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1 with Modelling and Visualization( 3 E)
Extra Notes on Slides Pre Calculus with Modelling and Visualization( 3 E) By Gary Rockswold

2 Ch 1.1 Numbers, Data, and Problem Solving
Important Definition Examples Natural 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, , 0.25 = ¼, = 1/3, and all fractions Irrational numbers Can be written as non-repeating, non-terminating decimals; cannot be rational numbers; a square root of A positive integer that is not an integer. Real Numbers – Represented by standard or decimal numbers, includes the rational numbers 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 = 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

3 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 = 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 the middle 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

4 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 (2, 7) (1, 6) (3, 5) Scatter Plot ( 1, 2) Connecting the data points with straight-line segments called a line graph

5 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). The mid point of the line segment connecting ( 3,4) and ( 1, -2) is M = ( 3 + 1, 4 – 2) =( 2, 1) 2 2 2 =

6 Determine whether the triangle is isosceles ?
(7, 1) 5 5 (0,0) (3, 4) 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 = = = = 5 ( Using distance formula ) d= ( 7 – 3) 2 + ( 1 – 4)2 = (-3) 2 = = = 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.

7 Midpoint Formula ( pg 27, no 46 ex 1.2)
The population of US was 151 million in 1950 179 million in 1960 What 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, the estimated population in 1970 is = 207 million. The midpoint formula is M = ( , y ) = (160, 179 ) M = y = 179 or y = 207 million 2 207 million population in 1970

8 Data involving two variables
Pg 27, No 62 ( Ex 1.2 ) { (1, 1), (3, 0), (-5, -5), (8, -2), (0, 3) } The Domain is D = {1,3, -5, 8, 0} and the range is R = {1, 0, -5, -2, 3} 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. The axes must include -5< x < 8 and -5 < y < 3 4 2 -2 -4 (0, 3) (1, 1) (3, 0) (8, -2) (-5, -5)

9 Function Notation Output Input
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

10 Representations of Functions
Verbal representation (words) – Words describe precisely what is Computed- 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 1 2 f(x) 4 1 2 1 2 3 5 7 9 5 6 7

11 Use the graph of the function f to estimate its domain and range
Use the graph of the function f to estimate its domain and range. Evaluate f(0) Ex 1.3 3 1 -1 -3 2 1 -1 -2 Domain (D) = {x/ -3 < x < 3} Range (R) = { y/ 0< y < 3} f(0) = 3 Domain D = { x/ -2 < x < 4 Range R = { y / -2 < y < 2 }; f(0) = -2 Domain D = all real numbers Range R = all real numbers; f(0) = 0

12 Symbolic expression : f(x) = 0.50x Numerical
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 Symbolic expression : f(x) = 0.50x Numerical Miles 1 2 3 4 5 6 cost 0.50 1.00 1.50 2.00 2.50 3.00 Graphical Cost of driving is $0.50 per mile

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

14 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 Example – a line passing through ( 1, 3) nand (2, 7) has slope m = = 4 2 - 1 This slope indicates that the line rises 4 units for each unit increase in x m = 2 > m = - ½ < m = m is undefined

15 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) = Square Root Function f(x) = x2 f(x) = x Absolute Value Function f(x) = x3 Cube Function

16 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

17 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

18 Correlation Coefficient r, (-1 < r < 1)
Value of r Comments Sample Scatterplot r = There is an exact linear fit. The line passes through all data points and has a positive slope r = There is an exactlinear fit. The line passes through all data points and has a negative slope 0<r< There is a positive correlation. As the x-values increase, so do the y-values. The fit is not exact -1 < r < There is a negative correlation. As the x-values increase, the y- values decrease. The fit is not exact r = There is no correlation. The data has no tendency toward being linear. A regression line predicts poorly

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

20 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 Table enter points [ 0, 5, 1] by [0, 5, 1] Hit GRAPH

21 Use Graphing Calculator
Linear Regression Example 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 (%) D( minutes) Let P correspond to x-values and D to y-values. Find the least-squares regression line that models these data. Plot the data and the regression line Estimates the delay for a 50% increase in the number of calls Solution Use 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

22 Choose STAT and CALC and LINEAR REGRESSION and enter
Hit Y and enter a= and b= 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 GRAPH Hit WINDOW and enter points [ -10, 110, 10 ] by [ 0, 5, 1]

23 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 and solve for x To find any y-intercepts, let x = 0 in the equation and solve for y Equations of Horizontal and vertical lines – An equation of the horizontal Line with y-intercept b is y = b. An equation of the vertical line with x-intercept k is x = k Perpendicular lines- Two lines with nonzero slopes m1 and m2 are perpendicular if and only if their slopes have product – 1, that is m1m2 = -1

24 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 linear Find the slope-intercept form of the line that passes through these data points. Predict y when x = -2.7 and 6.3. Decide if these calculations involve interpolation or extrapolation No 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 use any two points to determine the slope. Using the points (0, 6.8) and (1, 5.1), m = = ( slope formula , m = y2 – y1 1 – x2 – x1 Slope intercept form of the line is y = - 1.7x ( y = mx+ b) b) When x = - 2.7, y = - 1.7(-2.7) = This calculation involves extrapolation When x = 6.3, y = -1.7(6.3) = This calculation involves extrapolation x -2 -1 y 10.2 8.5 6.8 Y-intercept

25 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 = 0 where a and b are real numbers with a= 0 If 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 equality If a, b, and c are real numbers, then a = b is equivalent to a + c = b + c Multiplication Property of Equality If a, b, c are real numbers with c = 0 then a =b is equivalent to ac = bc

26 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

27 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

28 Solving Application problems
Step 1 : Read the problem and make sure you understand it. Assign a variable to what you are being asked to find. If necessary, write other quantities in terms of this variable Step 2 : Write an equation that relates the quantities described in the problem. 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 ?

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

30 Solve the linear equation with the intersection –of-graphs method
Solve the linear equation with the intersection –of-graphs method. Approximate the solution to the nearest thousandth whenever appropriate ( Ex 2.3) No No No – 2x = = 4x – – x = 2x – Enter Y1= 8-2x Y2 = 1.6 Enter Y1= Y2 = 4x - 6 EnterY1 = (6 – x) / 7 Y2 = (2x – 3) /3 Hit Graph and calc, you will get graph and point of intersection

31 Applications No 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 ft V = r2 h , 100 = 1 (3) 2 . h, 100 = 3 h = h, h = 10.6 ft

32 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) = 25 4x + 10x – 2 = 25 14x = ( Add 2) 14x = 27 x = 27 ft = 1.93 ft 14 5x – 1 = 5( 27/4) – 1 ( substitute x value ) = 121/14 = 8.6 ft

33 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

34 Interval Notation Inequality Interval Notation Number line Graph
- 2 < x < ( - 2, 2) - 1 < x < ( - 1, 3] - 3 < x < [ - 3, 2 ] x < - 1 or x > ( -  , - 1) U (2,  ) x > ( - 1,  ) x < ( - , 2 ] ( ) ( ] [ ] ( ) ) ] - < x < 

35 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

36 Compound Inequalities
 Two inequalities connected by the word and or or Examples x < 3 or x > 4 x> - 3 and x < 4 The inequality x> 5 and x < 20 can be written as the three-part inequality 5< x < 20

37 Ex 2.4 Solving Linear Inequalities Symbolically -2(x – 10) + 1 > 0
-2x > 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) 1 < 1 – 2t < 2 3 < 1 – 2t < 2 2 1 < - 2t < 1 - 1 > t >- 1 - 1 < t < 1 ( -1 , - 1 ]

38 Solving Linear Inequality Graphically ( Ex 2.4)
x + 2 > 2x x -2 > - 4 x < 1-x < 2 Y1 = x + 2, Y2 = 2x Y1 = -2, Y2 = 1-x, Y3 = 2 Y1 = 2x -2 , Y2 = - 4 x + 4 [ -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}

39 Solve the compound linear inequality graphically
Solve the compound linear inequality graphically. Write the solution set in interval notation, and approximate endpoints to the nearest tenth whenever appropriate Ex 2.4 66 3 < 5x – 17 < x < 2x -5 < 8 3 Y1 = 3, Y2 = 5x – 7 and Y3 = Y1 = 0.2x, Y2 = 2x – 5, Y3 = 8 [ -5, 15, 5] by [ -5, 20, 5] Y1 = 3, Y2 = 5x – 17, Y3 = 15 Y1 = 0.2x , Y2 = (2x – 5)/3 , Y3 = 8

40 f(x) = g(x) = 4 b) g(x) = h(x) = 2 c) f(x) < g(x) < h(x)
No 70 ( Ex 2.4)Use the figure to solve each equation or inequality a) f(x) = g(x) b) g(x) = h(x) f(x) < g(x) < h(x) d) g(x) > h(x) 700 600 500 400 300 200 100 y = g(x) y = h(x) y = f(x) f(x) = g(x) = 4 b) g(x) = h(x) = 2 c) f(x) < g(x) < h(x) 2 <x < 4 d) g(x) > h(x) 0 < x < 2

41 Estimate the years when production was between 56,000 and 75,000
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 Did the demand for Harley- Davidson motorcycles increase or decrease over this time period ? Explain your reasoning. Estimate the years when production was between 56,000 and 75,000 Solution N(x) = 6409(x – 1985) + 30,300 Slope= m Slope of the graph of N is 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 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 < 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 Y3 Y2 Y1

42 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 < x - 1 if 0 < x < 3 x + 3 if -1 < x < The domain of f { x/ - 3< x < 3} f(-2) = 2(-2) + 1= -3 The domain of f - 5 < x < f(0) = 0 -1 = -1 f( -2) = f(3)= 3 – 1 = 2 f(0) = f is not continuous f(3)= 3 + 3 = 6 f is continuous 8 6 4 2 -2 8 6 4 2 -2

43 V shaped Never go below the x-axis
Absolute value function f(x) = x , The output from the absolute value function is never negative, Examples f(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 = - 4 3x = or 3x = -2 x = or x = -2/3 If 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 >3 First solve x – 5 = -3 and x – 5 = 3 x = x = 8 The solutions are 2 to 8 The solution set to x – 5 < 3 is 2 < x < 8 And the solution set to x – 5 > 3 is x < 2 or x > 8 V shaped Never go below the x-axis

44 Alternative method for solving absolute value inequalities
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

45 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 days a) Estimate the initial and final amounts of water in the pool. b) When did the amount of water in the pool remain constant c) Approximate f(2) and f(4) d) At what rate was water being drained from the pool when 1 < x < 3 125 100 75 50 25 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 The water level remained constant during the first day and the fourth day, when 0 < x < 1 or 3 < x < 4 f(2) = 45 thousand and f(4) = 40 thousand 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 Gallons (thousands) Time ( days)

46 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

47 Solve the absolute value equation ( Ex 2.5)
51 - 3x – 2 = x – = 2 Then – 3x – 2 = x – 5 = -1, has no solution -3x= ( add 2) since the absolute value of 3x = 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)

48 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

49 No 95 ( Ex 2.5) Average temperature Analyze the temperature range in Boston Massachussetts The inequality T – 50 < 22 describes the range of monthly average temperatures T in degree Fahrenheit. Solve the inequality graphically and symbolically b) 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

50 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 Vertex Axis of Symmetry Axis of Symmetry

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

52 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 < 0 Let f(x) = 3(x – 1) Parabola opens upward: a > 0 Vertex : ( 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 +3 x = - 6/ 3(1) = - 2 Y-value of vertex f(- 2)= (-2) 2 – 6(-2) + 3 = = 19 Completing the square to find the vertex form To complete the square for x2 + kx, add (k/2) 2 to make a perfect square trinomial Y = x2 – 4x + 1 Y – 1 = x2 – 4x Y – 1 = x2 – 4x Add ( 4/2) 2 Y = (x – 2) 2 + 1

53 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 on a) Write a formula for a function R that gives the revenue for renting x rooms b) 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 R(x) = x(120 – 2x) The graph of R(x) = x(120 – 2x) or R(x) = - 2x x, 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)= 30 Maximum revenue occurs when 30 rooms are rented R(30) = - 2(30) (30) = = $ 1800 is the maximum revenue

54 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 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) = 3.125x x – 0.9 = 3.125(3.5) (3.5) – 0.9 = 44.56

55 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 500 Tower Tower 120 feet 20 feet Cable 300 The vertex is (2000, 20) and another point on the cable is (0, 500). Using vertex form, f(x) = a(x – h) 2+ k f(x) = a(x – 0) = ax2 + 20 F( 150) = 120 , f( 150) = a(150) , 120 = a + 20 100 = a a =1/225 The shape of the cable is given by the equation f(x) = 1/225 x = x2 + 20

56 Quadratic Equation A 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

57 Square Root Property x2 = k
Let k be a nonnegative number. Then the solutions to the equation. x2 = k are x = + k. If k < 0. Then this equation has no real solutions.

58 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 2a Discriminant D = b2 - 4ac If D > 0, the equation has two unequal real solutions If D = 0, the equation has one real solution of multiplicity two If D < 0, the equation has two complex (conjugate) solutions

59 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

60 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 = -2 Check 6, and – 2 are two solutions and satisfy the original equation And x-intercepts of the graph are 6, -2 By calculator, draw the graph

61 Quadratic Formula The solutions to ax 2 + bx + c = 0 with a = 0 are given by b + b2 – 4ac X = 2a

62 Vertical and Horizontal Translations
Translated upward and downward y2 = x y1= x2 y3 = x2 - 2 y1= x2 y1= (x-1)2 Translated horizontally to the right 1 unit y1= x2 y2= (x + 2 )2 Translated horizontally to the left 2 units

63 Ch 4.1 Polynomial 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

64 Applications No 118 ( Ex 4.1) Natural Gas The US consumption of natural gas from 1965 to 1980 can be modeled by f(x) = x4 – x x2 –0.5145x = 1.514, where x = 6 corresponds to 1966 and x = 20 to Consumption is measured in trillion cubic feet. Evaluate f(10) and interpret the result Graph f in [ 6, 20, 5] by [0.4, 0.8, 0.1] c) Determine the local extrema and interpret the results Local Maximum Local minimum a) f(10) = (10)4 – (10) (10)2 – (10) 0.706 in 1970 U.S consumption of natural gas was about trillion cubic feet.

65 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 90 70 60 50 Mar Jul Nov a) Even b) 83 degree Farenheit c) They are equal

66 Use interval notation to identify where f is increasing and when f is decreasing Ex 4.1
36 35 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,  ) The graph of this cubic equation has turning points ( -2, 20) and (1, -7)It is increasing ( -  , -1) or ( 1,  ) and decreasing [-1, 1]

67 Determine graphically any a) local extrema b) absolute extrema
Ex 4.1 No 68 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 Local maximum of 6 and occurs when x = -1. There is no local minima

68 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

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

70 Cubic Polynomial Functions
a < a > a> 0 Quartric Polynomial Functions a > a < a< 0

71 Piecewise- Defined Functions Evaluate f(x) at the given value of x y = f(x)
Ex 4.2 No 67 70 10 6 2 x = -2 and 1 f(-2) = 5 f(1) = 0 x = -2, 0, and 2 f(-2) = 0, f(0) = -3, f(2) = 2

72 Ex 5.2 No 76 f(x)= x2 if -2 < x < 0 x + 1 if 0 < x < 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)= x if -2 < x < 0 x if 0 < x < 2 The graph of f is discontinous at x = 0 x2 = 0 , x = 0, this is not the domain of the equation x + 1 = 0, x = -1, this is not the domain of the equation. So no solution

73 Ex 4.2 No 77 f(x)= 2x if -5 < x < -1 - 2 if -1 < x < 0
f is continuous x2 -2 = 0 x =

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

75 Ch 4.4 The Fundamental Theorem of Algebra
Properties of the imaginary unit i i= i2 = - 1 The expression If a > 0, then = i

76 Real Numbers Imaginary numbers
Complex Numbers a + bi Real Numbers Imaginary numbers a + bi , b = a + bi, b = 0 Fundamental theorem of Algebra A polynomial f(x) of degree n > 1 has atleast one complex zero Number of Zeros Theorem A polynomial of degree n has atmost n distinct zeros Conjugate zeros theorem If 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)

77 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 -

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

79 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

80 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

81 Ch 4.7 Power functions and Radical Equations Properties of Rational Exponents Let 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

82 Ex 4.7 No No 43 f(x) = x f(x) =

83 Ch 5.1 Combining Functions
Operations on Functions If f(x) and g(x) both exists, the sum, difference, product and quotient of two Functions 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

84 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”

85 Compositions of Function
Multiplication: (fg)(5) Composition: (fog)5 Input 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))

86 Ex 5. 1 No 117 Methane Emissions – Methane is a greenhouse gas
Ex 5.1 No 117 Methane Emissions – Methane is a greenhouse gas. It lets sunlight 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 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) x 1990 2000 2010 2020 2030 f(x) 27 28 29 30 31 g(x) 5 7.5 10 12.5 15 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

87 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) = = 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. 50 40 30 20 10 y = f(x) + g(x) y = f(x) y = g(x)

88 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 ) ( Substitute f(x) and g(x) ) = 0.1x +0.25x -172 – ( Combine like term) = 0.35x – 664.5

89 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 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

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

91 Use the graph of f to determine if f is One – to – One Functions ( Ex 5.2 )
3 1 -3 3 -3 One- to – One Not one –to-One

92 Graph y = f(x), and y = x. Then graph y = f -1(x)
Ex 5.2 f(x)= -1/2 x f(x) = y = f-1(x) = -2x y = f-1(x) = x3 + 1

93 Cost ( millions of Dollars
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 x c) Evaluate C-1(1) 2.4 2.0 1.6 1.2 0.8 0.4 Cost ( millions of Dollars Year Solution h(x) = g(x) –f(x) h(1995) = g(1995) – f(1995)= = 200; h(2000) = g(2000) – f(2000) = 950 – 700 = 250 c) The function h is linear and passes through the points (1995, 200) and (2000, 250) Therefore , its graph has a slope of m = = 10 2000 – 1995 Thus h(x) = 10(x – 1995) = 10x – 19,750

94 Exponential Functions and Models
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

95 5.3 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 g(x)= 2x f(x)= 2x

96 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

97 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( /12) 12(4) =

98 The number e is an irrational number and important in mathematics,
much like e = 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) =

99 Sketch the graph of y = f (x) Ex 5.3
44 f(x) = 4x f(x) = 2(1/3) x f(x) = 8(2) - x

100 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, The amount of cesium remaining after x years in an initial sample of 100 milligrams can be described by the formula A(x) = 100e x 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 Solution a) A(x) = 100e x A(50) = 100e – (50)= 31.7 milligrams Since 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^ ( x) Y2 = 50 The graphs intersect near (30.2, 50). Therefore, the half-life of cesium is approximately 30.2years [0, 50, 10] by [0, 100, 10]

101 Horizontal first five years decreasing
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( )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( )t5 = S(4) = 100( )1024 = 99.3 S(15) = 100( )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 Horizontal first five years 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 age Yes, y = 100 and y= 0, percentage cannot be larger than 100 not smaller than 0 decreasing

102 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 Input Output x= 10k log 104 k Common Logarithms

103 Inverse Properties Of The common Logarithm
The following inverse properties hold for the common logarithm log 10x = x for any real number x , and 10logx = 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 Properties The following inverse properties hold for logarithms with base a Logaakx = x for any real number x, and alogax = x for any positive number x

104 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

105 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) , ,000 Types of Insects 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 = 800 Solving the equations for a gives a = 500 – b and a = 800 – 2b, thus 500 – b = 800 – 2b , b = 300 Since a + b = 500 and b = 300, a = 500 a = 200, The function that models the given data is f(x) = logx b) f(x) = 1200, 1200 = logx, 1000 = 300logx, logx = 10/3 10logx = 10 10/3, x = 10 10/3 = The acreage would be about 2154 acres

106 (Ex 5.4) No 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) a) Evaluate f(0) and f(100) . Interpret the results b) 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 mercury Solution - a) f(x) = 0.48 ln (x + 1) + 27 , f(0) = 0.48 ln (0 + 1) + 27 = 0.48 ln = 27 and f(100) = 0.48 ln( ) + 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 mercury b) Graph Y1= 0.48 ln (X + 1) 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 miles c) f(x) = 28, 28 = 0.48 ln ( x + 1) + 27, ln (x + 1) = 1/0.48 = eln(x + 1) = e 1/0.48 x + 1 = e1/0.48 , x = e1/0.48 – 1 = 7.03 The air pressure is 28 inches of mercury about 7 miles from the eye of the hurricane [0, 250, 50] by [25, 30, 1]

107 Properties of Logaritms
Logam + logan = loga(mn) Logam – logan = loga (m/n) 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)

108 f(x) = log2 + logx, g(x) = log 2x
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

109 5.6 Exponential and Logarithmic Functions
Exponential equations Typical form , Cax = k Solve for ax. Then take a base a logarithm of each side. Use the inverse property: Logaax = x Example 4ex = 24, ex = 6 Ln ex = ln6 X = ln6 = 1.79 Logarithmic equations Equation : C logax = k Solve for logax. Then exponentiate each side with base a. Use the inverse property alogax = x 4logx = 10 Logx = 2.5 10logx = x = = 316 Equation : logabx + logacx = k When more than one logarithm with the same base occurs, use properties of logarithms to combine logarithms. Be sure to check any solutions Example Logx + log4x = 2 Log4x2 = 2 4x2 = 102 x2 = 25 x = + 5 - The only solution is 5

110 Applications N0 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 Determine symbolically the year when this amount reached $ 55 billion Solve part (a) graphically or numerically Solution – We must solve the equation f(x) = 55 36.2e0.14x = 55 e0.14x = 55/36.2 Ln e0.14x =ln 55/36.2 x = ln 55/36.2/0.14 = 2.99 Since x = 0 corresponds to 1987, x = 3 represents = 1990, rounded to the nearest year b) Graph Y1 = 36.2e^(0.14x) and Y2 = 55 The graph intersect near (3, 55) Therefore , Christmas credit card spending was approximately $ 55 billion in 1990

111 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 steps Step 1 : Choose a variable in one of the two equations. Solve the equation for that variable Step 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 equations Linear Equation Non Linear Equation Solve 32. -2x – y = x2 + y2 = x2 – y = 5 3x + 4y = x + y = x2 + 2y = 10 Solve the first equation Solve the second equation Solve the first equation -y = 2x – 2 ( Add + 2x both sides) y = 3 - x x2 - y = 5 y = - 2x + 2 ( multiply by -1) Substitute this into first equation y = -2x2 + 5 Substitute y in second equation x2 + y2 = y = 2x2 - 5 3x + 4( -2x + 2) = x2 + (3 – x) 2 = Substitute y into the second equation 3x -8x + 8 = x2 + 9 – 6x + x2 = x2+ 2( 2x2 – 5) = -10 - 5x + 8 = x2 – 6x = x2 + 4x2 - 10 - 5x = - 8 – 7 ( add – 8 both sides) x ( x – 3) = = -10 - 5x = x = 0 or x = There are infinitely many x = -15/-5 = 3 ( Divide by -5 both sides) When x = 0, y = 3 – 0 = solutions y = - 2 (3) + 2 = = and when x = 3, y = 3 – 3 = 0 x = 3, The solutions are ( 0, 3) and (3, 0) y = - 4

112 Graphical method for two equations
Solve both equations for the same variable. Then apply the intersection-of-graphs method 47 ( Ex 9.1) 2x + y = 1 x – 2y = 3 y= 1- 2x -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) = 3 x – 2 + 4x = 3 5x = 5 x = 1 If x = 1, then y = 1 – 2(1) = -1 The solution is (1, -1) Numerically

113 Approximate to the nearest thousandth any solutions to the system of nonlinear equations graphically ( Ex 9.1) No No 78 e2x + y = x2 + y= 3 y = 4 – e2x y = 3 – 3x2 ln x – 2y = (0.3)x + 4y = 1 Y = lnx/ y = 1-(0.3)x/ 4 Graph Y1 = 4 – e^ (2X) Graph Y1 = 3 – 3X^2 Graph Y2 = ln (X) / Graph Y2 = (1 – 0.3^X) /4 Solution = ( 0.714, ) Solution ( , ), (0.971, 0.172)

114 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 y H = x y 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 y y = 1/ ( 180 – 0.491x – 11.2) 180 = x y y= 1/ ( x – 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]

115 Ch 9.2 Consistent, Dependent, Independent
Consistent 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. y y y x x x Inconsistent dependent Unique

116 Linear Programming In 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 problem Step 1 : Read the problem carefully. Consider making a table to display the information given Step 2 : Use the table to write the objective function and all the constraints Step 3 : Sketch a graph of the region of feasible solutions. Identify all vertices or corner problems Step 4 : Evaluate the objective function at each vertex. A maximum ( or a minimum) occurs at a vertex. Note : If the region is unbounded, a maximum ( or minimum) may not exist

117 Linear Programming Problems Shade the region of feasible solutions
Ex (9.2) No No 88 3x + 2y < x + 2y < 8 2x + 3y < x + y > 2 x > 0, y > x > 0, y> 0

118 9.3 Solving a system of linear equations in three variables
Step 1 : Eliminate one variable, such as x, from two of the equations Step 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 z Note : 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 3 Step 3: Substitute the values for y and z in one of the given equations to find x. The solution is (x, y, z)

119 Solve the system of Linear Equations
No 12 ( Ex 9.3) x – y + z = - 2 3x – 2y + z = -1 x + y = - 3 Solution Subtract the first two equations x – y + z = - 2 -2x + y = 3 Subtract this equation from the third 2x+ y = 3 3x = - 6 x = - 2 Substitute x = -2 into x + y = - 3 ( -2) + y = - 3 y = - 1 Substitute x = - 2 and y = -1 into x – y + z = 2 ( -2) – (-1) + z = 2 1 + z = 2 z= 3 So the solution is ( - 2, - 1, 3)

120 No 25 Solve the system of linear equations
2x + 6y + 2z = 6 2x + 7y + 4z = 13 4x = y + 2z = 7 Multiply the second equation by 2 and subtract the third equation 4x + 14y + 8z = 26 4x + y + 2z = 7 13y + 6z = 19 Subtracting the two new equations we get 0 = 0 Therefore, we have infinitely many solutions 13y = - 6z + 19 y = -6z + 19 13 Multiply the third equation by 3 and subtract from the first equation x + 3y + z = 3 12x + 3y + 6z =21 -13x - 5z = -18 x = - 5z + 18 We have infinitely many solutions ( - 5z , - 6z + 19, z )

121 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

122 Minors and Cofactors The 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

123 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 b a2 c a2 b2

124 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 , F = det = 37, D = det Thus x = E/D = -13/-44 = 13/44 and y = F/D = 37/-44 = - 37/44. The solution is ( 13/44 , - 37/44)

125 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 = F = det = D = = 0.53 Thus x = E/D = /0.53= 0.45 y = F/D = /0.53= 0.67 The solution is ( 0.45, 0.67)

126 10.1 Parabolas A 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 Vertical Axis Horizontal Axis P ( x, y) c x = - p d2 F ( 0, p) d1 P ( x, y) F ( p, 0) d2 V (0,0) V (0, 0) y = - p Vertical axis The parabola with a focus at (0, p) and directrix y = - p has equation x2 = 4py The parabola opens upward if p> 0 and downward if p < 0 Horizontal Axis The parabola with a focus at (p, 0) and directrix x= - p has equation y2 = 4px

127 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 downward Focus : ( 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 left Focus: (h + p, k) ; directrix: x = h - p (x – h) 2 = 4p(y – k) c F ( h, k + p) c V ( h, k) y = k - p (y – k) 2 = 4p(x – h) V(h, k) F ( h + p, k)

128 Graph the parabola. Label the vertex, focus and directrix
15 16y = x y= - 2x 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 = Thus - ½ = 4p or p = - 1/ The vertex is v (0, 0). Thus The focus is F ( 0, 4), the equation The focus is F (0, - 1/8) = 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 y2 y = 1/8 x = -2 V (0, 0) F (2, 0) F (0, 4) V (0, 0) V (0, 0) F ( 0, - 1/8 ) y = - 4

129 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 equation x2/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 equation x2/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

130 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) ( y – k) 2 = 1 ( Major axis : horizontal foci: ( h+ c, k) a b Vertices : ( h, k + e) ( x – h) ( y – k) 2 = 1 Major axis : vertical , foci ( h, k + e) b a Vertices : ( h, k+ e) Standard equation of circle The standard equation of a circle with center ( h, k) and radius r is ( x – h) 2 + (y – k) 2 = r2 Area inside an Ellipse Given the standard equation of an ellipse, the area A of the region contained inside is given by A = ab

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

132 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) (y – k)2 = Transverse axes: horizontal a b Vertices: (h + a, k); ( h + c, k) Asymptotes: y = + b/a ( x – h) + k (y – k) (x – h) 2 = Transverse axes: vertical a b Vertices: ( h + a); foci : ( h, k + c) Asymptotes : y = + a/b ( x – h)


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