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slide1
2
  • 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

Functions and Their Graphs

slide2

2.1

The Cartesian Coordinate System and Straight lines

the cartesian coordinate system
The Cartesian Coordinate System
  • We can represent real numbers geometrically by points on a real number, or coordinate, line:
the cartesian coordinate system1
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

– 1

–2

–3

–4

the cartesian coordinate system2
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

– 1

–2

–3

–4

x

the cartesian coordinate system3
The Cartesian Coordinate System
  • The point where these two lines intersect is called the origin.

y

4

3

2

1

– 1

–2

–3

–4

Origin

x

the cartesian coordinate system4
The Cartesian Coordinate System
  • In thex-axis, positive numbers are to the right and negative numbers are to the left of the origin.

y

4

3

2

1

– 1

–2

–3

–4

Negative Direction

Positive Direction

x

the cartesian coordinate system5
The Cartesian Coordinate System
  • In they-axis, positive numbers are above and negativenumbers are below the origin.

y

4

3

2

1

– 1

–2

–3

–4

Positive Direction

x

Negative Direction

the cartesian coordinate system6
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

3

2

1

– 1

–2

–3

–4

(4, 3)

x

(3, –1)

(–1, –2)

the cartesian coordinate system7
The Cartesian Coordinate System
  • The axes divide the plane into four quadrants as shown below.

y

4

3

2

1

– 1

–2

–3

–4

Quadrant II

(–, +)

Quadrant I

(+, +)

x

Quadrant III

(–, –)

Quadrant IV

(+, –)

slope of a vertical line
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
Slope of a Nonvertical Line
  • If (x1, y1) and (x2, y2) are two distinct points on a nonvertical lineL, then the slopem of L is given by

y

L

(x2, y2)

y2 – y1 = y

(x1, y1)

x2 – x1 = x

x

slope of a nonvertical line1
Slope of a Nonvertical Line
  • If m > 0, the line slants upwardfromleft to right.

y

L

m = 1

y = 1

x = 1

x

slope of a nonvertical line2
Slope of a Nonvertical Line
  • If m > 0, the line slants upwardfromleft to right.

y

L

m = 2

y = 2

x = 1

x

slope of a nonvertical line3
Slope of a Nonvertical Line
  • If m < 0, the line slants downwardfromleft to right.

y

m = –1

x = 1

y = –1

x

L

slope of a nonvertical line4
Slope of a Nonvertical Line
  • If m < 0, the line slants downwardfromleft to right.

y

m = –2

x = 1

y = –2

x

L

examples
Examples
  • Sketch the straight line that passes through the point (2, 5) and has slope –4/3.

Solution

  • Plot the point(2, 5).
  • A slope of –4/3 means that if xincreases by 3, ydecreases by 4.
  • Plot the resulting point(5, 1).
  • Draw a line through the two points.

y

6

5

4

3

2

1

x = 3

(2, 5)

y = –4

(5, 1)

x

1 2 3 4 5 6

L

examples1
Examples
  • Find the slopem 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
examples2
Examples
  • Find the slopem 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
examples3
Examples
  • Find the slopem of the line that goes through the points(–2, 5) and (3, 5).

Solution

  • The slope of a horizontal line is zero:

y

6

4

3

2

1

(3, 5)

(–2, 5)

L

m = 0

x

–2 –1 1 2 3 4

parallel lines
Parallel Lines
  • Two distinct lines are parallel if and only if their slopes are equal or their slopes are undefined.
example
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 slopem1 of L1 is given by
  • The slopem2 of L2 is given by
  • Since m1=m2, the lines L1 and L2 are in fact parallel.
slide23

2.2

Equations of Lines

equations of lines
Equations of Lines
  • Let L be a straight lineparallel to the y-axis.
  • Then Lcrosses 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, ), where is an appropriate number.
  • The vertical lineL can therefore be described as

x = a

y

L

(a, )

(a, 0)

x

equations of lines1
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
  • Multiplying both sides by x – x1 we get
point slope form
Point-Slope Form
  • An equation of the line that has slope m and passes through point (x1, y1) is given by
examples4
Examples
  • Find an equation of the line that passes through the point (1, 3) and has slope 2.

Solution

  • Use the point-slope form
  • Substituting for point(1, 3) and slopem = 2, we obtain
  • Simplifying we get
examples5
Examples
  • Find an equation of the line that passes through the points (–3, 2) and (4, –1).

Solution

  • The slope is given by
  • Substituting in the point-slope form for point (4, –1) and slope m = – 3/7, we obtain
perpendicular lines
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
example1
Example
  • Find the equation of the line L1 that passes through the point (3, 1) and is perpendicular to the line L2 described by

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
crossing the axis
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

x

(a, 0)

L

slope intercept form
Slope-Intercept Form
  • An equation of the line that has slopem and intersects the y-axis at the point(0, b) is given by

y = mx + b

examples6
Examples
  • Find the equation of the line that has slope3 and y-intercept of –4.

Solution

  • We substitute m = 3 and b = –4 into y = mx + b and get

y = 3x – 4

examples7
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.
  • Comparing to y = mx + b, we find that m = ¾ and b = –2.
  • So, the slope is ¾ and the y-interceptis –2.
applied example
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 value3 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
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 variablesx and y.

general form of a linear equation1
General Form of a Linear Equation
  • An equation of a straight line is a linear equation; conversely, every linear equation represents a straight line.
example2
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).
example3
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

–1

–2

–3

–4

(4, 0)

x

1 2 3 4 5 6

(0, –3)

equations of straight lines
Equations of Straight Lines

Vertical line:x = a

Horizontal line:y = b

Point-slope form:y – y1 = m(x – x1)

Slope-intercept form:y = mx + b

General Form:Ax + By + C = 0

slide41

2.3

Functions and Their Graphs

functions
Functions
  • A function f is a rule that assigns to each element in a setAone and only one element in a setB.
  • The setA 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 setB 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.
example4
Example
  • Let the function fbe defined by the rule
  • Find: f(1)

Solution:

example5
Example
  • Let the function fbe defined by the rule
  • Find: f(–2)

Solution:

example6
Example
  • Let the function fbe defined by the rule
  • Find: f(a)

Solution:

example7
Example
  • Let the function fbe defined by the rule
  • Find: f(a + h)

Solution:

applied example1
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
  • Find ThermoMaster’s weekly profit if its level of production is:
    • 1000 thermometers per week.
    • 2000 thermometers per week.
applied example2
Applied Example

Solution

  • We have
    • The weekly profit by producing1000 thermometers is

or $2,000.

    • The weekly profit by producing2000 thermometers is

or $7,000.

determining the domain of a function
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
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

x

16

applied example packaging1
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.
    • Find the expression that gives the volumeV of the box as a function of x.
    • What is the domain of the function?
  • The dimensions of the resulting box are:

x

10 – 2x

16 – 2x

applied example packaging2
Applied Example: Packaging

Solution

a. The volume of the box is given by multiplying its dimensions(length ☓ width ☓ height), so:

x

10 – 2x

16 – 2x

applied example packaging3
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

must be satisfied simultaneously. Simplified:

All three are satisfied simultaneously provided that:

Thus, the domain of the function f is the interval [0, 5].

more examples
More Examples
  • Find the domain of the function:

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 examples1
More Examples
  • Find the domain of the function:

Solution

  • Our only constraint is that you cannot divide by zero, so
  • Which means that
  • Or more specifically x ≠–2 and x ≠ 2.
  • Thus the domain of f consists of the intervals (–, –2), (–2, 2), (2, ).
more examples2
More Examples
  • Find the domain of the function:

Solution

  • Here, any real number satisfies the equation, so the domain of f is the set of all real numbers.
graphs of functions
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 functionf 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).
example8
Example
  • The graph of a function f is shown below:

y

y

(x, y)

Range

x

x

Domain

example9
Example
  • The graph of a function f is shown below:
    • What is the value of f(2)?

y

4

3

2

1

–1

–2

x

1 2 3 4 5 6 7 8

(2,–2)

example10
Example
  • The graph of a function f is shown below:
    • What is the value of f(5)?

y

4

3

2

1

–1

–2

(5,3)

x

1 2 3 4 5 6 7 8

example11
Example
  • The graph of a function f is shown below:
    • What is the domain of f(x)?

y

4

3

2

1

–1

–2

x

1 2 3 4 5 6 7 8

Domain: [1,8]

example12
Example
  • The graph of a function f is shown below:
    • What is the range of f(x)?

y

4

3

2

1

–1

–2

Range:

[–2,4]

x

1 2 3 4 5 6 7 8

example sketching a graph
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:
example sketching a graph1
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

10

8

6

4

2

x

– 3 – 2 – 1 1 2 3

example sketching a graph2
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

10

8

6

4

2

x

– 3 – 2 – 1 1 2 3

example sketching a graph3
Example: Sketching a Graph
  • Sketch the graph of the function defined by the equation

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
  • In the subdomain[0, ), the rule for f is given by
example sketching a graph4
Example: Sketching a Graph
  • Sketch the graph of the function defined by the equation

Solution

  • Substituting negative values for x into , while

substituting zero and positive values into we get:

example sketching a graph5
Example: Sketching a Graph
  • Sketch the graph of the function defined by the equation

Solution

  • Plotting these data and graphing we get:

y

3

2

1

x

– 3 – 2 – 1 1 2 3

the vertical line test
The Vertical Line Test
  • A curve in the xy-plane is the graph of a functiony = f(x)if and only if each vertical line intersects it in at mostone point.
examples8
Examples
  • Determine if the curve in the graph is a function of 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.

y

x

examples9
Examples
  • Determine if the curve in the graph is a function of x:

Solution

  • The curve is not a function of x, because there is more than one value of y for some values of x.

y

x

examples10
Examples
  • Determine if the curve in the graph is a function of 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.

y

x

slide73

2.4

The Algebra of Functions

the sum difference product and quotient of functions
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

2000

1800

1600

1400

1200

1000

y = R(t)

y = S(t)

S(t)

Billions of Dollars

R(t)

t

1990 1992 1994 1996 1998 2000

t

Year

the sum difference product and quotient of functions1
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

2000

1800

1600

1400

1200

1000

y = R(t)

y = S(t)

S(t)

Billions of Dollars

D(t) = R(t) – S(t)

R(t)

t

1990 1992 1994 1996 1998 2000

t

Year

the sum difference product and quotient of functions2
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 asR(t) and S(t).

y

400

200

0

–200

–400

y = D(t)

t

Billions of Dollars

t

1992 1994 1996 1998 2000

D(t)

Year

the sum difference product and quotient of functions3
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 functions4
The Sum, Difference, Product and Quotient of Functions
  • Let f and g be functions with domains A and B, respectively.
  • The sumf + g, the differencef – g, and the productfg of f and g are functions with domain A ∩ Band rule given by

(f + g)(x) = f(x) + g(x)Sum

(f–g)(x) = f(x) –g(x)Difference

(fg)(x) = f(x)g(x) Product

  • The quotientf/g of f and g has domain A ∩ Bexcluding all numbers x such that g(x)= 0 and rule given by

Quotient

example13
Example
  • Let and g(x) = 2x + 1.
  • Find the sums, the differenced, the productp, and the quotientq 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:
example14
Example
  • Let and g(x) = 2x + 1.
  • Find the sums, the differenced, the productp, and the quotientq 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:
applied example cost functions
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 functions1
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 functions2
Applied Example: Cost Functions

Let’s now consider profits

  • Suppose that the total revenueR realized by Puritron from the sale of x water filters is given by

R(x) = –0.0005x2 + 20x (0 ≤x≤ 40,000)

  • Find
    • The total profit function for Puritron.
    • The total profit when Puritron produces 10,000 filters per month.
applied example cost functions3
Applied Example: Cost Functions

Solution

  • The totalprofitP realized by the firm is the difference between the total revenueR and the total costC:

P(x) = R(x) – C(x)

= (–0.0005x2 + 20x)– (–0.0001x2 + 10x + 10,000)

= –0.0004x2 + 10x – 10,000

  • The totalprofit realized by Puritron when producing10,000 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
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:
  • Evaluating the function gat the pointf(x), we find that:
  • This is an entirely new function, which we could call h:
the composition of two functions1
The Composition of Two Functions
  • Let f and g be functions.
  • Then the composition of g and f is the function ggf(read “gcirclef ”) defined by

(ggf)(x) = g(f(x))

  • The domain of ggfis the set of all x in the domain of f such that f(x) lies in the domain of g.
example15
Example
  • Let
  • Find:
    • The rule for the composite function ggf.
    • The rule for the composite function fgg.

Solution

  • To find ggf, evaluate the function g at f(x):
  • To find fgg, evaluate the function f at g(x):
applied example automobile pollution
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 ofmotor vehiclesisx 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:
    • An expression for the concentration of CO in the air due to automobile exhaust t years from now.
    • The level of concentration 5years from now.
applied example automobile pollution1
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 pollution2
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.

slide91

2.5

Linear Functions

linear function
Linear Function
  • The function f defined by

where m and b are constants, is called a linear function.

applied example linear depreciation
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 depreciation1
Applied Example: Linear Depreciation

Solution

  • Let V(t) denote the Web server’s book value at the end of thetth 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
applied example linear depreciation2
Applied Example: Linear Depreciation

Solution

  • Using the point-slope form of the equation of a line with point(0, 10,000) and slopem = –1400, we obtain the required expression
  • The book value at the end of the second year is given by

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 depreciation3
Applied Example: Linear Depreciation

Solution

  • The graph of V is:

V

(0, 10,000)

10,000

(5, 3000)

3000

t

1 2 3 4 5 6

cost revenue and profit functions
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 manufacturingxunits of the product

  • The revenue function is

R(x) = Total revenue realized from the sale ofxunits of the product

  • The profit function is

P(x) = Total profit realized from manufacturing and selling xunits of the product

applied example profit function
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,
finding the point of intersection
Finding the Point of Intersection
  • Suppose we are given two straight linesL1 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 equationy = m1x + b1.
  • The point P(x0, y0) also lies on the line L2 and so satisfiesy = m2x + b2 as well.
  • Therefore, to find the point of intersectionP(x0, y0) of the lines L1 and L2, we solve for x and ythe system composed of the two equations

y = m1x + b1 and y = m2x + b2

example16
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
  • Substituting this value of x into either one of the given equations yieldsy = 2.
  • Therefore, the required point of intersection is (1, 2).
example17
Example
  • Find the point of intersection of the straight lines that have equations

y = x + 1 and y = –2x + 4

Solution

  • The graph shows the point of intersection(1, 2) of the two lines:

y

5

4

3

2

1

L1

(1, 2)

x

–1 1 2 3 4 5

L2

applied example break even level
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 functionR and the cost functionC are given respectively by
  • Setting R(x) = C(x), we obtain
applied example break even level1
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 = 2000into R(x) =10x gives
  • So, Prescott’s break-even point is 2000 units of the product, resulting in a break-even revenue of $20,000 per month.
slide104

2.6

Quadratic Functions

quadratic functions
Quadratic Functions
  • A quadratic function is one of the form

where a, b, and c are constants and a ≠0.

  • For example, the function

is quadratic, with a = 2, b = –4, and c = 3.

quadratic functions1
Quadratic Functions
  • Below is the graph of the quadratic function
  • The graph of a quadratic function is a curve called a parabola that opens upward or downward.

y

10

8

6

4

2

Parabola

x

–2 –1 1 2 3 4

quadratic functions2
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 highestpoint of the parabola, which is called the vertex of the parabola.

Axis of symmetry

y

10

8

6

4

2

Parabola

Vertex(1, 1)

x

–2 –1 1 2 3 4

quadratic functions3
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
  • If we complete the square in x, we obtain
  • Note that (x – 1)2is 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 functions4
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

4

2

–2

x

–2 2 4

Vertex(1, –2)

properties of quadratic functions
Properties of Quadratic Functions

Given f(x) = ax2 + bx +c (a ≠0)

  • The domain of f is the set of all real numbers.
  • If a> 0, the parabolaopens upward, and if a< 0, it opens downward.
  • The vertex of the parabola is
  • The axis of symmetry of the parabola is
  • The x-intercepts (if any) are found by solving f(x)= 0. The y-intercept is f(0) =c.
example18
Example
  • Given the quadratic function f(x) = –2x2 + 5x – 2
    • Find the vertex of the parabola.
    • Find the x-intercepts (if any) of the parabola.
    • Sketch the parabola.

Solution

  • Here a = –2, b = 5, and c = –2. therefore, the x-coordinate of the vertex of the parabola is

The y-coordinate of the vertexis therefore given by

Thus, the vertex of the parabola is the point

example19
Example
  • Given the quadratic function f(x) = –2x2 + 5x – 2
    • Find the vertex of the parabola.
    • Find the x-intercepts (if any) of the parabola.
    • Sketch the parabola.

Solution

  • For the x-intercepts of the parabola, we solve the equation

using the quadratic formula with a = –2, b = 5, and c = –2.

We find

Thus, the x-intercepts of the parabola are 1/2 and 2.

example20
Example
  • Given the quadratic function f(x) = –2x2 + 5x – 2
    • Find the vertex of the parabola.
    • Find the x-intercepts (if any) of the parabola.
    • Sketch the parabola.

Solution

  • The sketch:

y

Vertex

1

–1

–2

x-intercepts

x

–1 1 2

y-intercept

some economic models
Some Economic Models
  • People’s decision on how much to demand or purchase of a given product depends on the price of the product:
    • Thehigherthe price thelessthey want to buy of it.
    • A demand functionp = d(x) can be used to describe this.
some economic models1
Some Economic Models
  • Similarly, firms’ decision on how much to supply or produce of a product depends on the price of the product:
    • Thehigherthe price, themorethey want to produce of it.
    • A supply functionp = s(x) can be used to describe this.
some economic models2
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 curveintersect: where d(x)= s(x).
applied example supply and demand
Applied Example: Supply and Demand
  • The demandfunction for a certain brand of bluetooth wireless headset is given by
  • The corresponding supply function is given by

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 demand1
Applied Example: Supply and Demand

Solution

  • We solve the following system of equations:
  • Substituting the second equation into the first yields:
  • 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 demand2
Applied Example: Supply and Demand

Solution

  • The equilibrium price is given by:

or $40per headset.

slide120

2.7

Functions and Mathematical Models

mathematical models
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
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

Mathematical model

Solve

Test

Solution of real-world problem

Solution of mathematical model

Interpret

modeling with polynomial functions
Modeling With Polynomial Functions
  • A polynomial function of degree n is a function of the form

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:
modeling with polynomial functions1
Modeling With Polynomial Functions
  • A polynomial function of degree n is a function of the form

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:
applied example global warming
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:
applied example global warming1
Applied Example:Global Warming
  • Below is a scatter plot associated with these data:

y (ppmv)

380

360

340

320

t (years)

10 20 30 40 50

applied example global warming2
Applied Example:Global Warming
  • A mathematical model giving the approximate amount of CO2 is given by:

y (ppmv)

380

360

340

320

t (years)

10 20 30 40 50

applied example global warming3
Applied Example:Global Warming
  • Use the model to estimate the average amount of atmospheric CO2 in 1980(t= 23).
  • Assume that the trend continued and use the model to predict the average amount of atmospheric CO2 in 2010.
applied example global warming4
Applied Example:Global Warming

Solution

  • The average amount of atmospheric CO2 in 1980 is given by

or approximately 338 ppmv.

  • Assuming that the trend will continue, the average amount of atmospheric CO2 in 2010 will be
applied example social security trust fund
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:
  • The scatter plot associated with these data is:

y ($trillion)

6

4

2

t (years)

5 10 15 20 25 30

applied example social security trust fund1
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:
  • A mathematical model giving the approximate value of assets in the trust fund (in trillions of dollars) is:

y ($trillion)

6

4

2

t (years)

5 10 15 20 25 30

applied example social security trust fund2
Applied Example: Social Security Trust Fund
  • The first baby boomers will turn 65 in 2011. What will be the assets of the Social Security trust fund at that time?
  • The last of the baby boomers will turn 65 in 2029. What will the assets of the trust fund be at the time?
  • Use the graph of function A(t) to estimatethe year in which the current Social Security system will go broke.
applied example social security trust fund3
Applied Example: Social Security Trust Fund

Solution

  • The assets of the Social Security fund in 2011(t = 3) will be:

or approximately $3.18 trillion.

The assets of the Social Security fund in 2029(t = 21) will be:

or approximately $5.59 trillion.

applied example social security trust fund4
Applied Example: Social Security Trust Fund

Solution

  • The graph shows that function Acrosses the t-axis at about t = 32, suggesting the system will go broke by 2040:

y ($trillion)

6

4

2

t (years)

5 10 15 20 25 30

rational and power functions
Rational and Power Functions
  • A rational function is simply the quotient of two polynomials.
  • In general, a rational function has the form

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 functions1
Rational and Power Functions
  • Examples of rational functions:
rational and power functions2
Rational and Power Functions
  • Functions of the form

where r is any real number, are called power functions.

  • We encountered examples of power functions earlier in our work.
  • Examples of power functions:
rational and power functions3
Rational and Power Functions
  • Many functions involve combinations of rational and power functions.
  • Examples:
applied example driving costs
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:
  • A mathematical model giving the average cost in cents per mile is:

where x(in thousands) denotes the number of miles the car is driven in 1 year.

applied example driving costs1
Applied Example: Driving Costs
  • Below is the scatter plot associated with this data:

y (¢)

140

120

100

80

60

40

20

C(x)

x (☓1000 miles/year)

5 10 15 20 25

applied example driving costs2
Applied Example: Driving Costs
  • Using this model, estimate the average cost of driving a 2007 medium-sized sedan 8,000 miles per year and 18,000miles per year.

Solution

  • The average cost for driving a car 8,000 miles per year is

or approximately 68.8¢/mile.

applied example driving costs3
Applied Example: Driving Costs
  • Using this model, estimate the average cost of driving a 2007 medium-sized sedan 8,000 miles per year and 18,000miles per year.

Solution

  • The average cost for driving a car 18,000 miles per year is

or approximately 48.95¢/mile.

constructing mathematical models
Constructing Mathematical Models
  • Some mathematical models can be constructed using elementary geometric and algebraic arguments.

Guidelines for constructing mathematical models:

    • Assign a letter to each variable mentioned in the problem. If appropriate, draw and label a figure.
    • Find an expression for the quantity sought.
    • Use the conditions given in the problem to write the quantity sought as a functionf 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
Applied Example: Enclosing an Area
  • The owner of the Rancho Los Feliz has 3000yards 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 functionf in the variablex giving the area of the grazing land if she uses all of the fencing.
applied example enclosing an area1
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
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 $1for each additional person.
  • Letting x denote the number of passengers above200, find a function giving the revenue realized by the company.
applied example charter flight revenue1
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)