y
  goes to infinity). It is the same as a vertical
  asymptote (as long as a is a finite number).
• A limit at infinity is a limit of the form(x
  goes to infinity). It is the same as a horizontal
  asymptote (as long as L is a finite number).

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The Rate of Change
For y f (x), the average rate of change from x
a to x a h is
The above expression is also called a difference
quotient. See Chiang 6.1. It can be interpreted
as the slope of a secant. See the picture on the
next slide for illustration.
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Visual Interpretation
Q
slope
f (a h) f (a)
Average rate of change slope of the secant line
through P and Q
P
h
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Example 1
The revenue generated by producing and selling
widgets is given by R(x) x (75 3x) for 0 ? x
? 20. What is the change in revenue if production
changes from 9 to 12?
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Example 1
The revenue generated by producing and selling
widgets is given by R(x) x (75 3x) for 0 ? x
? 20. What is the change in revenue if production
changes from 9 to 12? R(12) R(9) 468 432
36. Increasing production from 9 to 12 will
increase revenue by 36.
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Example 1 (continued)
The revenue is R(x) x (75 3x) for 0 ? x ?
20. What is the average rate of change in revenue
(per unit change in x) if production changes from
9 to 12?
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Example 1 (continued)
The revenue is R(x) x (75 3x) for 0 ? x ?
20. What is the average rate of change in revenue
(per unit change in x) if production changes from
9 to 12? To find the average rate of change we
divide the change in revenue by the change in
production Thus the average change in revenue
is 12 when production is increased from 9 to
12. Like Change in Y over change in X.
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The Instantaneous Rate of Change
Consider the function y f (x) only near the
point P (a, f (a)). The difference quotient
gives the average rate of change of f over the
interval a, ah. If we make h smaller and
smaller, in the limit we obtain the instantaneous
rate of change of the function at the point P
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Visual Interpretation
Q
Tangent
Slope of tangent instantaneous rate of change.
f (a h) f (a)
P
Let h approach 0
h
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Instantaneous Rate of Change
Given y f (x), the instantaneous rate of change
at x a is provided that the limit exists. It
can be interpreted as the slope of the tangent at
the point (a, f (a)). See illustration on
previous slide.
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The Derivative
For y f (x), we define the derivative of f at
x, denoted f (x), to be
if the limit exists. If f (a) exists, we call
f differentiable at a. If f (x) exist for each
x in the open interval (a, b), then f is said
to be differentiable over (a, b).
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Interpretations of the Derivative

• If f is a function, then f is a new
  function with the following interpretations
• For each x in the domain of f , f (x) is the
  slope of the line tangent to the graph of f at
  the point (x, f (x)).
• For each x in the domain of f , f (x) is the
  instantaneous rate of change of y f (x) with
  respect to x.
• If f (x) is the position of a moving object at
  time x, then v f (x) is the velocity of the
  object at that time.

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Finding the Derivative
To find f (x), we use a four-step
process Step 1. Find f (x h) Step 2. Find
f (x h) f (x) Step 3. Find Step 4. Find
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Example 2
Find the derivative of f (x) x 2 3x. Step 1.
f (x h) (x h)2 3(x h) x2 2xh h2
3x 3h Step 2. Find f (x h) f (x) 2xh
h2 3h Step 3. Find Step 4. Find
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Example 3
Find the slope of the tangent to the graph of f
(x) x 2 3x at x 0, x 2, and x
3. Solution In example 2 we found the derivative
of this function at x to be f (x) 2x
3 Hence f (0) -3 f (2) 1, and f
(3) 3
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Example 4
We know that the derivative of f (x) x 2 3x
is f (x) 2x 3. Verify this for x 2 using
Mathematica. D is the derivative function
tangent equation
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Example 5
Find the derivative of f (x) 2x 3x2 using
Mathematica with a symbolic differentiation
routine.
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Example 6
Find the derivative of f (x) 2x 3x2 using
the four-step process. Step 1. f (x h) 2(x
h) 3(x h)2 Step 2. f (x h) f (x) 2h
6xh - 3h2 Step 3. Step 4.
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Nonexistence of the Derivative
The existence of a derivative at x a depends on
the existence of the limit
If the limit does not exist, we say that the
function is nondifferentiable at x a, or f
(a) does not exist.
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Nonexistence of the Derivative(continued)

• Some of the reasons why the derivative of a
  function may not exist at x a are
• The graph of f has a hole or break at x a,
  or
• The graph of f has a sharp corner at x a, or
• The graph of f has a vertical tangent at x a.

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Summary

• For y f (x), we defined the derivative of f
  at x, denoted f (x), to be
• if the limit exists.
• We have seen how to find the derivative
  algebraically, using the four-step process.

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Derivative Notation

• In the preceding section we defined the
  derivative of a function. There are several
  widely used symbols to represent the derivative.
  Given y f (x), the derivative may be
  represented by any of the following
• f (x)
• y
• dy/dx

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Example 1 (continued)
What is the slope of a constant function?
The graph of f (x) C is a horizontal line with
slope 0, so we would expect f (x) 0.
Theorem 1. Let y f (x) C be a constant
function, then y f (x) 0.
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Power Rule
A function of the form f (x) xn is called a
power function. This includes f (x) x (where n
1) and radical functions (fractional n).
Theorem 2. (Power Rule) Let y xn be a power
function, then y f (x) n xn 1.
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Example 2
Differentiate f (x) x5. Solution By the
power rule, the derivative of xn is n xn1. In
our case n 5, so we get f (x) 5 x4.
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Example 3
Differentiate Solution Rewrite f (x) as a
power function, and apply the power rule
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Constant Multiple Property
Theorem 3. Let y f (x) k? u(x) be a
constant k times a function u(x). Then
y f (x) k ? u(x). In words The
derivative of a constant times a function is the
constant times the derivative of the function.
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Example 4
Differentiate f (x) 7x4. Solution Apply
the constant multiple property and the power
rule. f (x) 7?(4x3) 28 x3.
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Sum and Difference Properties

• Theorem 5. If
• y f (x) u(x) v(x),
• then
• y f (x) u(x) v(x).
• In words
• The derivative of the sum of two differentiable
  functions is the sum of the derivatives.
• The derivative of the difference of two
  differentiable functions is the difference of the
  derivatives.

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Example 5
Differentiate f (x) 3x5 x4 2x3 5x2 7x
4. Solution Apply the sum and difference
rules, as well as the constant multiple property
and the power rule. f (x) 15x4 4x3 6x2
10x 7.
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Applications

• Remember that the derivative gives the
  instantaneous rate of change of the function with
  respect to x. That might be
• Instantaneous velocity.
• Tangent line slope at a point on the curve of
  the function.
• Marginal Cost. If C(x) is the cost function,
  that is, the total cost of producing x items,
  then C(x) approximates the cost of producing one
  more item at a production level of x items. C(x)
  is called the marginal cost.

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Tangent Line Example
Let f (x) x4 - 6x2 10. (a) Find f (x) (b)
Find the equation of the tangent line at x 1
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Tangent Line Example (continued)

• Let f (x) x4 - 6x2 10.
• (a) Find f (x)
• (b) Find the equation of the tangent line at x
  1
• Solution
• f (x) 4x3 - 12x
• Slope f (1) 4(13) - 12(1) -8.Point If
  x 1, then y f (1) 1 - 6 10 5.
  Point-slope form y - y1 m(x -
  x1) y - 5 -8(x -1) y -8x 13

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Application Example

• The total cost (in dollars) of producing x
  portable radios per day is
• C(x) 1000 100x 0.5x2
• for 0 x 100.
• Find the marginal cost at a production level of x
  radios.

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Example (continued)

• The total cost (in dollars) of producing x
  portable radios per day is
• C(x) 1000 100x 0.5x2
• for 0 x 100.
• Find the marginal cost at a production level of x
  radios.
• Solution The marginal cost will be
• C(x) 100 x.

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Example (continued)

• Find the marginal cost at a production level of
  80 radios and interpret the result.
• Solution C(80) 100 80 20.
• It will cost approximately 20 to produce the
  81st radio.
• Find the actual cost of producing the 81st radio
  and compare this with the marginal cost.
• Solution The actual cost of the 81st radio will
  be
• C(81) C(80) 5819.50 5800 19.50.
• This is approximately equal to the marginal cost.

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Example (continued) for you
4. If the price of the radio is 11.00, how many
radios do you decide to produce? 5. Calculate the
firms total profits. 6. Calculate the firms
break-even point 7. Show and plot the firms
profit function
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Summary

• If f (x) C, then f (x) 0
• If f (x) xn, then f (x) n xn-1
• If f (x) k?u(x), then f (x) k?u(x)
• If f (x) u(x) v(x), then f (x) u(x)
  v(x).

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Increments
In a previous section we defined the derivative
of f at x as the limit of the difference
quotient
Increment notation will enable us to interpret
the numerator and the denominator of the
difference quotient separately.
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Example
Let y f (x) x3. If x changes from 2 to
2.1, then y will change from y f (2) 8 to y
f (2.1) 9.261. We can write this using
increment notation. The change in x is called the
increment in x and is denoted by ?x. ? is the
Greek letter delta, which often stands for a
difference or change. Similarly, the change in y
is called the increment in y and is denoted by
?y. In our example, ?x 2.1 2 0.1 ?y
f (2.1) f (2) 9.261 8 1.261.
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Graphical Illustration of Increments
For y f (x) ?x x2 - x1 ?y y2 - y1 x2
x1 ?x f (x2) f (x1) f (x1 ?x) f
(x1)
(x2, f (x2))

• ?y represents the change in y corresponding to a
  ?x change in x.
• ?x can be either positive or negative.

?y
(x1, f (x1))
x1
x2
?x
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Differentials
Assume that the limit
exists. For small ?x, Multiplying both
sides of this equation by ?x gives us ?y ? f
(x) ?x. Here the increments ?x and ?y represent
the actual changes in x and y.
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Differentials (continued)
One of the notations for the derivative is If we
pretend that dx and dy are actual quantities, we
get We treat this equation as a definition, and
call dx and dy differentials.
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Interpretation of Differentials
?x and dx are the same, and represent the
change in x. The increment ?y stands for the
actual change in y resulting from the change in
x. The differential dy stands for the approximate
change in y, estimated by using derivatives. In
applications, we use dy (which is easy to
calculate) to estimate ?y (which is what we want).
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Example 1
Find dy for f (x) x2 3x and evaluate dy for
x 2 and dx 0.1. Solution dy f (x) dx
(2x 3) dx When x 2 and dx 0.1, dy 2(2)
3 0.1 0.7.
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Example 2 Cost-Revenue
A company manufactures and sells x transistor
radios per week. If the weekly cost and revenue
equations are
find the approximate changes in revenue and
profit if production is increased from 2,000 to
2,010 units/week.
85
Example 2 Solution
The profit is We will approximate ?R and ?P
with dR and dP, respectively, using x 2,000 and
dx 2,010 2,000 10.
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Marginal Cost
Remember that marginal refers to an instantaneous
rate of change, that is, a derivative.
Definition If x is the number of units of a
product produced in some time interval, then
Total cost C(x) Marginal cost C(x)
87
Marginal Revenue andMarginal Profit
Definition If x is the number of units of a
product sold in some time interval, then Total
revenue R(x) Marginal revenue R(x) If
x is the number of units of a product produced
and sold in some time interval, then Total
profit P(x) R(x) C(x) Marginal profit
P(x) R(x) C(x)
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Marginal Cost and Exact Cost
Assume C(x) is the total cost of producing x
items. Then the exact cost of producing the (x
1)st item is C(x 1) C(x). The marginal
cost is an approximation of the exact cost.
C(x) C(x 1) C(x). Similar statements
are true for revenue and profit.
89
Example

• The total cost of producing x electric guitars is
  C(x) 1,000 100x 0.25x2.
• Find the exact cost of producing the 51st guitar.
  
• The exact cost is C(x 1) C(x).
• C(51) C(50) 5,449.75 5375 74.75.
• Use the marginal cost to approximate the cost of
  producing the 51st guitar.
• The marginal cost is C(x) 100 0.5x
• C(50) 75.

90
Marginal Average Cost
Definition If x is the number of units of a
product produced in some time interval,
then Average cost per unit Marginal average
cost
91
Marginal Average Revenue Marginal Average Profit

• If x is the number of units of a product sold in
  some time interval, then
• Average revenue per unit Marginal average
  revenue
• If x is the number of units of a product produced
  and sold in some time interval, then
• Average profit per unit Marginal average
  profit

92
Warning!
To calculate the marginal averages you must
calculate the average first (divide by x), and
then the derivative. If you change this order you
will get no useful economic interpretations.
STOP
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Example 2
The total cost of printing x dictionaries is
20,000 to start your business, and there are 10
in variable costs Thus, C(x) 20,000
10x 1. Find the average cost per unit if 1,000
dictionaries are produced.
94
Example 2 (continued)
The total cost of printing x dictionaries is
C(x) 20,000 10x 1. Find the average cost
per unit if 1,000 dictionaries are produced.
30
95
Example 2 (continued)

• Find the marginal average cost at a production
  level of 1,000 dictionaries, and interpret the
  results.
• Marginal average cost

This means that if you raise production from
1,000 to 1,001 dictionaries, the price per book
will fall approximately 2 cents.
96
Example 2 (continued)
3. Use the results from above to estimate the
average cost per dictionary if 1,001 dictionaries
are produced. Average cost for 1000 dictionaries
30.00 Marginal average cost - 0.02 The
average cost per dictionary for 1001 dictionaries
would be the average for 1000, plus the marginal
average cost, or 30.00 (- 0.02) 29.98
97
Example 3

• The price-demand equation and the cost function
  for the production of television sets are given
  by
• where x is the number of sets that can be sold at
  a price of p per set, and C(x) is the total cost
  of producing x sets.
• Find the marginal cost.
• Solution The marginal cost is C(x) 30.

98
Example 3 (continued)

• Find the revenue function in terms of x.
• The revenue function is
• 3. Find the marginal revenue.
• The marginal revenue is
• Find R(1500) and interpret the results.
• At a production rate of 1,500, each additional
  set increases revenue by approximately 200.

99
Example 3 (continued)
5. Graph the cost function and the revenue
function on the same coordinate. Find the
break-even point.
0 lt x lt 9,000
0 lt y lt 700,000
100
Example 3 (continued)
5. Graph the cost function and the revenue
function on the same coordinate. Find the
break-even point.
0 lt x lt 9,000
R(x)
0 lt y lt 700,000
Solution There are two break-even points.
C(x)
(600,168,000)
(7500, 375,000)
101
Example 3 (continued)

• Find the profit function in terms of x.
• The profit is revenue minus cost, so
• Find the marginal profit.
• 8. Find P(1500) and interpret the results.
• At a production level of 1500 sets, profit is
  increasing at a rate of about 170 per set.

102
Example 3 (continued)

1. Find Maximum Profits without finding MR or MC