Transcript Chapter 2 Lectures

2
Functions and Their Graphs
 The Cartesian Coordinate System






and Straight lines
Equations of Lines
Functions and Their Graphs
The Algebra of Functions
Linear Functions
Quadratic Functions
Functions and Mathematical Models
2.1
The Cartesian Coordinate System
and Straight lines
y
6
5
4
x = 3
(2, 5)
y = – 4
3
2
(5, 1)
1
x
1
2
3
4
5
6
L
The Cartesian Coordinate System
 We can represent real numbers geometrically by points on
a real number, or coordinate, line:
–4
–3
–2
–1
0
1
2
3
4
The Cartesian Coordinate System
 The Cartesian coordinate system extends this concept to a
plane (two dimensional space) by adding a vertical axis.
4
3
2
1
–4
–3
–2
–1
–1
–2
–3
–4
1
2
3
4
The Cartesian Coordinate System
 The horizontal line is called the x-axis, and the vertical line is
called the y-axis.
y
4
3
2
1
–4
–3
–2
–1
–1
–2
–3
–4
1
2
3
4
x
The Cartesian Coordinate System
 The point where these two lines intersect is called the origin.
y
4
3
2
1
–4
–3
–2
–1
–1
–2
–3
–4
Origin
1
2
3
4
x
The Cartesian Coordinate System
 In the x-axis, positive numbers are to the right and negative
numbers are to the left of the origin.
y
4
3
2
Negative Direction
–4
–3
–2
–1
1
–1
–2
–3
–4
Positive Direction
1
2
3
4
x
The Cartesian Coordinate System
 In the y-axis, positive numbers are above and negative
numbers are below the origin.
3
2
1
–4
–3
–2
–1
–1
–2
–3
–4
1
Negative Direction
4
Positive Direction
y
2
3
4
x
The Cartesian Coordinate System
 A point in the plane can now be represented uniquely in this
coordinate system by an ordered pair of numbers (x, y).
y
(– 2, 4)
4
(4, 3)
3
2
1
–4
–3
–2
(–1, – 2)
–1
–1
–2
–3
–4
1
2
3
4
(3, –1)
x
The Cartesian Coordinate System
 The axes divide the plane into four quadrants as shown below.
y
4
Quadrant II
(–, +)
Quadrant I
(+, +)
3
2
1
–4
–3
–2
Quadrant III
(–, –)
–1
–1
–2
–3
–4
1
2
3
4
Quadrant IV
(+, –)
x
Slope of a Vertical Line
 Let L denote the unique straight line that passes through
the two distinct points (x1, y1) and (x2, y2).
 If x1 = x2, then L is a vertical line, and the slope is
undefined.
y
L
(x1, y1)
(x2, y2)
x
Slope of a Nonvertical Line
 If (x1, y1) and (x2, y2) are two distinct points on a
nonvertical line L, then the slope m of L is given by
m
y y2  y1

x x2  x1
y
L
(x2, y2)
y2 – y1 = y
(x1, y1)
x2 – x1 = x
x
Slope of a Nonvertical Line
 If m > 0, the line slants upward from left to right.
y
L
m=1
y = 1
x = 1
x
Slope of a Nonvertical Line
 If m > 0, the line slants upward from left to right.
y
L
m=2
y = 2
x = 1
x
Slope of a Nonvertical Line
 If m < 0, the line slants downward from left to right.
y
m = –1
x = 1
y = –1
x
L
Slope of a Nonvertical Line
 If m < 0, the line slants downward from left to right.
y
m = –2
x = 1
y = –2
x
L
Examples
 Sketch the straight line that passes through the point
(2, 5) and has slope –4/3.
Solution
1. Plot the point (2, 5).
2. A slope of –4/3 means
that if x increases by 3,
y decreases by 4.
3. Plot the resulting
point (5, 1).
4. Draw a line through
the two points.
y
6
5
4
x = 3
(2, 5)
y = –4
3
2
(5, 1)
1
x
1
2
3
4
5
6
L
Examples
 Find the slope m of the line that goes through the points
(–1, 1) and (5, 3).
Solution
 Choose (x1, y1) to be (–1, 1) and (x2, y2) to be (5, 3).
 With x1 = –1, y1 = 1, x2 = 5, y2 = 3, we find
y2  y1
3 1
2 1
m

 
x2  x1 5  ( 1) 6 3
Examples
 Find the slope m of the line that goes through the points
(–2, 5) and (3, 5).
Solution
 Choose (x1, y1) to be (–2, 5) and (x2, y2) to be (3, 5).
 With x1 = –2, y1 = 5, x2 = 3, y2 = 5, we find
y2  y1
55
0
m

 0
x2  x1 3  ( 2) 5
Examples
 Find the slope m of the line that goes through the points
(–2, 5) and (3, 5).
Solution
 The slope of a horizontal line is zero:
y
6
(–2, 5)
(3, 5)
L
4
m=0
3
2
1
–2
–1
x
1
2
3
4
Parallel Lines
 Two distinct lines are parallel if and only if their
slopes are equal or their slopes are undefined.
Example
 Let L1 be a line that passes through the points (–2, 9) and
(1, 3), and let L2 be the line that passes through the points
(– 4, 10) and (3, – 4).
 Determine whether L1 and L2 are parallel.
Solution
 The slope m1 of L1 is given by
3 9
m1 
 2
1  ( 2)
 The slope m2 of L2 is given by
4  10
m2 
 2
3  ( 4)
 Since m1 = m2, the lines L1 and L2 are in fact parallel.
2.2
Equations of Lines
y
L
1
(4, 0)
x
–1
1
–2
–3
–4
(0, – 3)
2
3
4
5
6
Equations of Lines
 Let L be a straight line
parallel to the y-axis.
 Then L crosses the x-axis at
some point (a, 0) , with the
x-coordinate given by x = a,
where a is a real number.
 Any other point on L has
the form (a, y ), where y
is an appropriate number.
 The vertical line L can
therefore be described as
x=a
y
L
(a, y )
(a, 0)
x
Equations of Lines
 Let L be a nonvertical line with a slope m.
 Let (x1, y1) be a fixed point lying on L, and let (x, y) be a
variable point on L distinct from (x1, y1).
 Using the slope formula by letting (x, y) = (x2, y2), we get
y  y1
m
x  x1
 Multiplying both sides by x – x1 we get
y  y1  m( x  x1 )
Point-Slope Form
 An equation of the line that has slope m and
passes through point (x1, y1) is given by
y  y1  m( x  x1 )
Examples
 Find an equation of the line that passes through the point
(1, 3) and has slope 2.
Solution
 Use the point-slope form
y  y1  m( x  x1 )
 Substituting for point (1, 3) and slope m = 2, we obtain
y  3  2( x  1)
 Simplifying we get
2x  y  1  0
Examples
 Find an equation of the line that passes through the points
(–3, 2) and (4, –1).
Solution
 The slope is given by
y y
1  2
3
m 2 1 

x2  x1 4  (3)
7
 Substituting in the point-slope form for point (4, –1) and
slope m = – 3/7, we obtain
3
y  1   ( x  4)
7
7 y  7  3x  12
3x  7 y  5  0
Perpendicular Lines
 If L1 and L2 are two distinct nonvertical lines that
have slopes m1 and m2, respectively, then L1 is
perpendicular to L2 (written L1 ┴ L2) if and only if
1
m1  
m2
Example
 Find the equation of the line L1 that passes through the
point (3, 1) and is perpendicular to the line L2 described by
y  3  2( x  1)
Solution
 L2 is described in point-slope form, so its slope is m2 = 2.
 Since the lines are perpendicular, the slope of L1 must be
m1 = –1/2
 Using the point-slope form of the equation for L1 we obtain
1
y  1   ( x  3)
2
2 y  2  x  3
x  2y 5  0
Crossing the Axis
 A straight line L that is neither horizontal nor vertical
cuts the x-axis and the y-axis at, say, points (a, 0) and
(0, b), respectively.
 The numbers a and b are called the x-intercept and
y-intercept, respectively, of L.
y
y-intercept
(0, b)
x-intercept
(a, 0)
x
L
Slope-Intercept Form
 An equation of the line that has slope m and
intersects the y-axis at the point (0, b) is given by
y = mx + b
Examples
 Find the equation of the line that has slope 3 and
y-intercept of –4.
Solution
 We substitute m = 3 and b = –4 into y = mx + b and get
y = 3x – 4
Examples
 Determine the slope and y-intercept of the line whose
equation is 3x – 4y = 8.
Solution
 Rewrite the given equation in the slope-intercept form.
3x  4 y  8
4 y  8  3x
y
3
x2
4
 Comparing to y = mx + b, we find that m = ¾ and b = – 2.
 So, the slope is ¾ and the y-intercept is – 2.
Applied Example
 Suppose an art object purchased for $50,000 is expected to
appreciate in value at a constant rate of $5000 per year for
the next 5 years.
 Write an equation predicting the value of the art object for
any given year.
 What will be its value 3 years after the purchase?
Solution
 Let
x = time (in years) since the object was purchased
y = value of object (in dollars)
 Then, y = 50,000 when x = 0, so the y-intercept is b = 50,000.
 Every year the value rises by 5000, so the slope is m = 5000.
 Thus, the equation must be y = 5000x + 50,000.
 After 3 years the value of the object will be $65,000:
y = 5000(3) + 50,000 = 65,000
General Form of a Linear Equation
 The equation
Ax + By + C = 0
where A, B, and C are constants and A and B
are not both zero, is called the general form
of a linear equation in the variables x and y.
General Form of a Linear Equation
 An equation of a straight line is a linear
equation; conversely, every linear equation
represents a straight line.
Example
 Sketch the straight line represented by the equation
3x – 4y – 12 = 0
Solution
 Since every straight line is uniquely determined by two
distinct points, we need find only two such points through
which the line passes in order to sketch it.
 For convenience, let’s compute the x- and y-intercepts:
✦ Setting y = 0, we find x = 4; so the x-intercept is 4.
✦ Setting x = 0, we find y = –3; so the y-intercept is –3.
 Thus, the line goes through the points (4, 0) and (0, –3).
Example
 Sketch the straight line represented by the equation
3x – 4y – 12 = 0
Solution
 Graph the line going through the points (4, 0) and (0, –3).
y
L
1
(4, 0)
x
–1
1
–2
–3
–4
(0, – 3)
2
3
4
5
6
Equations of Straight Lines
Vertical line:
Horizontal line:
Point-slope form:
Slope-intercept form:
General Form:
x=a
y=b
y – y1 = m(x – x1)
y = mx + b
Ax + By + C = 0
2.3
Functions and Their Graphs
y
f  x  x
3
f  x  x
2
1
–3
–2
–1
1
2
3
x
Functions
 A function f is a rule that assigns to each element in a




set A one and only one element in a set B.
The set A is called the domain of the function.
It is customary to denote a function by a letter of the
alphabet, such as the letter f.
If x is an element in the domain of a function f, then the
element in B that f associates with x is written f(x) (read
“f of x”) and is called the value of f at x.
The set B comprising all the values assumed by y = f(x)
as x takes on all possible values in its domain is called
the range of the function f.
Example
 Let the function f be defined by the rule
f  x   2 x2  x  1
 Find: f(1)
Solution:
f 1  2 1  1  1  2  1  1  2
2
Example
 Let the function f be defined by the rule
f  x   2 x2  x  1
 Find: f( – 2)
Solution:
f  2   2  2    2   1  8  2  1  11
2
Example
 Let the function f be defined by the rule
f  x   2 x2  x  1
 Find: f(a)
Solution:
f  a   2  a    a   1  2a 2  a  1
2
Example
 Let the function f be defined by the rule
f  x   2 x2  x  1
 Find: f(a + h)
Solution:
f  a  h   2  a  h    a  h   1  2a 2  4ah  2h 2  a  h  1
2
Applied Example
 ThermoMaster manufactures an indoor-outdoor
thermometer at its Mexican subsidiary.
 Management estimates that the profit (in dollars)
realizable by ThermoMaster in the manufacture and sale
of x thermometers per week is
P  x   0.001x 2  8x  5000
 Find ThermoMaster’s weekly profit if its level of
production is:
a. 1000 thermometers per week.
b. 2000 thermometers per week.
Applied Example
Solution
 We have
P  x   0.001x 2  8x  5000
a. The weekly profit by producing 1000 thermometers is
P 1000   0.001 1000   8 1000   5000  2000 =
2
or $2,000.
b. The weekly profit by producing 2000 thermometers is
P  2000   0.001  2000   8  2000   5000  7000 =
2
or $7,000.
Determining the Domain of a Function
 Suppose we are given the function y = f(x).
 Then, the variable x is called the independent variable.
 The variable y, whose value depends on x, is called the
dependent variable.
 To determine the domain of a function, we need to find
what restrictions, if any, are to be placed on the
independent variable x.
 In many practical problems, the domain of a function is
dictated by the nature of the problem.
Applied Example: Packaging
 An open box is to be made from a rectangular piece of
cardboard 16 inches wide by cutting away identical
squares (x inches by x inches) from each corner and
folding up the resulting flaps.
x
10 10 – 2x
x
x
16 – 2x
16
x
Applied Example: Packaging
 An open box is to be made from a rectangular piece of
cardboard 16 inches wide by cutting away identical
squares (x inches by x inches) from each corner and
folding up the resulting flaps.
 The dimensions of the
resulting box are:
x
10 – 2x
16 – 2x
a. Find the expression that gives the volume V of the box as
a function of x.
b. What is the domain of the function?
Applied Example: Packaging
Solution
a. The volume of the box is given by multiplying its
dimensions (length ☓ width ☓ height), so:
V  f  x   16  2 x   10  2 x   x
 160  52 x  4 x 2  x
 4 x 3  52 x 2  160 x
x
10 – 2x
16 – 2x
Applied Example: Packaging
Solution
b. Since the length of each side of the box must be greater
than or equal to zero, we see that
16  2 x  0
10  2 x  0
x0
must be satisfied simultaneously. Simplified:
x 8
x 5
x  0=
All three are satisfied simultaneously provided that:
0 x5
Thus, the domain of the function f is the interval [0, 5].
More Examples
 Find the domain of the function:
f  x  x 1
Solution
 Since the square root of a negative number is undefined, it
is necessary that x – 1  0.
 Thus the domain of the function is [1,).
More Examples
 Find the domain of the function:
1
f  x  2
x 4
Solution
 Our only constraint is that you cannot divide by zero, so
x2  4  0
 Which means that
x 2  4   x  2 x  2  0
 Or more specifically x ≠ –2 and x ≠ 2.
 Thus the domain of f consists of the intervals (– , –2),
(–2, 2), (2, ).
More Examples
 Find the domain of the function:
f  x   x2  3
Solution
 Here, any real number satisfies the equation, so the
domain of f is the set of all real numbers.
Graphs of Functions
 If f is a function with domain A, then corresponding to
each real number x in A there is precisely one real
number f(x).
 Thus, a function f with domain A can also be defined as
the set of all ordered pairs (x, f(x)) where x belongs to A.
 The graph of a function f is the set of all points (x, y) in
the xy-plane such that x is in the domain of f and y = f(x).
Example
 The graph of a function f is shown below:
y
y
(x, y)
Range
x
Domain
x
Example
 The graph of a function f is shown below:
✦ What is the value of f(2)?
y
4
3
2
1
1
2
3
–1
–2
(2, –2)
4
5
6
7
8
x
Example
 The graph of a function f is shown below:
✦ What is the value of f(5)?
y
4
3
(5, 3)
2
1
1
–1
–2
2
3
4
5
6
7
8
x
Example
 The graph of a function f is shown below:
✦ What is the domain of f(x)?
y
4
3
2
1
1
2
3
4
5
6
–1
–2
Domain: [1,8]
7
8
x
Example
 The graph of a function f is shown below:
✦ What is the range of f(x)?
y
4
3
2
Range:
[–2,4]
1
1
–1
–2
2
3
4
5
6
7
8
x
Example: Sketching a Graph
 Sketch the graph of the function defined by the equation
y = x2 + 1
Solution
 The domain of the function is the set of all real numbers.
 Assign several values to the variable x and compute the
corresponding values for y:
x
–3
–2
–1
0
1
2
3
y
10
5
2
1
2
5
10
Example: Sketching a Graph
 Sketch the graph of the function defined by the equation
y = x2 + 1
Solution
 The domain of the function is the set of all real numbers.
 Then plot these values in a graph:
y
x
–3
–2
–1
0
1
2
3
y
10
5
2
1
2
5
10
10
8
6
4
2
–3
–2
–1
1
2
3 x
Example: Sketching a Graph
 Sketch the graph of the function defined by the equation
y = x2 + 1
Solution
 The domain of the function is the set of all real numbers.
 And finally, connect the dots:
y
x
–3
–2
–1
0
1
2
3
y
10
5
2
1
2
5
10
10
8
6
4
2
–3
–2
–1
1
2
3 x
Example: Sketching a Graph
 Sketch the graph of the function defined by the equation
  x
f  x  
 x
if x  0
if x  0
Solution
 The function f is defined in a piecewise fashion on the set
of all real numbers.
 In the subdomain (–, 0), the rule for f is given by
f  x  x
 In the subdomain [0, ), the rule for f is given by
f  x  x
Example: Sketching a Graph
 Sketch the graph of the function defined by the equation
  x
f  x  
 x
if x  0
if x  0
Solution
 Substituting negative values for x into f  x    x, while
substituting zero and positive values into f  x   x we get:
x
–3
–2
–1
0
1
2
3
y
3
2
1
0
1
1.41
1.73
Example: Sketching a Graph
 Sketch the graph of the function defined by the equation
  x
f  x  
 x
if x  0
if x  0
Solution
 Plotting these data and graphing we get:
x
–3
–2
–1
0
1
2
3
y
3
2
1
0
1
1.41
1.73
y
f  x  x
3
f  x  x
2
1
–3
–2
–1
1
2
3
x
The Vertical Line Test
 A curve in the xy-plane is the graph of a function
y = f(x) if and only if each vertical line intersects it
in at most one point.
Examples
 Determine if the curve in the graph is a function of x:
y
x
Solution
 The curve is indeed a function of x, because there is one
and only one value of y for any given value of x.
Examples
 Determine if the curve in the graph is a function of x:
y
x
Solution
 The curve is not a function of x, because there is more than
one value of y for some values of x.
Examples
 Determine if the curve in the graph is a function of x:
y
x
Solution
 The curve is indeed a function of x, because there is one
and only one value of y for any given value of x.
2.4
The Algebra of Functions
y
Billions of Dollars
2000
y = R(t)
1800
y = S(t)
1600
S(t)
1400
R(t)
1200
1000
1990
1992
1994
t 1996
Year
1998
2000
t
The Sum, Difference, Product and Quotient
of Functions
 Consider the graph below:
✦ R(t) denotes the federal government revenue at any time t.
✦ S(t) denotes the federal government spending at any time t.
y
Billions of Dollars
2000
y = R(t)
1800
y = S(t)
1600
S(t)
1400
R(t)
1200
1000
1990
1992
1994
t
Year
1996
1998
2000
t
The Sum, Difference, Product and Quotient
of Functions
 Consider the graph below:
✦ The difference R(t) – S(t) gives the budget deficit (if negative)
or surplus (if positive) in billions of dollars at any time t.
y
Billions of Dollars
2000
y = R(t)
1800
y = S(t)
1600
S(t)
D(t) = R(t) – S(t)
1400
R(t)
1200
1000
1990
1992
1994
t
Year
1996
1998
2000
t
The Sum, Difference, Product and Quotient
of Functions
 The budget balance
D(t) is shown below:
✦ D(t) is also a function that denotes the federal government
deficit (surplus) at any time t.
✦ This function is the difference of the two functions R and S.
✦ D(t) has the same domain as R(t) and S(t).
y
Billions of Dollars
400
200
0
D(t)
–200
–400
y = D(t)
1992
1994
t 1996
Year
1998
2000
t
The Sum, Difference, Product and Quotient
of Functions
 Most functions are built up from other, generally
simpler functions.
 For example, we may view the function f(x) = 2x + 4
as the sum of the two functions g(x) = 2x and h(x) = 4.
The Sum, Difference, Product and Quotient of Functions
 Let f and g be functions with domains A and B, respectively.
 The sum f + g, the difference f – g, and the product fg of f
and g are functions with domain A ∩ B and rule given by
(f + g)(x) = f(x) + g(x)
Sum
(f – g)(x) = f(x) – g(x)
(fg)(x) = f(x)g(x)
Difference
Product
 The quotient f/g of f and g has domain A ∩ B excluding all
numbers x such that g(x) = 0 and rule given by
 f 
f ( x)
 g   x   g ( x)
 
Quotient
Example
 Let f ( x ) 
x  1 and g(x) = 2x + 1.
 Find the sum s, the difference d, the product p, and the
quotient q of the functions f and g.
Solution
 Since the domain of f is A = [–1,) and the domain of g
is B = (– , ), we see that the domain of s, d, and p is
A ∩ B = [–1,).
 The rules are as follows:
s( x )  ( f  g )( x )  f ( x )  g ( x )  x  1  2 x  1
d ( x )  ( f  g )( x )  f ( x )  g ( x )  x  1  2 x  1
p( x )  ( fg )( x )  f ( x ) g ( x )   2 x  1 x  1
Example
 Let f ( x ) 
x  1 and g(x) = 2x + 1.
 Find the sum s, the difference d, the product p, and the
quotient q of the functions f and g.
Solution
 The domain of the quotient function is [–1,) together
with the restriction x ≠ – ½.
 Thus, the domain is [–1, – ½) U (– ½,).
 The rule is as follows:
 f 
f ( x)
x 1
q( x )    ( x ) 

g ( x) 2 x  1
g
Applied Example: Cost Functions
 Suppose Puritron, a manufacturer of water filters, has a
monthly fixed cost of $10,000 and a variable cost of
– 0.0001x2 + 10x
(0  x  40,000)
dollars, where x denotes the number of filters
manufactured per month.
 Find a function C that gives the total monthly cost
incurred by Puritron in the manufacture of x filters.
Applied Example: Cost Functions
Solution
 Puritron’s monthly fixed cost is always $10,000, so it can
be described by the constant function:
F(x) = 10,000
 The variable cost can be described by the function:
V(x) = – 0.0001x2 + 10x
 The total cost is the sum of the fixed cost F and the
variable cost V:
C(x) = V(x) + F(x)
= – 0.0001x2 + 10x + 10,000
(0  x  40,000)
Applied Example: Cost Functions
Let’s now consider profits
 Suppose that the total revenue R realized by Puritron from
the sale of x water filters is given by
R(x) = – 0.0005x2 + 20x
(0 ≤ x ≤ 40,000)
 Find
a. The total profit function for Puritron.
b. The total profit when Puritron produces 10,000 filters per
month.
Applied Example: Cost Functions
Solution
a. The total profit P realized by the firm is the difference
between the total revenue R and the total cost C:
P(x) = R(x) – C(x)
= (– 0.0005x2 + 20x) – (– 0.0001x2 + 10x + 10,000)
= – 0.0004x2 + 10x – 10,000
b. The total profit realized by Puritron when producing 10,000
filters per month is
P(x) = – 0.0004(10,000)2 + 10(10,000) – 10,000
= 50,000
or $50,000 per month.
The Composition of Two Functions
 Another way to build a function from other functions is
through a process known as the composition of functions.
 Consider the functions f and g:
g ( x)  x
f ( x)  x 2  1
 Evaluating the function g at the point f(x), we find that:
g  f ( x)  
f ( x)  x 2  1
 This is an entirely new function, which we could call h:
h( x )  x 2  1
The Composition of Two Functions
 Let f and g be functions.
 Then the composition of g and f is the function
ggf (read “g circle f ”) defined by
(ggf )(x) = g(f(x))
 The domain of ggf is the set of all x in the
domain of f such that f(x) lies in the domain of g.
Example
 Let f ( x )  x  1 and g ( x ) 
2
x  1.
 Find:
a. The rule for the composite function ggf.
b. The rule for the composite function fgg.
Solution
 To find ggf, evaluate the function g at f(x):
( g f )( x )  g ( f ( x )) 
f ( x)  1  x 2  1  1
 To find fgg, evaluate the function f at g(x):
( f g )( x )  f ( g ( x ))  ( g ( x ))2  1  ( x  1)2  1
 x  2 x 11  x  2 x
Applied Example: Automobile Pollution
 An environmental impact study conducted for the city of
Oxnard indicates that, under existing environmental
protection laws, the level of carbon monoxide (CO)
present in the air due to pollution from automobile
exhaust will be 0.01x2/3 parts per million when the
number of motor vehicles is x thousand.
 A separate study conducted by a state government agency
estimates that t years from now the number of motor
vehicles in Oxnard will be 0.2t2 + 4t + 64 thousand.
 Find:
a. An expression for the concentration of CO in the air due
to automobile exhaust t years from now.
b. The level of concentration 5 years from now.
Applied Example: Automobile Pollution
Solution
 Part (a):
✦ The level of CO is described by the function
g(x) = 0.01x2/3
where x is the number (in thousands) of motor vehicles.
✦ In turn, the number (in thousands) of motor vehicles is
described by the function
f(t) = 0.2t2 + 4t + 64
where t is the number of years from now.
✦ Therefore, the concentration of CO due to automobile
exhaust t years from now is given by
(ggf )(t) = g(f(t)) = 0.01(0.2t2 + 4t + 64)2/3
Applied Example: Automobile Pollution
Solution
 Part (b):
✦ The level of CO five years from now is:
(ggf )(5) = g(f(5)) = 0.01[0.2(5)2 + 4(5) + 64]2/3
= (0.01)892/3 ≈ 0.20
or approximately 0.20 parts per million.
2.5
Linear Functions
y
5
L1
4
3
2
(1, 2)
1
–1
1
2
L2
3
4
5
x
Linear Function
 The function f defined by
f ( x )  mx  b
where m and b are constants, is called a linear
function.
Applied Example: Linear Depreciation
 A Web server has an original value of $10,000 and is to
be depreciated linearly over 5 years with a $3000 scrap
value.
 Find an expression giving the book value at the end of
year t.
 What will be the book value of the server at the end of
the second year?
 What is the rate of depreciation of the server?
Applied Example: Linear Depreciation
Solution
 Let V(t) denote the Web server’s book value at the end of
the tth year. V is a linear function of t.
 To find an equation of the straight line that represents the
depreciation, observe that V = 10,000 when t = 0; this tells
us that the line passes through the point (0, 10,000).
 Similarly, the condition that V = 3000 when t = 5 says that
the line also passes through the point (5, 3000).
 Thus, the slope of the line is given by
10,000  3000
7000
m

 1400
05
5
Applied Example: Linear Depreciation
Solution
 Using the point-slope form of the equation of a line with
point (0, 10,000) and slope m = –1400, we obtain the
required expression
V  10,000  1400(t  0)
V  1400t  10,000
 The book value at the end of the second year is given by
V (2)  1400(2)  10,000  7200
or $7200.
 The rate of depreciation of the server is given by the
negative slope of the depreciation line m = –1400, so the
rate of depreciation is $1400 per year.
Applied Example: Linear Depreciation
Solution
 The graph of V is:
V
10,000
(0, 10,000)
(5, 3000)
3000
V  1400t  10,000
t
1
2
3
4
5
6
Cost, Revenue, and Profit Functions
 Let x denote the number of units of a product
manufactured or sold.
 Then, the total cost function is
C(x) = Total cost of manufacturing x units
of the product
 The revenue function is
R(x) = Total revenue realized from the sale
of x units of the product
 The profit function is
P(x) = Total profit realized from
manufacturing and selling x units of
the product
Applied Example: Profit Function
 Puritron, a manufacturer of water filters, has a monthly
fixed cost of $20,000, a production cost of $20 per unit, and
a selling price of $30 per unit.
 Find the cost function, the revenue function, and the
profit function for Puritron.
Solution
 Let x denote the number of units produced and sold.
 Then,
C ( x )  20 x  20,000
R( x )  30 x
P( x )  R( x )  C ( x )
 30 x  (20 x  20,000)
 10 x  20,000
Finding the Point of Intersection
 Suppose we are given two straight lines L1 and L2 with
equations
y = m1x + b1 and
y = m2x + b2
(where m1, b1, m2, and b2 are constants) that intersect at the
point P(x0, y0).
 The point P(x0, y0) lies on the line L1 and so satisfies the
equation y = m1x + b1.
 The point P(x0, y0) also lies on the line L2 and so satisfies
y = m2x + b2 as well.
 Therefore, to find the point of intersection P(x0, y0) of the
lines L1 and L2, we solve for x and y the system composed
of the two equations
y = m1x + b1 and y = m2x + b2
Example
 Find the point of intersection of the straight lines that have
equations
y=x+1
and
y = – 2x + 4
Solution
 Substituting the value y as given in the first equation into
the second equation, we obtain
x  1  2 x  4
3x  3
x 1
 Substituting this value of x into either one of the given
equations yields y = 2.
 Therefore, the required point of intersection is (1, 2).
Example
 Find the point of intersection of the straight lines that have
equations
y=x+1
y = – 2x + 4
and
Solution
 The graph shows the point of intersection (1, 2) of the two
lines:
y
5
L1
4
3
2
(1, 2)
1
–1
1
2
L2
3
4
5
x
Applied Example: Break-Even Level
 Prescott manufactures its products at a cost of $4 per unit
and sells them for $10 per unit.
 If the firm’s fixed cost is $12,000 per month, determine the
firm’s break-even point.
Solution
 The revenue function R and the cost function C are given
respectively by
R( x )  10 x
and C ( x )  4 x  12,000.
 Setting R(x) = C(x), we obtain
10 x  4 x  12,000
6 x  12,000
x  2000
Applied Example: Break-Even Level
 Prescott manufactures its products at a cost of $4 per unit
and sells them for $10 per unit.
 If the firm’s fixed cost is $12,000 per month, determine the
firm’s break-even point.
Solution
 Substituting x = 2000 into R(x) = 10x gives
R(2000)  10(2000)  20,000
 So, Prescott’s break-even point is 2000 units of the product,
resulting in a break-even revenue of $20,000 per month.
2.6
Quadratic Functions
y
5 9
Vertex  , 
4 8
1
x-intercepts
–1
11
22
1
x
f ( x )  2 x 2  5x  2
–1
–2
2
y-intercept
Quadratic Functions
 A quadratic function is one of the form
f ( x )  ax 2  bx  c
where a, b, and c are constants and a ≠ 0.
 For example, the function
f ( x)  2 x 2  4 x  3
is quadratic, with a = 2, b = – 4, and c = 3.
Quadratic Functions
 Below is the graph of the quadratic function
f ( x)  2 x 2  4 x  3
 The graph of a quadratic function is a curve called a
parabola that opens upward or downward.
y
f ( x)  2 x 2  4 x  3
10
8
6
Parabola
4
2
–2
–1
1
2
3
4
x
Quadratic Functions
 The parabola is symmetric with respect to a vertical
line called the axis of symmetry.
 The axis of symmetry also passes through the lowest or
highest point of the parabola, which is called the vertex
of the parabola.
Axis of
symmetry
y
f ( x)  2 x 2  4 x  3
10
8
6
Parabola
4
2
Vertex (1, 1)
–2
–1
1
2
3
4
x
Quadratic Functions
 We can use these properties to help us sketch the graph
of a quadratic function.
 Suppose we want to sketch the graph of
f ( x )  3x 2  6 x  1
 If we complete the square in x, we obtain
f ( x )  3( x 2  2 x )  1
 3[ x 2  2 x  ( 1) 2 ]  1  3
 3( x  1) 2  2
 Note that (x – 1)2 is nonnegative: it equals to zero when x = 1
and is greater than zero if x ≠ 1.
 Thus, we see that f(x)  – 2 for all values of x.
 This tells us the vertex of the parabola is the point (1, – 2).
Quadratic Functions
 We know the vertex of the parabola is the point (1, – 2) and
that it is the minimum point of the graph, since f(x)  – 2 for
all values of x.
 Thus, the graph of f(x) = 3x2 – 6x +1 looks as follows:
y
f ( x )  3x 2  6 x  1
4
2
–2
–2
2
4
Vertex (1, –2)
x
Properties of Quadratic Functions
Given f(x) = ax2 + bx +c
(a ≠ 0)
1. The domain of f is the set of all real numbers.
2. If a > 0, the parabola opens upward, and if a < 0,
it opens downward.
b 
 b

3. The vertex of the parabola is   , f     .
 2a  2a  
b
4. The axis of symmetry of the parabola is x   .
2a
5. The x-intercepts (if any) are found by solving
f(x) = 0. The y-intercept is f(0) = c.
Example
 Given the quadratic function f(x) = – 2x2 + 5x – 2
a. Find the vertex of the parabola.
b. Find the x-intercepts (if any) of the parabola.
c. Sketch the parabola.
Solution
a. Here a = – 2, b = 5, and c = – 2. therefore, the x-coordinate
of the vertex of the parabola is
b
5
5



2a
2( 2) 4
The y-coordinate of the vertex is therefore given by
2
9
5
5
5
f    2    5    2 
8
4
4
4
5 9
Thus, the vertex of the parabola is the point  ,  .
4 8
Example
 Given the quadratic function f(x) = – 2x2 + 5x – 2
a. Find the vertex of the parabola.
b. Find the x-intercepts (if any) of the parabola.
c. Sketch the parabola.
Solution
b. For the x-intercepts of the parabola, we solve the equation
2 x 2  5 x  2  0
using the quadratic formula with a = – 2, b = 5, and c = – 2.
We find
b  b2  4ac 5  25  4( 2)( 2) 5  3
x


2a
2( 2)
4
Thus, the x-intercepts of the parabola are 1/2 and 2.
Example
 Given the quadratic function f(x) = – 2x2 + 5x – 2
a. Find the vertex of the parabola.
b. Find the x-intercepts (if any) of the parabola.
c. Sketch the parabola.
Solution
y
c. The sketch:
5 9
Vertex  , 
4 8
1
x-intercepts
–1
1
2
1
–1
–2
2
x
f ( x )  2 x 2  5x  2
y-intercept
Some Economic Models
 People’s decision on how much to demand or purchase of a
given product depends on the price of the product:
✦ The higher the price the less they want to buy of it.
✦ A demand function p = d(x) can be used to describe this.
Some Economic Models
 Similarly, firms’ decision on how much to supply or
produce of a product depends on the price of the product:
✦ The higher the price, the more they want to produce of it.
✦ A supply function p = s(x) can be used to describe this.
Some Economic Models
 The interaction between demand and supply will ensure
the market settles to a market equilibrium:
✦ This is the situation at which quantity demanded equals
quantity supplied.
✦ Graphically, this situation occurs when the demand curve
and the supply curve intersect: where d(x) = s(x).
Applied Example: Supply and Demand
 The demand function for a certain brand of bluetooth
wireless headset is given by
p  d ( x )  0.025x 2  0.5x  60
 The corresponding supply function is given by
p  s( x )  0.02 x 2  0.6 x  20
where p is the expressed in dollars and x is measured in
units of a thousand.
 Find the equilibrium quantity and price.
Applied Example: Supply and Demand
Solution
 We solve the following system of equations:
p  0.025x 2  0.5x  60
p  0.02 x 2  0.6 x  20
 Substituting the second equation into the first yields:
0.02 x 2  0.6 x  20  0.025 x 2  0.5 x  60
0.045 x 2  1.1x  40  0
45 x 2  1100 x  40,000  0
9 x 2  220 x  8,000  0
 9 x  400  x  20  0
 Thus, either x = –400/9 (but this is not possible), or x = 20.
 So, the equilibrium quantity must be 20,000 headsets.
Applied Example: Supply and Demand
Solution
 The equilibrium price is given by:
p  0.02  20   0.6  20   20  40
2
or $40 per headset.
2.7
Functions and Mathematical Models
y ($trillion)
6
4
2
5
10
15
20
25
30
t (years)
Mathematical Models
 As we have seen, mathematics can be used to
solve real-world problems.
 We will now discuss a few more examples of
real-world phenomena, such as:
✦ The solvency of the U.S. Social Security
trust fund
✦ Global warming
Mathematical Modeling
 Regardless of the field from which the real-world problem
is drawn, the problem is analyzed using a process called
mathematical modeling.
 The four steps in this process are:
Real-world problem
Formulate
Solve
Test
Solution of realworld problem
Mathematical
model
Interpret
Solution of
mathematical model
Modeling With Polynomial Functions
 A polynomial function of degree n is a function of the form
f ( x )  an x n  an 1 x n 1    a2 x 2  a1 x  a0
(an  0)
where n is a nonnegative integer and the numbers
a0, a1, …. an are constants called the coefficients of the
polynomial function.
 Examples:
✦ The function below is polynomial function of degree 5:
f ( x)  2 x5  3x 4  12 x 3  2 x 2  6
Modeling With Polynomial Functions
 A polynomial function of degree n is a function of the form
f ( x )  an x n  an 1 x n 1    a2 x 2  a1 x  a0
(an  0)
where n is a nonnegative integer and the numbers
a0, a1, …. an are constants called the coefficients of the
polynomial function.
 Examples:
✦ The function below is polynomial function of degree 3:
g ( x )  0.001x 3  0.2 x 2  10 x  200
Applied Example: Global Warming
 The increase in carbon dioxide (CO2) in the atmosphere is
a major cause of global warming.
 Below is a table showing the average amount of CO2,
measured in parts per million volume (ppmv) for various
years from 1958 through 2007:
Year
1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount
315
325
330
335
345
355
365
375
380
Applied Example: Global Warming
Year
1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount
315
325
330
335
345
355
365
375
380
 Below is a scatter plot associated with these data:
380
y (ppmv)
360
340
320
10
20
30
40
50
t (years)
Applied Example: Global Warming
Year
1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount
315
325
330
335
345
355
365
375
380
 A mathematical model giving the approximate amount of
CO2 is given by:
380
y (ppmv)
A(t )  0.01076t 2  0.8212t  313.4
360
340
320
10
20
30
40
50
t (years)
Applied Example: Global Warming
Year
1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount
315
325
330
335
345
355
A(t )  0.010716t 2  0.8212t  313.4
365
375
380
(1  t  50)
a. Use the model to estimate the average amount of atmospheric
CO2 in 1980 (t = 23).
b. Assume that the trend continued and use the model to predict
the average amount of atmospheric CO2 in 2010.
Applied Example: Global Warming
Year
1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount
315
325
330
335
345
355
A(t )  0.010716t 2  0.8212t  313.4
365
375
380
(1  t  50)
Solution
a. The average amount of atmospheric CO2 in 1980 is given by
A(23)  0.010716  23  0.8212  23  313.4  337.96
2
or approximately 338 ppmv.
b. Assuming that the trend will continue, the average amount of
atmospheric CO2 in 2010 will be
A(53)  0.010716  53  0.8212  53  313.4  387.03
2
Applied Example: Social Security Trust Fund
 The projected assets of the Social Security trust fund (in
trillions of dollars) from 2008 through 2040 are given by:
Year
2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets
2.4
3.2
4.0
4.7
5.3
5.7
5.9
5.6
4.9
3.6
1.7
0
 The scatter plot associated with these data is:
y ($trillion)
6
4
2
5
10
15
20
25
30
t (years)
Applied Example: Social Security Trust Fund
 The projected assets of the Social Security trust fund (in
trillions of dollars) from 2008 through 2040 are given by:
Year
2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets
2.4
3.2
4.0
4.7
5.3
5.7
5.9
5.6
4.9
3.6
1.7
0
 A mathematical model giving the approximate value of assets
in the trust fund (in trillions of dollars) is:
y ($trillion)
A(t )  0.00000324t 4  0.000326t 3  0.00342t 2  0.254t  2.4
6
4
2
5
10
15
20
25
30
t (years)
Applied Example: Social Security Trust Fund
Year
Assets
2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
2.4
3.2
4.0
4.7
5.3
5.7
5.9
5.6
4.9
3.6
1.7
0
A(t )  0.00000324t 4  0.000326t 3  0.00342t 2  0.254t  2.4
a. The first baby boomers will turn 65 in 2011. What will be the
assets of the Social Security trust fund at that time?
b. The last of the baby boomers will turn 65 in 2029. What will
the assets of the trust fund be at the time?
c. Use the graph of function A(t) to estimate the year in which
the current Social Security system will go broke.
Applied Example: Social Security Trust Fund
Year
Assets
2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
2.4
3.2
4.0
4.7
5.3
5.7
5.9
5.6
4.9
3.6
1.7
0
A(t )  0.00000324t 4  0.000326t 3  0.00342t 2  0.254t  2.4
Solution
a. The assets of the Social Security fund in 2011 (t = 3) will be:
A(3)  0.00000324  3  0.000326  3  0.00342  3  0.254  3  2.4  3.18
4
3
2
or approximately $3.18 trillion.
The assets of the Social Security fund in 2029 (t = 21) will be:
A(21)  0.00000324  21  0.000326  21  0.00342  21  0.254  21  2.4  5.59
4
3
or approximately $5.59 trillion.
2
Applied Example: Social Security Trust Fund
Year
2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets
2.4
3.2
4.0
4.7
5.3
5.7
5.9
5.6
4.9
3.6
1.7
0
A(t )  0.00000324t 4  0.000326t 3  0.00342t 2  0.254t  2.4
Solution
b. The graph shows that function A crosses the t-axis at about
t = 32, suggesting the system will go broke by 2040:
y ($trillion)
6
4
Trust runs
out of funds
2
5
10
15
20
25
30
t (years)
Rational and Power Functions
 A rational function is simply the quotient of two
polynomials.
 In general, a rational function has the form
R( x) 
f ( x)
g ( x)
where f(x) and g(x) are polynomial functions.
 Since the division by zero is not allowed, we conclude that
the domain of a rational function is the set of all real
numbers except the zeros of g (the roots of the equation
g(x) = 0)
Rational and Power Functions
 Examples of rational functions:
3x 3  x 2  x  1
F ( x) 
x2
x2  1
G( x)  2
x 1
Rational and Power Functions
 Functions of the form
f ( x)  x r
where r is any real number, are called power functions.
 We encountered examples of power functions earlier in
our work.
 Examples of power functions:
f ( x)  x  x
1/2
and
1
g ( x )  2  x 2
x
Rational and Power Functions
 Many functions involve combinations of rational and
power functions.
 Examples:
1  x2
f ( x) 
1  x2
g ( x )  x 2  3x  4
h( x )  (1  2 x )
1/2
1
 2
( x  2)3/2
Applied Example: Driving Costs
 A study of driving costs based on a 2007 medium-sized
sedan found the following average costs (car payments,
gas, insurance, upkeep, and depreciation), measured in
cents per mile:
Miles/year, x
5000
10,000
15,000
20,000
Cost/mile, y (¢)
83.8
62.9
52.2
47.1
 A mathematical model giving the average cost in cents per
mile is:
C ( x) 
164.8
x 0.42
where x (in thousands) denotes the number of miles the car
is driven in 1 year.
Applied Example: Driving Costs
Miles/year, x
5000
10,000
15,000
20,000
Cost/mile, y (¢)
83.8
62.9
52.2
47.1
164.8
C ( x )  0.42
x
 Below is the scatter plot associated with this data:
y (¢)
140
120
100
80
60
40
20
C(x)
5
10
15
20
25
x (☓1000 miles/year)
Applied Example: Driving Costs
Miles/year, x
5000
10,000
15,000
20,000
Cost/mile, y (¢)
83.8
62.9
52.2
47.1
164.8
C ( x )  0.42
x
 Using this model, estimate the average cost of driving a
2007 medium-sized sedan 8,000 miles per year and 18,000
miles per year.
Solution
 The average cost for driving a car 8,000 miles per year is
C (8) 
164.8
 8
0.42
or approximately 68.8¢/mile.
 68.81
Applied Example: Driving Costs
Miles/year, x
5000
10,000
15,000
20,000
Cost/mile, y (¢)
83.8
62.9
52.2
47.1
164.8
C ( x )  0.42
x
 Using this model, estimate the average cost of driving a
2007 medium-sized sedan 8,000 miles per year and 18,000
miles per year.
Solution
 The average cost for driving a car 18,000 miles per year is
C (18) 
164.8
18
0.42
 48.95
or approximately 48.95¢/mile.
Constructing Mathematical Models
 Some mathematical models can be constructed using
elementary geometric and algebraic arguments.
Guidelines for constructing mathematical models:
1. Assign a letter to each variable mentioned in the
problem. If appropriate, draw and label a figure.
2. Find an expression for the quantity sought.
3. Use the conditions given in the problem to write
the quantity sought as a function f of one variable.
Note any restrictions to be placed on the domain
of f by the nature of the problem.
Applied Example: Enclosing an Area
 The owner of the Rancho Los Feliz has 3000 yards of
fencing with which to enclose a rectangular piece of
grazing land along the straight portion of a river.
 Fencing is not required along the river.
 Letting x denote the width of the rectangle, find a function
f in the variable x giving the area of the grazing land if she
uses all of the fencing.
Applied Example: Enclosing an Area
Solution
 This information was given:
✦ The area of the rectangular grazing land is A = xy.
✦ The amount of fencing is 2x + y which must equal 3000
(to use all the fencing), so:
2x + y = 3000
 Solving for y we get:
y = 3000 – 2x
 Substituting this value of y into the expression for A gives:
A = x(3000 – 2x) = 3000x – 2x2
 Finally, x and y represent distances, so they must be
nonnegative, so x  0 and y = 3000 – 2x  0 (or x  1500).
 Thus, the required function is:
f(x) = 3000x – 2x2
(0  x  1500)
Applied Example: Charter-Flight Revenue
 If exactly 200 people sign up for a charter flight, Leisure
World Travel Agency charges $300 per person.
 However, if more than 200 people sign up for the flight
(assume this is the case), then each fare is reduced by $1
for each additional person.
 Letting x denote the number of passengers above 200, find
a function giving the revenue realized by the company.
Applied Example: Charter-Flight Revenue
Solution
 This information was given.
✦ If there are x passengers above 200, then the number of
passengers signing up for the flight is 200 + x.
✦ The fare will be (300 – x) dollars per passenger.
 The revenue will be
R = (200 + x)(300 – x)
= – x2 + 100x + 60,000
 The quantities must be positive, so x  0 and 300 – x  0
(or x  300).
 So the required function is:
f(x) = – x2 + 100x + 60,000
(0  x  300)
End of
Chapter