Extra Notes on Slides

Ch 1.1 Numbers, Data, and Problem Solving Important Definition ExamplesNatural Numbers – referred as counting numbers N = { 1,2,3,4,…. }

Integers - Includes the natural numbers, their opposites, and 0 ….., -2, -1, 0, 1, 2,…..

Rational number -- Any number that can be expressed as the ratio of two integers p/q, where q = 0 , all repeating and all terminating decimals 2/1, 1/3, -1/4, 22/7, 1.2, 0.4369,

– 0.25 = ¼, 0.33 = 1/3, and all fractions

Irrational numbers Can be written as non-repeating, non-terminating decimals; cannotbe rational numbers; a square root of A positive integer that is not an integer. Real Numbers – Represented by standard or decimal numbers, includes the rationalnumbers and irrational numbers

Scientific notation A number in the form c x 10 n , where 1 < c < 10 and n is an integer,Used to represent numbers that are large or small in absolute value 0.789 = 7.89 x 10 -1

854,000 = 8.54 x 10 5

Percent change - If a quantity changes from c1 to c2, then the percent change equals c2 – c1 x 100 If the price of a gallon of gasoline changes from $1.00 to $ 1.40, then

c1

the percent change is 1.40 – 1 x 100 1

25

7 3 5

Ch 1.2 – Visualization of Data

Mean or average – To find the mean or average of n numbers, divide their sum by n. The mean of the four numbers -2, 3, 4,5, 6,8 is -2 + 3 + 4 + 5 + 6 + 8 = 4 6 Range - The range of a list of data is the difference between the maximum and the minimum of the numbers or values. The range of the data 3, -6, 2, 8, 10 is 10 – (-6) = 16 (Maximum – Minimum = Range)

Median – The median of a sorted list of numbers equals the value that is located in themiddle of the list. Half the data is greater than or equal to the median, and half the data is less than or equal to the median. The median of 1,3,5,9, the average of the two middle values 3, 5. Therefore 3+5 = 4 is the median 2

Ch 1.2

Relation is a set of ordered pairs. If the ordered pairs in a relation by (x, y). Then the set

of all x-values is called the domain( D) of the relation and the set of all y-values is called

the range (R). If the relation S = {(0, 3), ( 1,2), (1, 6), (2, 7), (3, 5) }

The relation has domain

D = { 0,1, 2, 3 }

R = { 2, 3, 5, 6, 7 }

Graphing a relation

( 1, 2)

(3, 5)

(1, 6)

Connecting the data points with straight-line segments called a line graph

(2, 7)

Scatter Plot

Distance Formula- The distance between (x1, y1) and (x2, y2) is d= (x2 = x1) 2+ (y2 – y1)2

Example - The distance between (1, -3) and (1, 4) d = = 7

Midpoint Formula – The midpoint of the line segment connecting ( x1, y1) and (x2, y2) is

M= (x1 + x2, y1 + y2).

2 2

The mid point of the line segment connecting ( 3,4) and ( 1, -2) is

M = ( 3 + 1, 4 – 2) =( 2, 1)

2 2

(-3))– ((4 1)-1 ( 2 = 4922

Determine whether the triangle is isosceles ?

Isosceles triangle has at least two equal sides

The vertices of the given triangle are (0,0), (7,1), (3,4)

d = ( 3- 0)2 + (4 – 0)2 = 3 2 + 4 2 = 9 + 16 = 25 = 5 ( Using distance formula )

d= ( 7 – 3) 2 + ( 1 – 4)2 = 4 2 + (-3) 2 = 16 + 9 = 25 = 5

The side between ( 0, 0) and (7, 1)and the side between

(3, 4) and (7, 1) have equal length, so the triangle is isosceles.

(0,0) (3, 4)

(7, 1)

5 5

Midpoint Formula ( pg 27, no 46 ex 1.2)

The population of US was 151 million in 1950 179 million in 1960What will be the population in 1970 ?

Solution

Let y represent the population in 1970. The data point ( 1960, 179) as the midpoint between the data point (1950, 151)and (1970, y).

The increase between 1950 and 1960 was 28 million. Therefore, theestimated population in 1970 is 179 + 28 = 207 million.

The midpoint formula isM = ( 1950 + 1970 , 151 + y ) = (160, 179 ) 2 2

M = 151 + y = 179 or y = 207 million 2

207 million population in 1970

Data involving two variables

Pg 27, No 62 ( Ex 1.2 )

{ (1, 1), (3, 0), (-5, -5), (8, -2), (0, 3) }

a) The Domain is D = {1,3, -5, 8, 0} and the range is R = {1, 0, -5, -2, 3}

b) The minimum x-value is -5, and the maximum x-value is 8. The minimum y-value is -5, and the maximum y-value is 3.

c) The axes must include -5< x < 8 and -5 < y < 3

(0, 3)

(1, 1)(3, 0)

(8, -2)

(-5, -5)

-8 -6 -4 -2 0 2 4 6 8

4

2

-2

-4

Function Notation

The notation y = f(x) is called function notation. The input is x, the output is y, and the name of the function is f.

Name

y = f(x)

Output Input

variable y is called independent variable and variable x is called dependent variable

A function computes exactly one output for each valid input.

Expression f(16) = 4 is read “f of 16 equals 4 and indicates that f outputs 4 when the input is 16

Representations of Functions

Verbal representation (words) – Words describe precisely what isComputed- Example- f(x) = x + 3, 3 is added with Input x to get output

Numerical representation (Table of values ) – A numerical representation is a table of values that lists input-output pairs for a Function- f(x) = 2x

Diagrammatic Representation (Diagram) – Functions are sometimes represented using diagrams

Symbolic Representation ( formula)- Mathematical formula – The function is defined by f(x) = x

Graphical Representation (Graph) – Graph of ordered pairs (x, y) that satisfy y = f(x) A graph of f(x) = 2x Each point on the graph satisfy y = 2x

Vertical line Test - If every vertical line intersects a graph at no more than one point, then the graph represents a function

x 0 1 2

f(x) 0 2 4

567

1

2

579

123

Use the graph of the function f to estimate its domain and range. Evaluate f(0)

Ex 1.3

- 3 -1 1 3

3

1

-1

-3

Domain (D) = {x/ -3 < x < 3}Range (R) = { y/ 0< y < 3}f(0) = 3

-2 -1 1 2 3 4

Domain D = { x/ -2 < x < 4Range R = { y / -2 < y < 2 }; f(0) = -2

2

1

-1

-2

-2 -1 1 2

Domain D = all real numbersRange R = all real numbers;f(0) = 0

43 45 48

No 74 ( Ex 1.3)In 2004 the average cost of driving a new car was about 50 cents per mile. Give symbolic, graphical, and numerical representations that compute the cost in dollars of driving x miles. For the numerical representations that compute the cost in dollars of driving x miles. For the numerical representation use a table with x = 1, 2, 3, 4, 5, 6

0

20

4

0

6

0

8

0

1

00

0 20 40 60 80 100

Symbolic expression : f(x) = 0.50x

Numerical

GraphicalCost of driving is $0.50 per mile

Miles 1 2 3 4 5 6

cost 0.50 1.00 1.50 2.00 2.50 3.00

Applications No. 105 ( Ex 1.3)When the relative humidity is 100%, air cools 5.80 F for every 1-mile increase in altitude. Give verbal, symbolic, graphical, and numerical representations of a function f that computes this change in temperature for an increase of x miles. Let the domain of f be 0< x < 3

Verbal : Multiply the input x by – 5.8 to obtain the change in temperature

Symbolic : f(x) = - 5.8 X Graphical : Y 1 = - 5.8XNumerical :Table Y1 = - 5.8X

Types of Functions and their rates of Change

Constant Function - A function f represented by f(x) = b, where b is constant ( fixed

number ), is a constant function

Linear Function – A function f represented by f(x) = ax + b, where a and b are

constants, is a linear function

Slope – The slope m of the line passing through the points (x1, y1 ) and (x2, y2 ) is

y = y2 – y1 where x1 = x2

x x2 – x1

m = 2 > 0 m = - ½ < 0 m = 0 m is undefined

Example – a line passing through ( 1, 3) nand (2, 7) has slope m = 7-3 = 4 2 - 1This slope indicates that the line rises 4 units for each unit increase in x

Non-Linear Functions

If a function is not linear, then the function is called a nonlinear function.

The graph of a nonlinear function is not a straight line

f(x) = x2

f(x) = x3

Cube Function

f(x) =Square Root Function

x

f(x) = xAbsolute Value Function

Average Rate of Change

Let (x1, y1) and (x2, y2 ) be distance points on the graph of a function f. The average

rate of change of f from x1 to x2 ) is y2 – y1

x2 – x1

That is, the average rate of change from x1 to x2 equals the slope of the line

passing through ( x1, y1) and (x2, y2)

Difference Quotient

The difference quotient of a function f is an expression of the form

f(x + h) – f(x) where h = 0 h

CH 2 Linear Functions and Equations

Modeling with a linear function

To model a quantity that is changing at a constant rate with a linear

function f, the following may be used

f(x) = (constant rate of change)x + (initial amount)

Note : The constant rate of change corresponds to the slope of the

graph of f and the initial amount corresponds to the y-intercept

Correlation Coefficient r, (-1 < r < 1)

Value of r Comments Sample Scatterplot

r = 1 There is an exact linear fit. The line passes

through all data points and has a positive slope

r = -1 There is an exactlinear fit. The line passes through

all data points and has a negative slope

0<r<1 There is a positive correlation. As the x-values increase,

so do the y-values. The fit is not exact

-1 < r < 0 There is a negative correlation. As the x-values increase,

the y- values decrease. The fit is not exact

r = 0 There is no correlation. The data has no tendency toward

being linear. A regression line predicts poorly

Choose STAT and CALC LINEAR REGRESSION and ENTER Hit Y and ENTER a= 1.5 and b = - 1/3

Find the line of least –squares fit for the data points ( 1, 1), (2, 3), and (3, 4). What is the correlation coefficient ? Plot the data and graph

Use Graphing Calculator

Hit STAT and EDIT Enter points (1, 1) , (2, 3), (3, 4) in the table

LINEAR REGRESSION

Hit STAT PLOT ON will blink, Clear OFF (If it is black by using side

arrows and enter)

and choose SCATTER PLOT and

MARK using down arrow keys

Hit WINDOW and Hit 2nd and Tableenter points [ 0, 5, 1] by [0, 5, 1] Hit GRAPH

Linear RegressionExample 2 Cellular Phones – One of the early problems with cellular phones was the delay involved

with placing a call when the system was busy. One study analysed this delay. The table shows that as the number of calls increased by P percent, the average delay time D to put through a call also increased.

P (%) 0 20 40 60 80 100

D( minutes) 1 1.6 2.4 3.2 3.8 4.4

Let P correspond to x-values and D to y-values. Find the least-squares regressionline that models these data. Plot the data and the regression line Estimates the delay for a 50% increase in the number of calls

SolutionUse Graphing Calculator

Hit STAT and EDIT Enter points (0, 1) , (20, 1.6), (40, 2.4), (60, 3.2), (80, 3.8), (100, 4.4) in the table

Choose STAT and CALC and LINEAR REGRESSION and enterHit Y and enter a= 0.0349 and b= 0.9905

Hit STAT PLOT ON will blink, Clear OFF (If it is black by using s arrows and enter) and choose SCATTER PLOT and MARK using down arrow keys

Hit WINDOW and enter points [ -10, 110, 10 ] by [ 0, 5, 1] Hit GRAPH

Point- Slope Form- the line with slope m passing through the point (x1, y1) has an equation y = m(x – x1) + y1 or y – y1 = m(x – x1)

Slope- intercept form- The line with slope m and y-intercept b is given by y = mx + b, the slope- intercept form of the equation of a line.

Finding intercepts – To find any x-intercepts, let y = 0 in the equation andsolve for xTo find any y-intercepts, let x = 0 in the equation and solve for y

Equations of Horizontal and vertical lines – An equation of the horizontalLine with y-intercept b is y = b. An equation of the vertical line with x-intercept kis x = k

Perpendicular lines- Two lines with nonzero slopes m1 and m2 areperpendicular if and only if their slopes have product – 1, that is m1m2 = -1

Interpolation –Estimates that are between two or more known data values

Extrapolation – Estimates values that are not between two known data values

The table lists data that are exactly lineara) Find the slope-intercept form of the line that passes through these data points.b) Predict y when x = -2.7 and 6.3. Decide if these calculations involve interpolation or extrapolationNo 54 ( Ex 2.2)

Since the point (0, 6.8) is on the graph, the y-intercept is 6.8. The data is exactly linear, so one can useany two points to determine the slope.

Using the points (0, 6.8) and (1, 5.1), m = 5.1- 6.8 = -1.7 ( slope formula , m = y2 – y1 1 – 0 x2 – x1

Slope intercept form of the line is y = - 1.7x + 6.8 ( y = mx+ b)b) When x = - 2.7, y = - 1.7(-2.7) + 6.8 = 11.39 This calculation involves extrapolationWhen x = 6.3, y = -1.7(6.3) + 6.8 = -3.91 This calculation involves extrapolation

x -2 -1 0 1 2

y 10.2 8.5 6.8 5.1 3.4

Y-intercept

2.3 – Linear Equations

Linear equation in one variable – A linear equation in one variable is an equation that can be written in the form ax + b = 0where a and b are real numbers with a= 0If the equation is not linear then the equation is nonlinear equation

x – 4 = 0, 2x – 3 = 0, x – 4 + 3(x – 1) = 0 These are examples of linear equations

Properties of Equality

Addition Property of equalityIf a, b, and c are real numbers, then a = b is equivalent to a + c = b + c

Multiplication Property of EqualityIf a, b, c are real numbers with c = 0 then a =b is equivalent to ac = bc

Intersection of Graphs Method

The intersection of graphs method can be used to solve the equation

graphically. To implement this procedure, follow these steps

Step 1 : Set y1 equal to the left side of the equation, and set y2 equal to

the right side of the equation

Step 2 : Graph y1 and y2

Step 3 : Locate any points of intersection. The x- coordinates of these

points correspond to solutions to the equation

Intermediate value Property

Let (x1, y1) and (x2, y2) with y1= y2 and x1< x2 be two points on the

graph of a continuous function f. Then , on the interval x1 < x < x2, f

assumes every value between y1 and y2 at least once

Solving Application problems

Step 1 : Read the problem and make sure you understand it. Assign avariable to what you are being asked to find. If necessary, write otherquantities in terms of this variable

Step 2 : Write an equation that relates the quantities described in theproblem. You may asked to sketch a diagram refer to known formulas

Step 3 : Solve the equation and determine the solution

Step 4 : Look back and check your solution. Does it seem reasonable ?

Use the intersection of graphs method to solve the equation by hand. Check your answer ( Ex 2.3) 51 ) x + 4 = 1 – 2x 52) 2x = 3x -1 y1 = x + 4, Y2 = 1 – 2x Let y1 = 2x y2 = 3x - 1

-4 -3 -2 -1 1 2 3 4

3

1

-1

-3

The line intersects at x = - 1 The line intersects at x = 1

-4 -3 -2 -1 1 2 3 4

3

2

1

-1

-2

-3

( 1, 2)

Solve the linear equation with the intersection –of-graphs method. Approximate the solution to the nearest thousandth whenever appropriate ( Ex 2.3)

No 58 No 60 No 63 8 – 2x = 1.6 = 4x – 6 6 – x = 2x – 3 7 3

2Enter Y1= 8-2xY2 = 1.6

Enter Y1= Y2 = 4x - 6

2

2

EnterY1 = (6 – x) / 7Y2 = (2x – 3) /3

Hit Graph and calc, you will get graph and point of intersection

ApplicationsNo 89 ( Ex 2.3 ) . Conical Water Tank A water tank in the shape of an inverted cone

has a height of 11 feet and a radius of 3.5 feet, as illustrated below, If the volume of the cone is V = 1/3 r2 h, Find the volume of the water in the tank when the water is 7 feet deep.

3.5 feet

11 feet

Use similar triangles to find the radius of the cone when the water is 7 feet deep; r = 7 , ( Multiply by 3.5 ) , r = 7 (3.5) , r = 2.23 ft3.5 11 11V = 1 r2 h , 100 = 1 (3) 2 . h, 100 = 3 h 100 = h, h = 10.6 ft 3 3 3

No 96 ( Ex 2.3)Perimeter Find the length of the longest side of the rectangle if its

perimeter is 25 feet

2x

5x - 1

Perimeter P = 2w + 2L = 25 ( given)2(2x) + 2(5x – 1) = 254x + 10x – 2 = 2514x = 25 + 2 ( Add 2)14x = 27x = 27 ft = 1.93 ft 14 5x – 1 = 5( 27/4) – 1 ( substitute x value )= 121/14 = 8.6 ft

2.4 Linear Inequalities

Linear Inequality in one variable

A linear inequality in one variable is an inequality that can be written in

the form ax +b > 0

Where a = 0 ( The symbol > may be replaced by > , <, or <

Examples of linear inequalities are

3x – 2 < 0, 2x + 5 > 6, x + 4 < 9 and 2x + 3 > -3x + 4

Interval Notation

Inequality Interval Notation Number line Graph

- 2 < x < 2 ( - 2, 2)

- 1 < x < 3 ( - 1, 3]

- 3 < x < 2 [ - 3, 2 ]

x < - 1 or x > 2 ( - , - 1) U (2, )

x > - 1 ( - 1, )

x < 2 ( - , 2 ]

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

]

][

( )

]

( )

(

- < x <

)

-4 -3 -2 -1 0 1 2 3 4

Properties of Inequalities

Let a, b, c be real numbers.  a < b and a+c < b+c are equivalent ( The same number may be added to or subtracted from both sides of an

inequality.)Example To solve x, x – 7 < 5, x < 12( add 7 both sides )

If c > 0, then a < b and ac < bc are equivalent.(Both sides of an inequality may be multiplied or divided by the same positive

number)Example : To solve x 3x > 6, divide each by 3 we get, x > 2

If c< 0, then a < b and ac > bc are equivalent.Each side of an inequality may be multiplied or divided by the same negative

number provided the inequality symbol is reversed. Example : To solve x,-7x > 21, divide by -7 we get x < - 3

Note : Replacing < with < and > with > results in similar properties  

Compound Inequalities

 Two inequalities connected by the word and or orExamplesx < 3 or x > 4x> - 3 and x < 4The inequality x> 5 and x < 20 can be written as the three-partinequality 5< x < 20

Ex 2.4

Solving Linear Inequalities Symbolically15) -2(x – 10) + 1 > 0 -2x + 20 + 1 > 0 ( distribute -2 to remove parenthesis) -2x > -21 ( add -21 both sides) x < 21/2= 10.5 (Divide by -2 both sides, the inequality sign will change) ( - , 10.5)

37) 1 < 1 – 2t < 2 2 3 3

3 < 1 – 2t < 2 2 1 < - 2t < 1 2 - 1 > t >- 1 4 2 - 1 < t < 1 2 -4 ( -1 , - 1 ] 2 4

Solving Linear Inequality Graphically ( Ex 2.4) 41 43 46

x + 2 > 2x 2x -2 > - 4 x + 4 -2 < 1-x < 2

3 3

Y1 = x + 2, Y2 = 2x Y 1 = -2, Y2 = 1-x, Y3 = 2Y1 = 2x -2 , Y2 = - 4 x + 4 3 3

[ -10, 10, 1] [ -10, 10, 1] [ -10, 10, 1]

Y1 > Y2, when the line of Y1 is above the line of Y2, which by the graph is left of the intersection point ( 2, 4) implies { x/x< 2}

Y1 > Y2, when the line of Y1 is above the line of Y2, which by the graph is left of the intersection point ( 3, 0) implies { x / x> 3}

Y1 > Y2, when the line of Y1 is above the line of Y2, which by the graph is left of the intersection point ( -1, 2) and (3, - 2) does not include { x/ -1 < x < 3}

Solve the compound linear inequality graphically. Write the solution set in interval notation, and approximate endpoints to the nearest tenth whenever appropriate

Ex 2.463 66

3 < 5x – 17 < 15 0.2x < 2x -5 < 8 3

Y1 = 3, Y2 = 5x – 7 and Y3 = 15 Y1 = 0.2x, Y2 = 2x – 5, Y3 = 8 3

Y1 = 3, Y2 = 5x – 17, Y3 = 15

[ -5, 15, 5] by [ -5, 20, 5]

Y1 = 0.2x , Y2 = (2x – 5)/3 , Y3 = 8

No 70 ( Ex 2.4)Use the figure to solve each equation or inequalitya) f(x) = g(x)b) g(x) = h(x)

f(x) < g(x) < h(x)d) g(x) > h(x)

0 1 2 3 4 5 6 7 8

700

600

500

400

300

200

100y = f(x)

y = h(x)

y = g(x)

a) f(x) = g(x) = 4b) g(x) = h(x) = 2c) f(x) < g(x) < h(x)

2 <x < 4d) g(x) > h(x)

0 < x < 2

No 92 ( Ex 2.4) Motorcycles The number of Harley-Davidson motorcycles manufactured between 1985 and 1995 can be approximated by N(x) =6409(x – 1985) + 30,300, where x is the year

a) Did the demand for Harley- Davidson motorcycles increase or decrease over this time period ? Explain your reasoning.

b) Estimate the years when production was between 56,000 and 75,000

Solution

a) N(x) = 6409(x – 1985) + 30,300

Slope= m

Slope of the graph of N is 6409. This means that Harley- Davidson has sold

approximately 6409 more motorcycles each year. Demand has increased during

this time period because the slope of N is positive

b) Graph Y1 = 6409(x – 1985) + 30,300, Y2 = 56,000, Y3 = 75,000

The points of intersection occur near (1989, 56,000) and (1992, 75,000).

For 1989< x < 1992. Sales were between 56,000 and 75,000 motorcycles per year.

That is, from 1989 to 1992 there were from 56,000 to 75,000 motorcycles sold per

year

Y1

Y2

Y3

2.5 Piece wise- defined function

A function is piecewise defined if it has different formulas on different

intervals of its domain. Many times the domain is restricted

Examples f(x) = 2x + 1 if -3 < x < 0

f(x) = 2 if - 5 < x < - 1 x - 1 if 0 < x < 3

x + 3 if -1 < x < 5 The domain of f { x/ - 3< x < 3}

f(-2) = 2(-2) + 1= -3

The domain of f - 5 < x < 5 f(0) = 0 -1 = -1

f( -2) = 2 f(3)= 3 – 1 = 2

f(0) = 0 + 3 f is not continuous

f(3)= 3 + 3

= 6

f is continuous

-6 -4 -2 2 4 6

8

6

4

2

-2

8

6

4

2

-2

-6 -4 -2 2 4 6

Absolute value function f(x) = x , The output from the absolute value function isnever negative, Examples f(3) = 3 = 3 f(6) = - 6 = 6

Absolute value equations ax + b = k with k > 0 is equivalent to ax + b = + k

Example 3x – 2 = 4 3x – 2 = 4 or 3x – 2 = - 43x = 6 or 3x = -2x = 2 or x = -2/3If k < 0 , then the absolute value equation has no solution

Absolute value inequalities Let the solutions to ax + b = k, k > 0 be s1 and s2 with s1 < s2 and the solution set to ax + b > k is x < s1 or x > s2

Examples To solve x – 5 < 3 or x – 5 >3First solve x – 5 = -3 and x – 5 = 3 x = 2 x = 8The solutions are 2 to 8The solution set to x – 5 < 3 is 2 < x < 8And the solution set to x – 5 > 3 is x < 2 or x > 8

V shaped Never go below the x-axis

Alternative method for solving absolute value inequalities

1. ax + b < k is equivalent to –k < ax + b < k

x - 1 < 4 is solved as follows

4 < x – 1 < 4

- 4 < x < 5

2. ax + b > k is equivalent to ax + b < -k or ax + b > k

x – 1 > 3 is solved as follows

x – 1 < -3 or x – 1 > 3

x < - 3 or x > 4

Swimming Pool Levels The graph of y = f(x) shows the amount of water in thousands of gallons remaining in a swimming pool after x daysa) Estimate the initial and final amounts of water in the pool.b) When did the amount of water in the pool remain constantc) Approximate f(2) and f(4)d) At what rate was water being drained from the pool when 1 < x < 3

a) Initial amount in the pool occurs when x = 0. f(0) = 50or 50,000 gallons. The final amount of water in the pool occurs when x = 5. Then f(5) = 30 or 30,000 gallons

b) The water level remained constant during the first day and the fourth day, when 0 < x < 1 or 3 < x < 4

c) f(2) = 45 thousand and f(4) = 40 thousandd) During the second and third days, the amount of water changed from 50,000

gallons to 40,000 gallons. This represents 10,000 gallons in 2 days or 5000 gallons per day were being pumped out of the pool

0 1 2 3 4 5 6 Time ( days)

125

100

75

50

25

0

Gal

lons

(th

ousa

nds)

No 33 ( Ex 2.5) Graph y = f(x)Use the graph of y = f(x) to sketch a graph of the equation y = f(x)c) Determine the x intercept for the graph of the equation y = f(x)

• A) The graph of Y1 = 2x

• B) The graph of y = 2x is similar to the graph of y = 2x except that is reflected across x-axis whenever 2x < 0 The graph of Y1 = 2x

• C) The x-intercept occurs when 2x = 0 or when x = 0. The x-intercept is located at (0,00

Solve the absolute value equation ( Ex 2.5)

42 51

- 3x – 2 = 5 4x – 5 + 3 = 2

Then – 3x – 2 = -5 4x – 5 = -1, has no solution

-3x= - 5 + 2 ( add 2) since the absolute value of

3x = -3 any quantity is always greater

x = 1 ( Divide by – 1) than or equal to 0

Or -3x – 2 = 5

-3x = 7 ( Add 2)

x = - 7/3 ( Divide by – 3)

No 61 ( Ex 2.5) Solve the equation graphically, numerically and symbolically

a) Graph Y1 = abs(2x – 5) and Y2 = 10

b) Table Y1 = abs(2x -5) starting at -5, incrementing by 2.5

c) 2x – 5 = 10

2x – 5 = -10

Or 2x – 5 = 10

x = -2.5 or 7.5

From each method, the solution to 2x – 5 < 10 lies between – 2.5 and 7.5,

-2.5 < x < 7.5

No 95 ( Ex 2.5) Average temperature Analyze the temperature range in Boston MassachussettsThe inequality T – 50 < 22 describes the range of monthly average temperatures T in degree Fahrenheit.Solve the inequality graphically and symbolicallyb) The high and low monthly average temperatures satisfy the absolute value equation T – 50 = 22

• Graph Y1 = abs (T – 50) and Y2 = 22• The V- shaped graph of y1 intersects the horizontal line at the points • T – 50 = -22 or T – 50 = 22 T = 28 or 72• The average monthly temperature range is 28 0 F• B) The monthly average temperatures in Boston vary between a low

of 280 F and a high of 720 F

3.1 Quadratic Function and Models

Quadratic Function- Let a, b and c be real numbers with a= 0. A function represented by f(x) = ax2 + bx + c is a quadratic function

VertexAxis of Symmetry

Axis of Symmetry

y = ax2 , a> 0 y = ax2 a< 0

Vertex Form- The parabola graph of f(x) = a(x – h) 2 + k with a = 0 has vertex (h,k).Its graph opens upward when a > 0 and opens downward when a < 0Let f(x) = 3(x – 1) 2 + 4 Parabola opens upward: a > 0Vertex : ( 1, 4) Axis of symmetry: x = 1

Vertex Formula The vertex of the graph of f(x) = ax2 + bx + c with a = 0 is the point ( -b/2a, f( -b/2a))

Example – The x –value of the vertex on the graph of f(x) = x2 – 6x +3x = - 6/ 3(1) = - 2Y-value of vertex f(- 2)= (-2) 2 – 6(-2) + 3 = 4 + 12 + 3= 19

Completing the square to find the vertex formTo complete the square for x2 + kx, add (k/2) 2 to make a perfect square trinomialY = x2 – 4x + 1Y – 1 = x2 – 4xY – 1 = x2 – 4x + 22 Add ( 4/2) 2

Y = (x – 2) 2 + 1

82 Maximizing Revenue- A large hotel is considering the following group

No 82 Ex 3.1- Maximizing Revenue A large hotel is considering the following group discount on room rates. The regular price of a room is $120, but for each room rented the price deceases by $2, for example, one room costs $118, two rooms cost $116 x 2 = $232, three rooms cost $114 x 3 = $ 342, and so ona) Write a formula for a function R that gives the revenue for renting x roomsb) Sketch a graph of R. What is a reasonable domain of R ?C) Determine the maximum revenue and the number of rooms that should be rented

a) R(x) = x(120 – 2x)b) The graph of R(x) = x(120 – 2x) or R(x) = - 2x2 + 120x, since when renting 60 rooms the price is $0 a reasonable domain d = { 0 < x < 60 }c) The graph is a parabola opening downward, by the vertex formula, maximum

area occurs when R= -b/2a R = 120/2(-2)= 30Maximum revenue occurs when 30 rooms are rentedR(30) = - 2(30)2 + 120(30) = -1800 + 3600= $ 1800 is the maximum revenue

10 20 30 40 50 60

200

4

0 0

60

0

800

180

0

Quadratic Regression

No 104, Ex 3 .1 Quadratic Models Use least-squares regression to find a quadratic function f that

Models the data given in th etable. Then estimate f(3, 5) to the nearest hundreth

x 0 2 4 6

f(x) -1 16 57 124

Enter the data into your calculator, and then select quadratic regression from the menu. The modeling function f is given by f(3.5) = 3.125x2 + 2.05x – 0.9= 3.125(3.5)2 + 2.05(3.5) – 0.9 = 44.56

No 102 Ex 3.1Suspension bridge- The cables that support a suspension bridge, such as the Golden Gate

Bridge, can be modeled by parabolas. Suppose that a 300- foot long suspension bridge has towers at its ends that are

120 feet tall. If the cable comes within 20 feet of the road in the center of the bridge, find a quadratic function that

models the height of the cable above the road a distance of x feet from the center of the bridge

300

Cable

500Tower

Tower

The vertex is (2000, 20) and another point on the cable is (0, 500).

Using vertex form, f(x) = a(x – h) 2+ kf(x) = a(x – 0) 2 + 20 = ax2 + 20F( 150) = 120 , f( 150) = a(150)2 + 20, 120 = 22.500a + 20

100 = 22.500a a =1/225 The shape of the cable is given by the equation f(x) = 1/225 x 2 + 20 = 0.0044x2 + 20

20 feet

120 feet

Quadratic EquationA quadratic equation in one variable is an equivalent that can be written

in the form ax2 +bx +c= 0, where a, b, c are real numbers with a = 0

Square Root Property

Let k be a nonnegative number. Then the solutions to

the equation.

x2 = kare x = + k. If k < 0. Then this equation has no

real

solutions.

Quadratic Formula and the discriminant

The solutions of the equation ax 2 + bx + c = 0 with

a = 0 are given by - b + b2 – 4ac

• Complex Numbers i2 = - 1 or i =

• For a > 0, = = i

1

2a

Discriminant D = b2 - 4acIf D > 0, the equation has two unequal real solutionsIf D = 0, the equation has one real solution of multiplicity twoIf D < 0, the equation has two complex (conjugate) solutions

a 1 a a

Quadratic Equation

The solutions of the quadratic equation

ax2 + bx + c = 0, where a, b, c are real numbers with a = 0

No x intercepts One x – intercepts Two x - intercepts

Ex 1

Solving Quadratic Equations by Factoring

Zero Factor Principle

The product of two factors equals zero if and only if one or both of the factors equals zero.

In symbols ab = 0 if and only if a = o or b = 0

Example (x – 6) (x + 2) = 0 x – 6 = 0 or x + 2 = 0 x = 6 or x = -2Check 6, and – 2 are two solutions and satisfy the original equationAnd x-intercepts of the graph are 6, -2

By calculator, draw the graph

Quadratic Formula

• The solutions to ax 2 + bx + c = 0 with a = 0 are given by - b + b2 – 4ac

2aX =

Vertical and Horizontal Translations

y2 = x2 + 1

y1= x2

y3 = x2 - 2

y1= x2

y1= (x-1)2

y1= x2

y2= (x + 2 )2

Translated horizontally to the left 2 units

Translated horizontally to the right 1 unit

Translated upward and downward

Ch 4.1Polynomial function f of degree n in the variable x can be represented by

f(x) = an xn + an-1xn-1 + …… + a2 x2 + a1x + a0 where each coefficient ak is a

real number, an = 0, and n is a nonnegative integer. The leading coefficient is

an an the degree is n

Increasing functions – f increases on an interval, if whenever x1 < x2, then

f(x1) < f(x2)

Decreasing functions – f decreases on an interval, if whenever x1 < x2, then

f(x1) > f(x2)Even Function f(-x) = f(x), The graph is symmetric with respect to the y-axis

Odd function f(-x) = -f(x) The graph is symmetric with respect to the origin

Absolute, or global maximum (minimum) – The maximum ( minimum) y-value on the graph y = f(x)

Local, or relative, maximum( minimum)- A maximum (minimum) y-value on the graph y = f(x) in an open interval of the domain of f

Applications No 118 ( Ex 4.1)

Natural Gas The US consumption of natural gas from 1965 to 1980 can bemodeled by f(x) =0.0001234x4 – 0.005689x3 + 0.08792x2 –0.5145x = 1.514, where x = 6 corresponds to 1966and x = 20 to 1980. Consumption is measured intrillion cubic feet. a) Evaluate f(10) and interpret the resultb) Graph f in [ 6, 20, 5] by [0.4, 0.8, 0.1]c) Determine the local extrema and interpret the results

a) f(10) = 0.0001234(10)4 – 0.005689(10)3 + 0.08792(10)2 – 0.5145(10) + 1.5140.706 in 1970 U.S consumption of natural gas was

about 0.706 trillion cubic feet.

Local Maximum

Local minimum

Ex 4.1 No 121. Average temperature The accompanying graph approximates the monthly average temperatures in degrees Farenheit in Austin, Texas. In this graph x represents the month, where x = 0 corresponds to July

a) Is this a graph of an odd or even function ?B) June corresponds to x = -1 and August to x = 1. The average temperature in June is 83degree F . What is the average temperature in August ?C) March corresponds to x= 4. According to the graph, how do their average temperature compare ? D) Interpret what this type of symmetry implies about average temperatures in Austin

-8 -4 0 4 8

90

70

60

50

Mar Jul Nov a) Even b) 83 degree Farenheit c) They are equal

Use interval notation to identify where f is increasing and when f is decreasing Ex 4.1

35 36

The graph of this cubic equation has turning points ( -2, 20) and (1, -7)It is increasing ( - , -1) or ( 1, ) and decreasing [-1, 1]

The graph of this cubic equation has a negative lead coefficient therefore is reflected through the x-axis has turning points(-1, -15) and (3, 17). It is increasing [-1, 3] and decreasing: ( - , -1] or [3, )

Determine graphically anya) local extremab) absolute extrema

Ex 4.1 No 64

The local maximum of 0 occurs at x= 0 and x= 2. A local minimum of – 1 occurs when x = 1

Ex 4.1 No 68

Local maximum of 6 and occurs when x = -1. There is no local minima

Polynomial Functions and Models

A polynomial function f of degree n can be expressed as f(x) = a0 xn + ….+ a2x2 + a1x + a0, where each coefficient ak is a real number, ak = 0 and n is a nonnegative integer. The leading coefficient is ak

where n is the largest exponent of x

f(x) is a polynomial and f(x) = 0

f(x) = x3 – 3x2 +x – 5 is a polynomial function

And f(x) = x3 – 3x2 +x – 5 = 0 is a polynomial equation

A turning point occurs whenever the graph of a polynomial function

changes from increasing to decreasing or from decreasing to

increasing. Turning points are associated with ‘hills’ or ‘valleys’ on a

graph

Quadratic Polynomial Function

The solutions of the quadratic equation

ax2 + bx + c = 0, where a, b, c are real numbers with a = 0

No x intercepts One x – intercepts Two x - intercepts

Quadratic, a> 0 Quadratic, a > 0 Quadratic a < 0Both sides graph go up End behavier tends to both sides End behavier is switched and tends to -

Cubic Polynomial Functions

Quartric Polynomial Functions

a > 0 a < 0 a< 0

a < 0 a > 0 a> 0

Piecewise- Defined FunctionsEvaluate f(x) at the given value of x

y = f(x)Ex 4.2 No 67

x = -2 and 1f(-2) = 5f(1) = 0

-3 -1 1 3

10

6

2

70

x = -2, 0, and 2f(-2) = 0, f(0) = -3, f(2) = 2

Sketch the graph of f and determine whether f is continuous on its domain and solve f(x)= 0

Ex 5.2 No 76

f(x)= x2 if -2 < x < 0

x + 1 if 0 < x < 2

The graph of f is discontinous at x = 0x2 = 0 , x = 0, this is not the domain of the equationx + 1 = 0, x = -1, this is not the domain of the equation. So no solution

Ex 4.2 No 77

f(x)= 2x if -5 < x < -1

- 2 if -1 < x < 0

x2 – 2 if 0 < x < 2

f is continuousx2 -2 = 0x = 2

Ex 4.2 No 88.The table lists the number in thousands of Americans over 100 years old for selected yearsa) Use least-squares regression to find a function that models the data. Let x = 0 corresponds to 1994b) Estimate the number of Americans over 100 in the year 2008

Ch 4.4 The Fundamental Theorem of Algebra

Properties of the imaginary unit i

i= i2 = - 1

The expression

If a > 0, then = i

1

a a

Complex Numbers

a + biReal Numbers Imaginary numbers

a + bi , b = 0 a + bi, b = 0 Fundamental theorem of AlgebraA polynomial f(x) of degree n > 1 has atleast one complex zero

Number of Zeros TheoremA polynomial of degree n has atmost n distinct zerosConjugate zeros theoremIf a polynomial f(x) has only real coefficients and if a + bi is a zero of f(x), then the conjugate a – bi is also a zero of f(x)

4.5 Rational Functions and Models

Rational Function A function f represented by f(x) =

p(x)/q(x), where p(x) and q(x) are polynomials and q(x) = 0

, is a rational function

Vertical Asymptote The line x = k is a vertical asymptote

of the graph of f if f(x) or f(x) as x

approaches k from either left or the right

Horizontal Asymptote The line y = b is a horizontal asymptote of the

graph of f, if f(x) b as x approaches either or -

Variation

Direct Variation as the nth power- Let x and y denote two quantities and n be a positive number. Then y is directly proportional to the nth power of x, or,y varies directly as the nth power of x, if there exists a nonzero number k such that y = kxn

Inverse variation as the nth powerLet x and y denote two quantities and n be a positive number. Then y is inversely proportional to the nth power of x, or y varies inversely as the nth power of x, if there exists a nonzero number k such thaty = k/xn

If y = k/xn, then y is inversely proportional to x or y varies inversely as x

4.5 Solving Polynomial Inequalities

• Step1 If necessary, write the inequality as p(x) < 0, where p(x) is a polynomial and < may be replaced by >, <, or >

• Step 2 Solve p(x) = 0 either symbolically or graphically. The solutions are called boundary numbers

• Step 3 Use the boundary numbers to separate the number line into disjoint intervals. On each interval, p(x) is either always positive or always negative

• Step 4 To solve the inequality, either make a table of test values for p(x) or use a graph of p(x). For example, the solution set for p(x) < 0 corresponds to intervals where test values result in negative outputs or to intervals where the graph of p(x) is below the x axis

Solving Rational Inequalities

Step 1 If necessary, write the inequality in the form p(x)/q(x) > 0, where p(x) and q(x) are polynomials. Note that > may be replaced by <,< or >

Step 2 Solve the equations p(x) = 0 and q(x) = 0 either symbolically or graphically. The solutions are boundary numbers

Step 3 Use the boundary numbers to separate the number line into disjoint intervals. On each interval, p(x)/q(x) is either always positive or always negative

Step 4 Use a table of test values or a graph to solve the inequality in step `1

Ch 4.7 Power functions and Radical Equations Properties of Rational ExponentsLet m and n be positive integers with m/n in lowest terms and n> 2.Let r and p be rational numbers. Assume that b is a nonzero real number and that each expression is a real number.

Property

1 b m/n = (bm)1/n = (b1/n)m

2 b m/n = n bm = (n b)m

3 (br)p = brp

4 b-r = 1/br

5 brbp = br+p

6 br/bp = b r-p

Ex 4.7

No 41 No 43

• f(x) = x - 1 f(x) = - 13

2

2x

Ch 5.1 Combining Functions

Operations on FunctionsIf f(x) and g(x) both exists, the sum,difference, product and quotient of twoFunctions f and g are defined by (f + g) (x) = f(x) + g(x),(f – g) (x) = f(x) –g(x)(fg)x = f(x).g(x), and (f/g)(x) = f(x)/g(x), where g(x) = 0

Composite Functions

• If f and g are functions, then the composite functions g0f, or composition of g and f is defined by

• (gof)(x) = g(f(x))

• Note : we read g(f(x)) as “g of f of x”

Compositions of FunctionMultiplication: (fg)(5) Composition: (fog)5

Input 5 Input 5

f(x) g(x) g(x)

Output f(5) Output g(5) Output g(5)

(g(5) becomes the input for f)

(fg)(x) = f(x).g(x) (f0g)(x) = f(g(x))

OUTPUT f(5)og(5) OUTPUT f(g(5))

Ex 5.1 No 117 Methane Emissions – Methane is a greenhouse gas. It letssunlight into the atmosphere but blocks heat from escaping the earth’s atmosphere. Methane is a by-product of burning fossil fuels. In the table, f models the predicted methane emissions in millions of tons produced by developed countries. The function g models the same emissions for developing countries

x 1990 2000 2010 2020 2030

f(x) 27 28 29 30 31

g(x) 5 7.5 10 12.5 15

(a) Make a table for a function h that models the total predicted methane emissions for developed and developing countries(b) Write an equation that relates f(x), g(x), and h(x)

(a) To find the total emissions we must add the developed and developing Countries emissions for each year

(b) h(x) = f(x) + g(x)

x 1990 2000 2010 2020 2030

f(x) 27 28 29 30 31

g(x) 5 7.5 10 12.5 15

Ex 5.1 No 118 Methane Emissions The accompanying figure shows graphs of the functions f and g that model methane emissions. Use their graphs to sketch a graph of the function h.( referred to 117)

• A graphical solution can be found by adding the corresponding y-values. For example when x = 1990, f(1990) = 27 and g(1990) = 5, so h(1990) = 27 + 5 = 32. Since f and g appear to be linear functions, their sum will also be linear. A graph of h = f + g, together with f and g is shown in the figure.

1990 2000 2010 2020 2030

50

40

30

20

10

y = f(x) + g(x)

y = f(x)

y = g(x)

Ex 5.1 no 119. Methane Emissions (Refer to 117 and 118) Symbolic representations for f and g are f (x) = 0.1x – 172 and g (x) = 0.25x – 492.5, where x is the year. Find a symbolic representation for h

h(x) = f(x) + g(x)

= (0.1x-172) + ( 0.25x -492.5) ( Substitute f(x) and g(x) )

= 0.1x +0.25x -172 – 492.5 ( Combine like term)

= 0.35x – 664.5

5.2 Inverse Functions and Their Representations

One- to- one Function

A function f is a one – to-one function if, for elements c and d in the

domain of f, c = d implies f(c) = f(d)

That is different inputs always result in different outputs

Horizontal Line Test If every horizontal line intersects the graph of a

function f atmost once, then f is a one-to-one function

Inverse Function Let f be a one-to-one function. Then f -1 is the

inverse function of f, if

(f -1 0 f)(x)= f -1 ( (f(x)) = x for every x in the domain of f and

(f 0 f -1 )(x) = f(f -1 (x)) = x for every x in the domain of f -1

Finding a symbolic Representation for f -1

To find a formula for f -1, perform the following steps

Step 1 Verify that f is a one to one function

Step 2 Solve the equation y = f(x) for x, resulting in the equation x = f -1 (y)

Step 3 Interchange x and y to obtain y = f -1 (x)

To verify f -1 (x) , show that (f -1 0f)(x)= x and (f 0 f -1 )(x) = x

Domains and Ranges of Inverse Functions

The domain of f equals the range of f -1

The range of f equals the domain of f -1

Use the graph of f to determine if f is One – to – One Functions ( Ex 5.2 )

16 17

-3 3

3

1

-3

-3 -2 -1 1 2 3

One- to – One Not one –to-One

3

-3

Graph y = f(x), and y = x. Then graph y = f -1(x)

112 116f(x)= -1/2 x + 1 f(x) = 1x

y = f-1(x) = -2x + 2 y = f -1(x) = x3 + 1

Ex 5.2

No 121 ( Ex 5.2) Advertising Costs – The line graph at the top of the next page represents a function C that computes the cost of a 30-second commercial during a Super Bowl telecast. Perform each calculation and interpret the results.a) Evaluate C(1995)b) Solve C(x) = 1 for xc) Evaluate C-1(1)

1991 1995 1999 2003

2.4

2.0

1.6

1.2

0.8

0.4

Year

Cos

t (

mill

ions

of

Dol

lars

Solutiona) h(x) = g(x) –f(x)b) h(1995) = g(1995) – f(1995)= 800 - 600 = 200; h(2000) = g(2000) – f(2000) = 950 – 700 = 250c) The function h is linear and passes through the points (1995, 200) and (2000, 250) Therefore , its graph has a slope of m = 250 -200 = 10 2000 – 1995Thus h(x) = 10(x – 1995) + 200 = 10x – 19,750

Exponential Functions and Models

Exponential Function

A function f represented by

f(x) = C ax, a> 0, a = 1, and C > 0, is an exponential function with

base a and coefficient C

Exponential FunctionA function f represented byf(x) = C ax, a> 0, a = 1, and C > 0, is an exponential function with base a and coefficient C

5.3 Exponential Functions and Models

f(x)= 2xg(x)= 2x

Modeling Radioactive Decay

If a radioactive sample containing C units has a half-life of k

years, then the amount A remaining after x years is given

by A = C(1/2) x/k

A 7 gram sample of radioactive material with a half-life of

400 years is modeled by

A(x) = 7(1/2) x/400

Compound interest

If a principal of P dollars in an account paying an annual

rate of interest r (expressed in decimal form) compounded

(paid) n times per year, then after t years the account will

contain A dollars, where

A = P(1+ r/n)nt

Example

$ 700 at 8% compounded monthly for 4 years yields

700( 1 + 0.08/12) 12(4) = 2274.299

The number e is an irrational number and important in mathematics,

much like e = 2.718282

Natural exponential function f(x) = ex

Interest compounded continuously A= Pert, where P is the principal, r is the interest

rate( expressed in decimal form), t is the number of years, and A is the amount after

t years

$ 300 at 8% compounded continuously for 2 years yields

300e 0.08(2) = 649.62

Sketch the graph of y = f (x)Ex 5.3

44 f(x) = 4x 47 f(x) = 2(1/3) x 51 f(x) = 8(2) - x

No 91 (Ex 5.3)

Radioactive Cesium – 137 Radioactive cesium -137 was emitted in large amounts in the Chenobyl nuclear power station accident in Russia on April 26, 1986. The amount of cesium remaining after x years in an initial sample of 100 milligrams can be described by the formula A(x) = 100e -0.02295x

a) How much is remaining after 50 years ? Is the half life of cesium more or less than 50 years?b) Estimate graphically the half-life of cesium-137

Solutiona) A(x) = 100e -0.02295x

A(50) = 100e – 0.02295(50)= 31.7 milligramsSince the original sample contained 100 milligrams, the half –life would be the time required for the sample to disintegrate into 50 milligrams. Since 31.7 is less than this, the half life must be less than 50 years

b) Graph Y1 = 100e^ ( -0.02295x)

Y2 = 50The graphs intersect near (30.2, 50). Therefore, the half-life of cesium is approximately 30.2years

[0, 50, 10] by [0, 100, 10]

No 106 ( Ex 5.3) Survival of Reindeer For all types of animals, the percentage that survive into the next year decreases. In one study, the survival rate of a sample of reindeer was modeled by S(t) = 100(0.99993)t5. The function S outputs the percentage of reindeer that survive t years

a) Evaluate S(4) and S(15). Interpret the graph. Does the graph have a horizontal asymptote ?b) Graph S in [0,15, 5] by [0, 110, 10]. Interpret the graph. Does the graph have a horizontal asymptote ?Solution

a) S(t) = 100(0.99993)t5

= S(4) = 100(0.99993)1024

= 99.3

S(15) = 100(0.99993)759,375

= 0.49

After 4 years approximately 99% of the reindeer are still alive, while after 15

years only about 0.5% are still alive. Evidently a 15 year old reindeer is quite

old

b) Initially graph is horizontal at 100. This means that during the first 5 years, very few reindeer . Then the graph begins to decrease very rapidly,until at 13 years, only about 7% are still alive. After 13 years the graph begins to level off slightly. This means that a few reindeer live to an old ageYes, y = 100 and y= 0, percentage cannot be larger than 100 not smaller than 0

Horizontal first five years

decreasing

5.4 Logarithmic Functions and Models

The common logarithm of a positive number x, denoted logx, is defined

by logx = k if and only if x = 10k , where k is a real number. The function

given by f(x) = logx is called the common logarithmic function

x= 10klog 104 k

Input Output

Common Logarithms

Inverse Properties Of The common Logarithm

The following inverse properties hold for the common logarithmlog 10x = x for any real number x , and10logx = x for any positive number x

Logarithm – The logarithm with base a of a positive number x, denoted

by loga x, is defined by

logax = k if and only if x = ak, where a> 0, a = 1, and k is a real number. The function, given by

f(x) = log ax is called the logarithmic function with base a

Inverse PropertiesThe following inverse properties hold for logarithms with base a

Logaakx = x for any real number x, and

alogax = x for any positive number x

Inverse Functions – The inverse function of f(x) = ax is f-1(x) = logax

Examples, f(x) = 10x f-1(x) = logx

g(x) = ex , g-1(x) = ex g-1(x) = ln x

h(x) = log2(x) h-1(x)

Exponential Equations – To solve ax = k, take the base-a logarithm of each

side

Examples, 10x = 15 ex = 20

log10x = log15 ln ex = ln 20

x = log 15 x = ln 20

Logarithmic equations- To solve logax = k, exponentiate each side; base a

Examples, logx = 3 lnx = 5

10logx = 103 elnx = e5

x= 1000, x = e5

Applications No 114 ( Ex 5.4)

Diversity of Insects – The accompanying table lists the number of types of insects found in wooded regions with various acreages. Find the values for a and b so that f(x) = a +b logx models these data. Then use f to estimate an acreage that might have 1200 types of insects.

• Area (acres) 10 100 1000 10,000 100,000

• Types of Insects 500 800 1100 1400 1700

Solution – f(x) = a + blogx and f(10) = 500 , 500 = a =b log 10 , 500 = a + b(1)a + b = 500, f(100) = 800 = a + blog 100, 800 = a + b(2), a + 2b = 800Solving the equations for a gives a = 500 – b and a = 800 – 2b, thus500 – b = 800 – 2b , b = 300Since a + b = 500 and b = 300, a + 300 = 500a = 200, The function that models the given data is f(x) = 200 + 300 logxb) f(x) = 1200, 1200 = 200 + 300logx, 1000 = 300logx, logx = 10/3 10logx = 10 10/3, x = 10 10/3 = 2154. The acreage would be about 2154 acres

(Ex 5.4) No 123 Hurricanes – Hurricanes are some of the largest storms on earth. They are very low pressure areas with diameters of over 500 miles. The barometric air pressure in inches of mercury at a distance of x miles from the eye of a severe hurricane is modeled by the formula f(x) = 0.48 ln (x + 1) + 27

a) Evaluate f(0) and f(100) . Interpret the resultsb) Graph f in [0, 250, 50] by [25, 30, 1]. Describe how air pressure changes as one moves away from the eye of the hurricane.c)At what distance from the eye of the hurricane is the air pressure 28 inches of mercurySolution - a) f(x) = 0.48 ln (x + 1) + 27 , f(0) = 0.48 ln (0 + 1) + 27 = 0.48 ln 1 + 27= 27 and f(100) = 0.48 ln( 100 + 1) + 27 = 29.2 inches. At the center , or eye, of the hurricane the pressure is 27 inches of mercury, while 100 miles from the eye the air pressure has risen to 29.2 inches of mercuryb) Graph Y1= 0.48 ln (X + 1) + 27. At first, the air pressure rises rapidly as one moves away from the eye. Then, the air pressure starts to level off and does not increase significantly for distances greater than 200 milesc) f(x) = 28, 28 = 0.48 ln ( x + 1) + 27, ln (x + 1) = 1/0.48 = eln(x + 1) = e 1/0.48x + 1 = e1/0.48 , x = e1/0.48 – 1 = 7.03The air pressure is 28 inches of mercury about 7 miles from the eye of the hurricane

[0, 250, 50] by [25, 30, 1]

Properties of Logaritms

1. loga1 = 0

2. Logam + logan = loga(mn)

3. Logam – logan = loga (m/n)

4. Loga(mr) = rlogam

Example ln 1 = 0 and log22= 1

Log3 + log 6 = log(3.6) = log 18

Log38 – log32 = log3 8/2 = log34

Log 67 = 7log 6

Change of base formula- Let x, a = 1 be positive real numbers. Then logax = logbx/logba

Example – log36 = log 6/log3 = ln 6/ln3 = 1.631

Graphing logarithmic functions – Use the change of base formula to graph y = logax,

whenever a = 10 or a = e

To graph y = log2x , let

Y1 = log(X)/log(2) or

Y1 = ln(X)/ln(2)

Properties of Logaritms

No 54 ( Ex 5.5)

Complete the following.A) Make a table of f(x) and g(x) starting at x = 1, incrementing by 1. Determine whether f(x) = g(x).B) If possible, use properties of logarithms to show that f(x) = g(x)

f(x) = log2 + logx,

g(x) = log 2x

5.6 Exponential and Logarithmic FunctionsExponential equations Typical form , Cax = kSolve for ax. Then take a base a logarithm of each side. Use the inverse property:Logaax = xExample 4ex = 24, ex = 6Ln ex = ln6X = ln6 = 1.79

Logarithmic equations Equation : C logax = kSolve for logax. Then exponentiate each side with base a. Use the inverse propertyalogax = x4logx = 10Logx = 2.510logx = 10 2.5

x = 102.5 = 316

Equation : logabx + logacx = kWhen more than one logarithm with the same base occurs, use properties of logarithms tocombine logarithms. Be sure to check any solutionsExample Logx + log4x = 2Log4x2 = 24x2 = 102

x2 = 25x = + 5 -The only solution is 5

ApplicationsN0 69 ( Ex 5.6) Credit Cards From 1987 to 1996 the number of Visa cards and Master Cards was up

80% to 376 million. The function given by f(x) = 36.2e 0.14x models the amount of credit card spending from Thanksgiving to Christmas in billions of dollars. In this formula x = 0 corresponds to 1987 and x= 9 to 1996

a) Determine symbolically the year when this amount reached $ 55 billionb) Solve part (a) graphically or numerically

Solution – We must solve the equation f(x) = 5536.2e0.14x = 55e0.14x = 55/36.2Ln e0.14x

=ln 55/36.2x = ln 55/36.2/0.14= 2.99Since x = 0 corresponds to 1987, x = 3 represents 1987 + 3 = 1990, rounded to the nearest yearb) Graph Y1 = 36.2e^(0.14x) and Y2 = 55The graph intersect near (3, 55)Therefore , Christmas credit card spending was approximately $ 55 billion in 1990

Ch 9.1 Functions and Equations in Two Variables

The method of Substitution

To use the method of substitution to solve a system of two equations in two variables, perform the following stepsStep 1 : Choose a variable in one of the two equations. Solve the equation for that variableStep 2 : Substitute the result fro Step 1 into the other equation and solve for the remaining variable.Step 3 : Use the value of the variable from Step 2 to determine the value of the other variable. To do this, you may want to use the equation you

found in step 1.Note : To check your answer, substitute the value of each variable into the given equations. These values should satisfy both equationsLinear Equation Non Linear Equation Solve 56. 61 32.-2x – y = -2 x2 + y2 = 9 2x2 – y = 53x + 4y = -7 x + y = 3 -4x2 + 2y = 10Solve the first equation Solve the second equation Solve the first equation-y = 2x – 2 ( Add + 2x both sides) y = 3 - x 2x2 - y = 5 y = - 2x + 2 ( multiply by -1) Substitute this into first equation -y = -2x2 + 5Substitute y in second equation x2 + y2 = 9 y = 2x2 - 53x + 4( -2x + 2) = -7 x2 + (3 – x) 2 = 9 Substitute y into

the second equation3x -8x + 8 = -7 x2 + 9 – 6x + x2 = 9 -4x2+ 2( 2x2 – 5) = -10- 5x + 8 = - 7 2x2 – 6x = 0 -4x2 + 4x2 - 10- 5x = - 8 – 7 ( add – 8 both sides) 2x ( x – 3) = 0 -10 = -10- 5x = - 15 x = 0 or x = 3 There are infinitely many x = -15/-5 = 3 ( Divide by -5 both sides) When x = 0, y = 3 – 0 = 3 solutionsy = - 2 (3) + 2 = -6 + 2= -4 and when x = 3, y = 3 – 3 = 0x = 3, The solutions are ( 0, 3) and (3, 0)y = - 4

Graphical method for two equationsSolve both equations for the same variable. Then apply theintersection-of-graphs method47 ( Ex 9.1)2x + y = 1x – 2y = 3y= 1- 2xx – 2y = 3 -2y = -x + 3

y = ½( x – 3) Graphically Symbolically

Graph Y1 = 1-2x Substituting y = 1-2x into the

Y2 = 0.5 (x – 3) equation x – 2y = 3

gives x – 2(1-2x) = 3x – 2 + 4x = 3

5x = 5x = 1

If x = 1, then y = 1 – 2(1) = -1The solution is (1, -1)

Numerically

No 77 No 78e2x + y = 4 3x2 + y= 3

y = 4 – e2x y = 3 – 3x2

ln x – 2y = 0 (0.3)x + 4y = 1

Y = lnx/2 y = 1-(0.3)x/ 4

Graph Y1 = 4 – e^ (2X) Graph Y1 = 3 – 3X^2

Graph Y2 = ln (X) / 2 Graph Y2 = (1 – 0.3^X) /4

Solution = ( 0.714, - 0.169) Solution ( - 1.111, - 0.702), (0.971, 0.172)

Approximate to the nearest thousandth any solutions to the system of nonlinear equations graphically ( Ex 9.1)

Applications ( No 91 , Ex 9.1)Heart Rate In one study the maximum heart rates of conditioned atheletes were

examined. A group of atheletes were exercised to exhaustion. Let x represent an

athelet’s heart rate 5 seconds after stopping exercise and y this rate after 10 seconds. It

was found that the maximum heart rate H for these athletes satisfied the following two

Equations

H = 0.491x + 0.468y + 11.2

H = - 0.981x + 1.872y + 26.4

If an athelet had a maximum heart rate of H = 180, determine x and y graphically. Interpret

your answer

Solution – Let H = 180 in each equation and solve for y

180 = 0.491x + 0.468y + 11.2

y = 1/0.468 ( 180 – 0.491x – 11.2)

180 = - 0.981x + 1.872y + 26.4

y= 1/1.872 ( 180 + 0.981x – 26.4)

Since we are solving this system graphically., it is not necessary to simplify them. The graph

of both equations and their point of intersection near (177.1, 174.9).

This means that an exhaustion, if an athelet achieves a maximum heart rate of 180 beats per

minute, then 5 seconds after stopping his or her heart rate would be approximately 177.1

and after 10 seconds it would be approximately 174.9

[0, 300, 50] by [ 0, 300, 50]

Ch 9.2 Consistent, Dependent, IndependentConsistent system of linear equations in two variables – A consistent linear

system has either one or infinitely many solutions. Its graph is either distinct,

intersecting lines or identical lines

Dependent system of linear equations in two variables – A dependent linear

system has infinitely many solutions. The graph consists of two identical lines

Inconsistent system of linear equations in two variables- An inconsistent

linear system has no solutions. The graph is two parallel lines.

Example

The equations x + y = 1 and 2x + 2y = 2 are equivalent.

If we divide the second equation by 2 we obtain the

first equation. As a result, their graphs are identical

and every point on the line represents a

solution. Thus there are infinitely many solutions,

and the system of equations is a dependent system.

dependentInconsistent Unique

y y y

x x x

Linear ProgrammingIn a linear programming problem, the maximum or minimum of an objective function is found, subject to constraints. If a solution exists, it occurs at a vertex in the region of feasible solutions

Solving a linear programming problemStep 1 : Read the problem carefully. Consider making a table to display the

information givenStep 2 : Use the table to write the objective function and all the constraintsStep 3 : Sketch a graph of the region of feasible solutions. Identify all verticesor corner problemsStep 4 : Evaluate the objective function at each vertex. A maximum ( or aminimum) occurs at a vertex. Note : If the region is unbounded, a maximum ( orminimum) may not exist

Linear Programming Problems

Shade the region of feasible solutions

1

2

3 4

5

6

1 2 3 4 5 6 1 2 3 4 5 6 7 8

1

2

3 4

5

6

3x + 2y < 12 x + 2y < 82x + 3y < 12 2x + y > 2x > 0, y > 0 x > 0, y> 0

No 89 No 88

Ex (9.2)

9.3 Solving a system of linear equations in three variables

Step 1 : Eliminate one variable, such as x, from two of the equationsStep 2:Apply the technique discussed in Sections 9.1 and 9.2 to solve the two resulting equations in two variables from step 1. If x is eliminated, then solve these equations to find y and zNote : If there are no solutions for y and z, then the given system also has no solutions. If there are infinitely many solutions for y and z, then write y in terms of z and proceed to step 3Step 3: Substitute the values for y and z in one of the given

equations to find x. The solution is (x, y, z)

Solve the system of Linear EquationsNo 12 ( Ex 9.3)x – y + z = - 23x – 2y + z = -1x + y = - 3SolutionSubtract the first two equationsx – y + z = - 23x – 2y + z = -1-2x + y = 3Subtract this equation from the third x + y = - 3- 2x+ y = 3 3x = - 6 x = - 2Substitute x = -2 into x + y = - 3( -2) + y = - 3y = - 1Substitute x = - 2 and y = -1 into x – y + z = 2( -2) – (-1) + z = 2- 1 + z = 2 z= 3So the solution is ( - 2, - 1, 3)

No 25 Solve the system of linear equations2x + 6y + 2z = 62x + 7y + 4z = 134x = y + 2z = 7Multiply the second equation by 2 and subtract the third equation4x + 14y + 8z = 264x + y + 2z = 713y + 6z = 19Subtracting the two new equations we get13y + 6z = 1913y + 6z = 19 0 = 0Therefore, we have infinitely many solutions13y + 6z = 1913y = - 6z + 19y = -6z + 19 13Multiply the third equation by 3 and subtract from the first equation x + 3y + z = 312x + 3y + 6z =21-13x - 5z = -18x = - 5z + 18 13 We have infinitely many solutions( - 5z + 18 , - 6z + 19, z ) 13 13

Determinants

Determinant of a 2x2 Matrix

The determinant of

A = a b

c d

Is a real number defined by

Det A = ad – cb

Invertible Matrix

A square matrix A is invertible if and only if det A = 0

Minors and CofactorsThe minor, denoted by Mij, for element aij in the square matrix A is the

real number computed by performing the following steps.

Step 1: Delete the ith row and jth column from the matrix A

Step 2: Mij is equal to the determinant of the resulting matrix.

The cofactor, denoted Aij, for a0 is defined by Aij = (-1)i+j M ij

Determinant of a Matrix using the method of cofactors

Multiply each element in any row or column of the matrix by its factor.

The sum of the products is equal to the determinant

Cramer’s rule for linear systems in two variables

The solution to the linear system

a1 x + b1y = c1

a2x + b2y = c2

Is given by x = E/D and y = F/D, where

E = det c1 b1 , F = det a1 c1 and det a1 b1 = 0

c2 b2 a2 c2 a2 b2

Use Cramer’s rule to solve the system of linear equations

No 32 ( Ex 9.7)

5x – 3y = 4- 3x – 7y = 5

By Cramer’s rule , the solution can be found as follows

E = det 4 -3 - 13, F = det 5 4 = 37, D = det 5 -3

5 -7 - 3 5 - 3 - 7

Thus x = E/D = -13/-44 = 13/44 and y = F/D = 37/-44

= - 37/44.

The solution is ( 13/44 , - 37/44)

Use Cramer’s rule to solve the system of linear equations

No 35 ( Ex 9.7)

1.7x – 2.5y = -0.91- 0.4x + 0.9y = 0.423

E= det -0.91 -2.5 = 0.2385

0.423 0.9

F = det 1.7 -0.91 = 0.3551

-0.4 0.423

D = 1.7 -2.5 = 0.53

-0.4 0.9

Thus x = E/D = 0.2385/0.53= 0.45

y = F/D = 0.3551/0.53= 0.67

The solution is ( 0.45, 0.67)

10.1 ParabolasA parabola is the set of points in a plane that are equidistant from a

fixed point and a fixed line. The fixed point is called the focus and the

fixed line is called the directrix

c

Vertical axisThe parabola with a focus at (0, p) and directrix y = - p has equationx2 = 4pyThe parabola opens upward if p> 0 and downward if p < 0Horizontal AxisThe parabola with a focus at (p, 0) and directrix x= - p has equationy2 = 4px

V (0,0)

F ( 0, p)

P ( x, y)

d1

d2

y = - p

x = - p

V (0, 0)

F ( p, 0)

P ( x, y)d2

Vertical Axis Horizontal Axis

Equation of a parabola with vertex (h, k)

( x – h) 2 = 4p(y - k) Vertical axis; vertex : (h, k)p > 0 : opens upward; p < 0: opens downwardFocus : ( h, k + p); directrix: y = k – p

( y – k) 2 = 4p(x –h)Horizontal axis; vertex: ( h, k)P > 0: opens to the right; p < 0: opens to the leftFocus: (h + p, k) ; directrix: x = h - p

c (x – h) 2 = 4p(y – k)

c F ( h, k + p)

V ( h, k)

y = k - p

(y – k) 2 = 4p(x – h)

V(h, k) F ( h + p, k)

Graph the parabola. Label the vertex, focus and directrix

15 16y = x2 16 y= - 2x2 17. x = 1/8 y2

In the form x2 = 4py Can be written as x2 = - 1/2 y can be written as y2 = 8x

And the vertex is V (0, 0) which is in the form x2 = 4py which is in the form y2 = 4px

Thus , 16 = 4p or p = 4. Thus - ½ = 4p or p = - 1/8 The vertex is v (0, 0). Thus

The focus is F ( 0, 4), the equation The focus is F (0, - 1/8) 8 = 4p or p = 2 The focus is F (2, 0)

Of the directrix is y = -4, and the the equation of the directrix the equation of the directrix is x = -2

Parabola opens upward. The graph is y = -1/8, and the parabola and the parabola opens to the

Of y = 1/16 x2 is shown opens downward. The graph right. The graph of x = 1/8 y 2 x = -2

F (2, 0)

V (0, 0)

y = - 4

F (0, 4)

y = 1/8V (0, 0)

F ( 0, - 1/8 )V (0, 0)

10.2 Ellipses

An ellipse is the set of points in a plane, the sum of whose distances from two fixed points is constant. Each fixed point is called a focus ( plural foci) of the ellipse

Standard Equations For Ellipses Centred at (0, 0)The ellipse with center at the origin, horizontal major axis, and equationx2/a2 + y2/b2 = 1 ( a> b > 0)

Has vertices ( + a, 0), end points of the minor axis ( 0, + b), and foci ( + c, 0), where c2 = a2 – b2 and c> 0

The ellipse with center at the origin, vertical major axis, and equationx2/b2 + y2/a2 = 1 ( a> b > 0)

Has vertices (0, + a) , endpoints of the minor axis ( + b, 0), and foci( 0, + c), where c2 = a2 – b2 and c> 0

Standard Equations for ellipses centered at (h, k)

An ellipse with center ( h, k) and either a horizontal or vertical major axis, satisfies one of the following equations, where a> b> 0 and c2 = a2 – b2 with c> 0

(x – h) 2 + ( y – k) 2 = 1 ( Major axis : horizontal foci: ( h+ c, k) a2 b2 Vertices : ( h, k + e)

( x – h) 2 + ( y – k) 2 = 1 Major axis : vertical , foci ( h, k + e) b2 a2 Vertices : ( h, k+ e)

Standard equation of circleThe standard equation of a circle with center ( h, k) and radius r is( x – h) 2 + (y – k) 2 = r2

Area inside an EllipseGiven the standard equation of an ellipse, the area A of the region contained inside is given by A = ab

10.3 Standard Equations For Hyperbolas Centered

The hyperbola with center at the origin, horizontal transverse, and equation

x2/a2 - y2/b2 = 1 has asymptotes y = + b/a x, vertices ( + a, 0) , and foci ( + c, 0) , where c2 = a2 + b2

The hyperbola with center at the origin, vertical transverse axis, and equationy2/a2 - x2/b2 = 1Has asymptotes y = + a/b x, vertices ( 0, + a) , and foci ( 0, + c) where c2 = a2 + b2

Standard Equations For Hyperbolas Centered

A hyperbola with center ( h, k) , and either a horizontal or vertical

transverse axis, satisfies one of the following equations, where c2 = a2 + b2

(x – h) 2 - (y – k)2 = 1 Transverse axes: horizontal

a2 b2 Vertices: (h + a, k); ( h + c, k)

Asymptotes: y = + b/a ( x – h) + k

(y – k) 2 - (x – h) 2 = 1 Transverse axes: vertical

a2 b2 Vertices: ( h + a); foci : ( h, k + c)

Asymptotes : y = + a/b ( x – h)