Transcript Math Review - University of Minnesota

Math Review
APEC 3001
Summer 2007
1
Objectives
• Review basic algebraic skills required to successfully solve homework,
quiz, and exam problems in this class.
• Review different ways of describe relationships, including
–
–
–
–
Functions
Tables
Surface Plots
Contour Plots
• Review the special class of linear functions.
– What is the slope?
– What is the intercept?
– Solutions to systems of linear equations.
• Review the basic differential calculus skills required to successfully solve
homework, quiz, and exam problems.
2
Types of Numbers
• Integers
– …,-3, -2, -1, 0, 1, 2, 3,…
• Rational Numbers
– An Integer Divided By An Integer (e.g. ½ = 0.5 & -4/3 = -1.333…)
– All Integers Are Rational Numbers
• Irrational Numbers
– Not Rational (e.g. 2 = 1.4142… &  = 3.1415…)
– Decimals Are Non-Repeating & Non-Terminating
– Any Irrational Number Lies Between Two Rational Numbers
• Real
– Includes All Rational & Irrational Numbers
3
Properties of Real Numbers
(a, b, & c are real numbers)
• Closure: a + b & a  b Are Real Numbers
– 1 + 2 = 3 & 1  2 = 2 Are Real
• Commutative: a + b = b + a & a  b = b  a
– 1+2=2+1=3&12=21=2
• Associative: (a + b) + c = a + (b + c) & (a  b)  c = a  (b  c)
– (1 + 2) + 3 = 1 + (2 + 3) = 6 & (1  2)  3 = 1  (2  3) = 6
• Distributive: a  (b + c) = a  b + a  c & (a + b)  c = a  c + b  c
– 1  (2 + 3) = 1  2 + 1  3 = 5 & (1 + 2)  3 = 1  3 + 2  3 = 9
• Identity: a + 0 = a & a  1 = a
– 1+0=1&11=1
4
Properties of Real Numbers Continued
(a, b, & c are real numbers)
• Inverse: b + (-b) = 0 & b  1/b = 1
– 2 + (-2) = 0 & 2  1/2 = 1
• Cancellation:
– If a + b = a + c, then b = c
– If a  b = a  c, then b = c (provided a is not 0)
• Zero Factor:
– a0=0a=0
– If a  b = 0, the a = 0, b = 0, or a = 0 & b = 0
• Negation:
– -1  (-a) = -(-a) = a
– (-a)  (-b) = a  b
– a  (-b) = (-a)  b = -(a  b)
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Exponents
(a & b are real numbers, m & n are integers)
• Exponent:
– an = a  a  …  a where a is multiplied by itself n times
– 24 = 2  2  2  2 = 16
• Properties of Exponents:
– an  am = an+m
– (an)m = an  m
– an  bn = (a  b)n
(e.g. 42  45 = 42+5 = 47 = 16,384)
(e.g. (42)5 = 42  5 = 410 = 1,048,576)
(e.g. 22  32 = (2  3)2 = 62 = 36)
– a-1 = 1/a
(e.g. 2-1 = ½ = 0.5)
Note that these properties will also hold if m & n are rational numbers,
but we can then end up with complex numbers.
In this class, we will choose rational number for m & n,
but we will never use complex numbers.
6
Functions
• What is a function?
– A function describes a definite relationship.
– For example, the relationship between the area and radius of a circle is y = f(x)
where
• y is the area
• x is the radius, &
• f(x) = x2
• Causal Relationship
– The way a function is written is sometimes meant to imply a causal
relationship.
– For example, y = f(x) means x determines y or x is the independent variable,
while y is the dependent variable.
– In this class, the way we write a function will not always imply causality.
7
Inverse Functions
Many of the functions we will deal with in this class will be invertible.
That is, we can write y = f(x) or x = f -1 (y).
This also implies, we can write y = f(f -1(y)).
For example, if y = f(x) = x2, then x  f
1
y 
y

,
2
 y
  y.
y   

 
8
Other Descriptions of Definite Relationships
Functions like y = f(x) = x2 provide a general and concise way
to describe definite relationships, but they are not the only way .
For example, we could have described the relationship
between the area and radius of a circle using a Table:
Area of Circle
3.14
12.57
28.27
50.27
78.54
Radius of Circle
1.0
2.0
3.0
4.0
5.0
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Other Descriptions of Definite Relationships
Area
y-axis: Often Used for
Dependent Variable
We could have also described the relationship
between the area and radius of a circle using a Figure.
80
70
60
50
40
30
20
10
0
0
1
2
3
4
5
Radius
x-axis: Often Used for Independent Variable
IMPORTANT NOTE: WE WILL NOT ALWAYS STICK TO THIS CONVENTION
10
Could Just As Well Drawn As
5
Radius
4
3
2
1
0
0
20
40
60
80
Area
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More Complex Relationships
• What is a function that describes the number of words in a book?
– T = f(p, n) where
•
•
•
•
T is the total number of words in a book
p is the number of pages in the book
n is the average number of words per page
f(p, n) = pn
• Many of the functions we will look at in this class characterize the
relationship between one variable and many others.
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Again, We Can Use Tables
Average
Number of
Words per
Page (n)
100
200
300
400
500
Number of Pages (p)
50
100
150
200
5,000 10,000 15,000 20,000
10,000 20,000 30,000 40,000
15,000 30,000 45,000 60,000
20,000 40,000 60,000 80,000
25,000 50,000 75,000 100,000
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But, What About Figures?
100,000
75,000
Total (T ) 50,000
500
25,000
300
0
Words Per
Page (n )
100
50
100
150
200
Pages (p )
Surface Plot
Not Very Easy to Draw
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There Has To Be An Easier Way
125,000
100,000
n = 500
Contour Plot
n = 400
75,000
Total (T )
n = 300
50,000
n = 200
25,000
n = 100
0
0
50
100
150
200
250
Pages (p )
Each contour shows the relationship between the total words and
the number of pages holding the words per page constant.
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Alternatively
125,000
p = 200
100,000
Contour Plot
p = 150
75,000
Total (T )
p = 100
50,000
p = 50
25,000
0
0
100
200
300
400
500
600
Words per Page (n )
Each contour shows the relationship between the total words and
the words per page holding the number of pages constant.
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Surface Versus Contour
• Surface plots become virtually impossible to draw once we are describing a
relationship with more than three variables.
• Contour plots are easy to draw regardless of how many variables you have.
– Plot the relationship between two variables holding all the other variables
constant.
– Change one of the constant variables and repeat.
– Repeat again as long as you need to in order to understand the relationship.
IMPORTANT NOTE: AN UNDERSTANDING OF
CONTOUR PLOTS IS ESSENTIAL IN THIS CLASS
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Linear Functions
• Looks Like y = b + mx
– b is referred to as the intercept
• The value of y when x = 0
– m is referred to as the slope
•
•
•
•
•
How much y changes as x increases by 1
m = y/x where y is the change in y (the rise) & x is the change in x (the run)
m > 0 implies a positive relationship
m < 0 implies a negative relationship
m = 0 implies no relationship
– Inverse is x = - (b/m) + (1/m)y
• - (b/m) is the intercept for the inverse function
• 1/m is the slope for the inverse function
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Graphical Example
y = b + mx = 10 + 2.5x
40
30
y
y = 10
20
x = 4
10
m = y/x = 10/4 = 2.5
0
0
2
4
6
8
10
x
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Some More Examples
40
y = 10 + 2.5x
y
30
y = 20 + 0 x
20
y = 30 - 1.5x
10
0
0
2
4
6
8
10
x
20
Solving Two Linear Functions
• Suppose we have two linear equations:
– y = b1 + m1x
– y = b2 + m2x
• Question:
– Is there a value for x that makes y the same for both functions?
• What if we take two functions from the previous slide:
– y = 10 + 2.5x
– y = 30 - 1.5x
• What is the answer to our question for these two functions?
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If y is equal for both equations when x = x*,
then we know
10 + 2.5x* = 30 - 1.5x*
10 - 10 + 2.5x* = 30 - 10 - 1.5x*
2.5x* = 20 - 1.5x*
2.5x* + 1.5x* = 20 - 1.5x* + 1.5x*
4x* = 20
4x* / 4 = 20 / 4
x* = 5
Is this right?
y* = 10 + 2.5  5 = 22.5
y* = 30 - 1.5  5 = 22.5
So, it would seem!
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Does this work in general?
b1 + m1x* = b2 + m2x*
b1 – b1 + m1x* = b2 – b1 + m2x*
m1x* = b2 – b1 + m2x*
m1x* - m2x* = b2 – b1 + m2x* - + m2x*
(m1 - m2)x* = b2 – b1
If (m1 - m2)  0, then
(m1 - m2)x* / (m1 - m2) = (b2 – b1) / (m1 - m2)
x* = (b2 – b1) / (m1 - m2)
Is this right?
y*  b1  m1
b2  b1
m b  m2b1 m1b2  m1b1 m1b2  m2b1
 11


m1  m2
m1  m2
m1  m2
m1  m2
y*  b2  m2
b2  b1
m b  m2b2 m2b2  m2b1 m1b2  m2b1
 1 2


m1  m2
m1  m2
m1  m2
m1  m2
So, it would seem, at least if (m1 - m2)  0!
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What is this thing we have found?
What we have found is the point
of intersection between two lines!
Later it will become obvious why
this point of intersection is of
interest to us!
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Example
40
y = 10 + 2.5x
30
y
22.5
20
y = 30 - 1.5x
10
0
0
2
4
5
6
8
10
x
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Why doesn’t it work with m1 = m2?
• If m1 = m2 and b1 = b2, then the two lines are identical and any value of x
will give you the same value of y.
• If m1 = m2 and b1  b2, then the two lines are different, but have the same
slope. They are parallel, which means there is no value x* that will give
you the same y*. There is no point of intersection!
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Elementary Calculus
• Calculus Has Many Uses
–
–
–
–
Compute Areas & Volumes
Compute Velocities
Describe Functional Relationships
Find Extremes
• In this class, we will typically be interested in
– describing functional relationships &
– finding extremes.
• Both these uses can be accomplished using what are called derivatives or
partial derivatives.
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Derivative of a Function
• Consider the function y = f(x)
• We will denote the derivative of a function with only one variable using ’:
f ’(x).
• The definition of a derivative is
f '  x   lim
h 0
f x  h   f x 
h
• What does this definition tell us?
– It tells us how the value of y is changing as x changes.
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Graphical Interpretation
60
f’(x) = slope as h goes to 0
or slope of line tangent to f(x) at x.
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y
f(x + h) - f(x)
= 38 – 32 = 6
Slope = (f(x + h) - f(x)) / h
=6/2=3
h=4–2=2
20
y = f(x) = 10+15x-2x2
0
0
2
4
6
8
10
x
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So why is this useful?
• What can we say about f(x) when f ’(x) > 0?
• What can we say about f(x) when f ’(x) < 0?
• What can we say about f(x) when f ’(x) = 0?
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When f ’(x) > 0, increasing x increases f(x) or y!
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y
40
20
0
0
2
4
6
8
10
x
31
When f ’(x) < 0, increasing x decreases f(x) or y!
60
y
40
20
0
0
2
4
6
8
10
x
32
When f ’(x) = 0, f(x) or y is at a maximum!
60
y
40
20
0
0
2
4
6
8
10
x
33
When f ’(x) = 0, f(x) or y is at a minimum!
60
y
40
20
0
0
2
4
6
8
10
x
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To Summarize
•
•
•
•
f ’(x) > 0 tells us there is a positive relationship between x and y given x.
f ’(x) < 0 tells us there is a negative relationship between x and y given x.
f ’(x) = 0 tells us y is at a maximum or minimum given x.
Note: There is a huge difference between being at a maximum vs. being at
a minimum! It would be nice to able to say which is the case.
– What is happening to f ’(x) when x is increasing for a maximum?
– What is happening to f ’(x) when x is increasing for a minimum?
35
For a maximum, f ’(x) is decreasing as x increases!
60
y
40
20
0
0
2
4
6
8
10
x
36
For a minimum, f ’(x) is increasing as x increases!
60
y
40
20
0
0
2
4
6
8
10
x
37
But how do we tell what is happening to f ’(x) as x
increase?
• This is precisely what a derivative tells us!
• Can we take the derivative of a derivative?
– Yes!
– It is called the second derivative, which we will denote with ’’: f ’’(x).
– The second derivative is defined as
f ' ' x   lim
h 0
f ' x  h   f ' x 
h
– It is just the derivative of the derivative!
– If f ’’(x) < 0, it tells us the derivative or slope of the tangent is decreasing as x
increases or that the x that solves f ’(x) = 0 is a maximum.
– If f ’’(x) > 0, it tells us the derivative or slope of the tangent is decreasing as x
increases or that the x that solves f ’(x) = 0 is a minimum.
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Useful Properties of Derivatives You Need to Know
• Suppose we have three functions: f(x), g(x), and h(x)
–
–
–
–
If f(x) = g(x) + h(x), f ’(x) = g’(x) + h’(x).
If f(x) = g(x)h(x), f ’(x) = g’(x)h(x) + g(x)h’(x).
If f(x) = h(g(x)), f ’(x) = h ’(g(x)) g’(x).
If f(x) = h(x)/g(x), f ’(x) = (h’(x) g(x)- h(x)g’(x))/g(x)2.
• Suppose a and n are real numbers:
– f(x) = a, f ’(x) = 0.
– f(x) = axn, f ’(x) = naxn - 1
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An Example
• Suppose f(x) = 10+15x-2x2
• f(x) = f1(x) + f2(x) + f3(x) where f1(x) = 10, f2(x) = 15x, and f3(x) = -2x2.
• f ’(x) = f1’(x) + f2’(x) + f3’(x) = 0 + 15 – 4x = 15 – 4x because
– for f1(x) = 10, f1’(x) = 0
– for f2(x) = 15x, f2’(x) = 1  15  x1 - 1 = 1  15  x0 = 15 and
– for f3(x) = -2x2, f3’(x) = 2  (-2)  x2 - 1 = -4 x1 = -4x
40
Another Example
• Suppose f(x) = x2x3
• f(x) = f1(x)f2(x) where f1(x) = x2 and f2(x) = x3.
• f’(x) = f1’(x) f2(x) + f1(x) f2’(x) = 2xx3 + x23x2 = 2x4 + 3x4 = 5x4 because
– for f1(x) = x2 , f1’(x) = 2x2 - 1 = 2x and
– for f2(x) = x3, f2’(x) = 3x3 - 1 = 3x2
• Notice that f(x) = x2x3 = x5, such that f ’(x) = 5x5 - 1 = 5x4, but things may not
always simplify so nicely.
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Yet Another Example
• Suppose f(x) = (x2 - 5)3
• f(x) = f1(f2(x)) where f1(x) = x3 and f2(x) = x2 - 5.
• f ’(x) = f1’(f2(x)) f2’(x) = 2(x2 - 5)2 2x = 4x(x2 - 5)2 because
– for f1(x) = x3 , f1’(x) = 3x3 - 1 = 3x2 and
– for f2(x) = x2 - 5, f2’(x) = 2x2 - 1 - 0 = 2x
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One Final Example Before Moving On
• Suppose f(x) = (x3 - 5) / x2
• f(x) = f1(x) / f2(x) where f1(x) = x3 - 5 and f2(x) = x2
• f ’(x) = (f1’(x) f2(x) – f1(x) f2’(x)) / f2(x)2 = (3x2x2 - (x3 - 5) 2x) / x4
= (3x4 - 2x4 + 10x) / x4 because
– for f1(x) = x3 - 5, f1’(x) = 3x3 - 1 - 0 = 3x2 and
– for f2(x) = x2, f2’(x) = 2x2 - 1 = 2x
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What about functions with more than one variable?
• Suppose we have a function y = f(x, z)
• The derivative of f(x, z) with respect to x is defined as
f x, z 
f  x  h, z   f x, z 
 f x x, z   lim
h 0
x
h
• The derivative of f(x, z) with respect to z is defined as
f x, z 
f  x, z  h   f  x, z 
 f z x, z   lim
h 0
z
h
• These are referred to as partial derivatives because we are taking the
derivative of the function with respect to a single variable, while holding
the other variable constant.
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What do these partial derivatives mean?
• fx(x,z) tells us the relationship between y and x on a contour holding z
constant.
• fz(x,z) tells us the relationship between y and z on a contour holding x
constant.
• Solving fx(x,z) = 0 and fz(x,z) = 0, we can find the x and z that maximizes or
minimizes y (there is also something called a saddle point that we will not
deal with in this class).
– Again, we can compute second partial derivatives to determine whether we are
at a maximum or minimum, but explaining how is more than what we need to
get into at this point.
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An Example
• Suppose y = f(x,z) = x0.4z0.6
– fx(x,z) = 0.4x0.4-1z0.6 = 0.4x-0.6z0.6
– fz(x,z) = x0.40.6z0.6-1 = 0.6x0.4z-0.4
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Surface Plot for y = f(x,z) = x0.4z0.6
35
30
25
20
y
15
10
5
0
20
5
10
20
30
40
z
30-35
25-30
20-25
15-20
10-15
5-10
0-5
50
x
47
Contour Plot for y = f(x,z) = x0.4z0.6 holding z constant.
40
fx(x,z) gives us slopes of tangents to contours
holding z constant.
z = 25
30
z = 20

z = 15
y20
z = 10
z=5
10
0
0
20
x
40
60
48
Contour Plot for y = f(x,z) = x0.4z0.6 holding x constant.
40
fz(x,z) gives us slopes of tangents to contours
holding x constant.
30
x = 50
x = 40
x = 30
x = 20

y20
x = 10
10
0
0
10
z
20
30
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Skills for Success
•
•
•
•
Manipulate Algebraic Expressions
Represent Functions Algebraically and Graphically
Find Solutions to Systems of Equations (Usually Two Equations)
Take Derivatives and Partial Derivative of Functions with Exponents
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