Chapter : January 26 January 30 Section.7: Inequalities As a diagnostic quiz, I want you to go through the first ten problems of the Chapter Test on page 32. These will test your knowledge of Sections. through.6. If you have trouble with any of these problems, review the sections given in the table at the end of the test. We re going to start with Section.7: Inequalities. An inequality looks just like an equation, except that in the place of the equal sign is one of the symbols <, >,, or. For example, the statement is an inequality. 3x + 5 < Just like an equation, there are values for x which make the statement true, like 0, and there are values for x which make the statement false, like 3. The solution set of a statement (an equation or an inequality) is the set of values which make the statement true. Solving an inequality is just like solving an equation. You re finding the values of x which make the statement true. Just as with equations, there are inequalities which say the same thing (have the same solution set). For example, the inequality x < 2 says the same thing as 3x + 5 <. Just as with equations, it is easier to read the solution set off of some inequalities than others. The book lists six operations we can do to get from one inequality to an equivalent one.:. We can add the same quantity to each side of an inequality. 3x 5 0x + 7 3x 5 + 5 0x + 7 + 5 3x 0x + 2 2. We can subtract the same quantity to each side of an inequality. 3x 5 0x + 7 3x + 3x 5 3x + 0x + 7 5 7x + 7 3. We can multiply each side of an inequality by the same positive quantity. Last edited 03/08/04, 0:08am. 5 7x + 7 ( ) ( ) 5 (7x + 7) 7 7 5 7 x +
4. We can multiply each side of an inequality by the same negative number AND THEN REVERSE THE DIRECTION OF THE INEQUALITY. 3x 5 0x + 7 2(3x 5) 2(0x + 7) 6x + 0 20x 4 5. If both sides of an inequality are positive, we can take reciprocals of each side of an inequality and THEN REVERSE THE DIRECTION OF THE INEQUALITY. x 2 < x 2 + x 2 > x 2 + 6. You can add two inequalities if the signs of the two inequalities are the same. 3x 5 0x + 7 x 2 < x 2 + 3x 5 + x 2 < 0x + 7 + x 2 + If one inequality has a < or, and the other has a > or, then you cannot add them. You have to reverse the direction of one of them first. If either of the two inequalities you are adding is a < or a >, then the resulting inequality will have a < or a >. You get a ONLY if you add two inequalities with s. You get a ONLY if you add two inequalities with s. With the fourth and fifth rules, it s too easy to forget to reverse the direction of the inequality, or check that both sides of an inequality are positive. Stick to multiplying by positive numbers (not variables) and addition and subtraction. 2x 7 (2x 7) 2x (2x 7) + 7 7 2x 7 2 x x 7 2 To get around problems with the fifth rule, subtract the whole right side (or the whole left side) of the inequality from both sides, and then work from there. Often times, if you re dealing with rational expressions (fractions with the variable in the denominator), you re going to want to do that anyway. x < x 2 + x x 2 + < 0 x 2 + x(x 2 + ) x x(x 2 + ) < 0 x 2 x + x(x 2 + ) < 0 We ll see how to find the solution from here a bit later. 2
As you move from one inequality to the next to the next, write them one below the other, keeping the <, >,, and signs directly underneath each other. The above tips will insure that you re writing the same symbol each time. Solutions of inequalities are usually intervals on the real line or unions of intervals. Page has a table showing how to describe different intervals on the real line using interval notation, and how to graph these intervals on the real line. One thing to notice is the difference between using a parenthesis and a square bracket. A parenthesis means that the endpoint is included. A square bracket means that the endpoint is not. For example, the interval (3, 5] contains all the numbers between 3 and 5, excluding 3 and including 5. We use and to denote intervals that extend indefinitely to the left and right, respectively. For example, (, 4] contains all of the numbers less than or equal to 4. Pay attention to the third column of the table, showing how to graph these intervals on the real line. Remember to use hollow circles to denote endpoints that are NOT in the interval, and filled-in ones to denote endpoints that are. For example, to graph (3, 5], I would draw a real line, mark off where 3 and 5 are, put a hollow circle at 3, put a filled-in circle at 5, and then highlight the part of the line between the two marks. We re going to start by applying these rules and tips to linear inequalities. An inequality is linear if each term is a constant or a multiple of the variable. 8x < 23 3x ( 23 + 8x) + 8x < ( 23 + 8x) + 23 3x 4 < 5x 4 5 < x x > 4 5 Notice that you do the same thing you do with linear equations: Put the x on one side and the constant on the other. Because we read from left to right, there is an urge to put the constant on the left. My advice is to be patient and wait until the end. The solution to the above problem in interval notation is ( 0.8, ). In general, I don t care if you use fractions or decimals, just as long as you re being careful about rounding errors. In linear equations, you don t include the endpoint if the inequality has a < or a >, and you do include the endpoint if the inequality has a or a. In some circumstances, you ll have some expression that has to be between two values, one low and one high. 5 0 < F 60 < 00 60 + 0 < 5 F 60 + 60 < 00 + 60 60 5 < F < 060 60 5 < 5 F 5 < 060 5 32 < F < 22 In this case, the solution is (32, 22). It s relatively easy to convert linear inequalities into forms we can read the solutions from. Nonlinear inequalities, inequalities with terms other than constants and multiples of the variable, are a little more difficult. We re going to start by looking at polynomial and rational inequalities. 3
When would the inequality be true? a b c 0. If any of a, b or c are equal to 0, then the statement is true, since the inequality has a sign. 2. If all three factors are negative, then the product is negative and the statement is false. For example, if a, b and c are all equal to -, then the product is -. 3. If two factors are negative and one is positive, then the product is positive and the statement will be true. For example, if a =, b = 2 and c = 5, then abc = 0. 4. If one factor is negative and the other two are positive, then the product is negative and the statement is false. For example, if a = 2, b = 2 and c = 3, then abc = 2. 5. If all three factors are positive, then the product is positive and the statement is true. If none of the factors are equal to 0: If I have an odd number of negative factors, the product is negative. If I have an even number of negative factors, the product is positive. Now let s look at the inequality (x 2)(x 4)(x 6) 0. We can figure out when we have 3 negative factors, 2 negative factors, negative factor, and 0 negative factors. expression (, 2) (2, 4) (4, 6) (6, ) (x 2) negative positive positive positive (x 4) negative negative positive positive (x 6) negative negative negative positive (x 2)(x 4)(x 6) negative positive negative positive At 2, 4 and 6, the product is equal to 0, so the statement is true. The solution set is [2, 4] [6, ). I m using the to describe the solution set is the union of the two intervals. A number is in the solution set if it is in one interval or the other. Now let s try x 3 + 44x 2x 2 + 48. It looks ugly. However, x 3 2x 2 + 44x 48 = (x 2)(x 4)(x 6), so and we are back where we were before. x 3 + 44x 2x 2 + 48 x 3 + 44x (2x 2 48) 2x 2 + 48 (2x 2 48) x 3 2x 2 + 44x 48 0 (x 2)(x 4)(x 6) 0, Page 8 gives a five-step process for solving nonlinear inequalities.. Move all the terms to one side. Do whatever additions and subtractions are necessary to get one of the sides equal to 0. 2. Factor. Factor the nonzero side of the inequality. 3. Find the intervals. Use the factorization to find the endpoints of the intervals you will look at. 4
4. Make a table. The table will help you look at each interval one by one and figure out when which factors are positive or negative. From there, you can determine when the whole nonzero side is positive or negative. 5. Solve. Read the solution off the table. Based on whether you have a <, >, or, determine which intervals and which endpoints are in the solution set. A few notes: To find the endpoints of the intervals, go through all of the factors and solve the equations (factor) = 0. Each solution to each equation is an endpoint you should mark off on the real line. Use these marks to cut the real line up into your intervals. One way to figure out whether a factor is positive or negative over a particular interval is to pick a number into that interval and plug it in for x. For example, suppose that I want to know whether (x 3) is positive or negative between x = 0 and x = 2. I can plug in x =, get 3 = 2, and conclude that (x 3) will be negative between x = 0 and x = 2. You can use the same procedure when working with quotients. Think: the expression whenever (x ) is positive. Let s try 4x 2x+3 > 2. 4x 2x + 3 > 2 4x 2x + 3 2 > 2 2 4x 2x + 3 + 4x 6 2x + 3 > 0 6 2x + 3 > 0 expression (,.5) (.5, ) 6 negative negative /(2x + 3) negative positive 6/(2x + 3) positive negative The endpoint -.5 comes from solving the equation 2x + 3 = 0. The solution set is (,.5). x is positive Here is another example. x 2 + 2x > 3 x 2 + 2x 3 > 0 (x )(x + 3) > 0 expression (, 3) ( 3, ) (, ) (x + 3) negative positive positive (x ) negative negative positive (x )(x + 3) positive negative positive Because the inequalities have > s in them, we don t include the endpoints. The solution set is (, 3) (, ). Notice how I handle the (x ) 2 factor in this one. x 3 5x 2 + 7x 3 x 3 5x 2 + 7x 3 0 (x ) 2 (x 3) 0 5
expression (, ) (, 3) (3, ) (x ) 2 positive positive positive (x 3) negative negative positive (x ) 2 (x 3) negative negative positive The factor (x ) 2 is never negative. However, it is equal to zero at x =. It is important to check ALL the endpoints, just in case. In this case, x = satisfies the inequality, even though values of x close to don t. The solution set is {} [3, ). This one has the squared factor in the denominator. x 2 2 x 2 4x + 4 x 2 2 x 2 4x + 4 0 x 2 (x 2) 2 2 (x 2) 2 0 x 4 (x 2) 2 0 expression (, 2) (2, 4) (4, ) (x 4) negative negative positive /((x 2) 2 ) positive positive positive x 4 (x 2) 2 negative negative positive Last time, I made a point of including an endpoint. This time I* make a point of excluding an endpoint. At x = 2, the fractions on both sides of the original inequality are undefined. So, 2 cannot be in the solution set. The solution set is (, 2) (2, 4). To graph the solution set on the real line, you draw the interval from the far left to x = 4, and then draw a circle to show the hole at x = 2. Someone came in to ask about Problems 3 and 4 in Section.7. Here they are as added examples. First I ll do Problem 3. x 3 x + 0 expression (, ) (, 3) (3, ) (x 3) negative negative positive /(x + ) negative positive positive (x 3)/(x + ) positive negative positive We include the endpoint 3 because (3) 3 ( ) 3 (3)+ = 0. We don t include the endpoint - because ( )+ = 4 0, which is undefined. The solution is (, ) [3, ). Now I ll do Problem 4. To follow the steps, go down the first column of inequalities, and then go down the second. If the inequalities are too small to see, try zooming in in Acrobat Reader. 6x (x )x + 6(x ) x(x ) 6 x 6 x 6 x 6 x 0 + x(x ) x(x ) 0 6x x(x ) + 6x+6 x(x ) + x2 +x x(x ) 0 6x 6x+6 x 2 +x x(x ) 0 x 2 +x+6 x(x ) 0 (x 2 x 6) x(x ) 0 (x 3)(x+2) x(x ) 0 6
expression (, 2) ( 2, 0) (0, ) (, 3) (3, ) negative negative negative negative negative (x + 2) negative positive positive positive positive (x 3) negative negative negative negative positive x negative negative positive positive positive x negative negative negative positive positive (x 3)(x+2) x(x ) negative positive negative positive negative We include the endpoints -2 and 3, but not 0 and. The solution is [ 2, 0) (, 3]. The last part of Section.7 looks at simple inequalities that involve absolute value. We re going to handle absolute value inequalities by finding an equivalent statement that doesn t have absolute value in it. The table below (copied from Page 83) shows four different inequalities and their equivalent statements. Inequality x < c x c x > c x c Equivalent statement c < x < c c x c c > x or x > c c x or x c The easiest way to verify the information in the table is by graphing what the solution sets might look like on the real line. Here s an example. 2x 3 0.4 x.5 0.2 0.2 x.5 0.2 0.2 +.5 x.5 +.5 0.2 +.5.3 x.7 Notice that once we got to the equivalent form, we were doing things we d done before. The solution set is [.3,.7]. Here s an example with a > sign. After I split the statement into two separate inequalities, I find the solution set to each inequality separately. The union of these two solution sets will be the solution set for the original inequality for the absolute value. 6x 3 > 5 6x 3 > 5 or 6x 3 < 5 6x 3 + 3 > 5 + 3 or 6x 3 + 3 < 5 + 3 6x > 8 or 6x < 2 x > 3 or x < 2 The solution set is (, 2) (3, ). The (, 2) comes from the inequality on the right. The (3, ) comes from the inequality on the left. Now we re going to look at a couple of word problems. We ll start with Problem 75 on Page 88: A telephone company offers two long-distance plans. Plan A: $25 per month and 5 cents per minute. Plan B: $5 per month and 2 cents per minute. For how many minutes of long-distance calls per month would Plan B to be financially advantageous? Let x be the number of long-distance minutes you use per month. Under Plan A, you would pay 25+x 0.05 dollars per month. Under Plan B, you would pay 5 + x 0.2 dollars per month. Plan B is advantageous if you pay less with Plan B. So, we start with the inequality 25 + x 0.05 > 5 + x 0.2. 7
From here, we can find the solution: 25 + x 0.05 > 5 + x 0.2 25 + x 0.05 5 x 0.05 > 5 + x 0.2 5 x 0.05 20 > x 0.07 285.742857 2000 > x. 7 I m guessing that the plans round up to the next highest minute. So, if you use 285 minutes of long distance or less, Plan B costs less. If you use 286 or more, then Plan B costs more. Now let s try Problem 7. A gardener has 20 feet of deer-resistant fence. She wants to enclose a rectangular vegetable garden in her backyard, and she wants the area enclosed to be at least 800 square feet. What range of values is possible for the length of her garden? Let x be the length of one side of the rectangle in feet. Her rectangle is going to be x feet by (400 x) feet. That s a total of x+x+(60 x)+(60 x) = 20 feet of fence. The area of the garden is x(60 x) = 60x x 2. It has to be at least than 800. The at least tells us that we should use a rather than a >. This gives us the inequality 60x x 2 800. Solving the inequality gives x 2 + 60x > 800 x 2 + 60x + x 2 60x > 800 60x + x 2 0 > 800 60x + x 2 0 > (x 20)(x 40) Now we have to make a table. The endpoints of the intervals are 20 and 40. expression (, 20) (20, 40) (40, ) (x 20) negative positive positive (x 40) negative negative positive (x 20)(x 40) positive negative positive The solution set is [20, 40]. So, the garden has to be between 20 and 40 feet long. Section.: Using the Graphing Calculator A graphing calculator or computer displays a rectangular portion of the coordinate plane called the viewing rectangle. On the Texas Instruments graphing calculators, the graphing menu has a command RANGE which allows you to adjust the viewing rectangle. A similar menu option or command should exist on other manufacturers calculators. On the TI, there are six numbers you can adjust: xmin xmax xscl ymin ymax yscl minimum value for x maximum value for x interval between marks on the x-axis minimum value for y maximum value for y interval between marks on the y-axis 8
The calculator will show you the rectangle where x is between xmin and xmax. The book describes this rectangle as [xmin, xmax] [ymin, ymax]. When you re graphing a function, you want to choose a viewing rectangle that shows all the features you re interested in. Start with a rectangle that s too big. For example, I choose [ 00, 00] [ 00, 00]. Even if there s a lot of dead space, even if there s too much pixelization to see what s really going on, it s OK as long as you know where the function is. You can shrink the viewing rectangle until it graphs the features you re interested in. Choosing a viewing rectangle is more of an art than a science. Just make sure that you can see everything you want to see. Suppose that we want to graph the equation y = x 2 5x + 6, and that we want the values of x where x 2 5x + 6 = 0.3. I started with the rectangle [ 0, 0] [ 0, 0]. Eventually, I shrunk my viewing rectangle to [0, 5] [ 2.5, 2.5]. Graphing calculators should have a tracing function (It s called TRACE on the TI s.) that allows you to follow the curve and get coordinates of points on the graph. It looked like the lowest the graph goes is y = 0.25 but I zoomed once more to make sure. Sure enough, y never reaches y = 0.30. Still, sometimes you have to zoom in, just to make sure. Solving an equation graphically, using the trace and zoom functions, will give you approximate answers, accurate to as many digits as you want to work for. Solving it algebraically, working with the equation using the rules of algebra, will give you exact answers. However, solving equations algebraically can be time-consuming, difficult, or just plain impossible. To solve an inequality graphically, zoom in on the parts of the graph close to the x-axis. If you have an inequality of the form (expression involving x) < 0, you want the values of x corresponding to parts of the graph below the x-axis. If you have an inequality of the form (expression involving x) > 0, you want the values of x corresponding to parts of the graph above the x-axis. There are similar rules to inequalities of the form (expression involving x) > 0 and (expression involving x) > 0 Suppose that I wanted to know the values of x where x 2 5x + 6 0. y = x 2 5x + 6 in the viewing rectangle [0, 5] [ 2.5, 2.5]. I can graph the function I know that the graph touches the x-axis twice: once between x =.8 and x = 2.02, and once between x = 2.7 and x = 3.0. So, I know that the x 2 5x + 6 is negative between x = 2.02 and x = 2.7. Any more accuracy would require zooming in in both areas. Solving it algebraically, I know that the function touches at x = 2 and x = 3. The expression x 2 5x+6 is less than or equal to 0 for x-values in [2, 3]. x. Sometimes you ll have an equation or inequality like f (x) = f 2 (x), where each side is some function of
You could graph both functions, and then use the trace and zoom functions to figure out where the graphs cross. The other option is to graph (f (x) f 2 (x)), and then figure out where (f (x) f 2 (x)) is above or below the x-axis. The context of a problem may limit what x-values are possible.. For example, in the problem with the gardener and her rectangular garden, we knew that the length of the fence had to be between 0 and 60 feet. If we were solving the inequality graphically, we could start with xmin = 0 and xmax = 60. From there, we can do trace or zoom as needed. Section.0: Lines Now we re going to work on straight lines in the coordinate plane. The first thing to think about is the steepness or pitch. In math, we refer to the slope of a line. The slope of the line is the rise divided by the run. Pick two points (x, y ) and (x 2, y 2 ) on the line. The rise is (y 2 y ). The run is (x 2 x ). The slope is y 2 y x 2 x. For example, if I have the line going through the points (2, 2) and (4, 6), then the rise is 6-2=4, the run is 4-2=2, and the slope is 4/2=2. The slope is the same no matter which points you choose. Lines that go southwest to northeast have positive slopes. Lines that go southeast to northwest have negative slopes. If we know the slope of a line m and a point on the line (x, y ), we can write down the equation for the line in slope-point form: y = m(x x ) + y. For example, consider the conversion from temperature in degrees Fahrenheit to temperature in degrees Celsius. The boiling point of water is 22 degrees Fahrenheit and 00 degrees Celsius. Raising the temperature degree Fahrenheit is raising the temperature 5 degrees Celsius. The equation for the conversion is If we multiply through and simplify, we get C = 5 (F 22) + 00. C = 5 F 22 5 + 00 = 5 F 60 = 5 ( F + 60 ). If we know the slope of the line m and the point (0, b) where the line crosses the y-axis, we can write the line in slope-intercept form y = mx + b. For example, the line has slope 5 and goes through ( ) 0, 60. C = 5 ( F + 60 ) 0
The graph of every linear equation Ax + By + C = 0, where A and B are not both 0, is a line. Also, every line is the graph of a linear equation. The equation for a horizontal line has the form y = b. The slope is 0. The equation for a vertical line has the form x = a. The slope is undefined, since the run is always 0. Two nonvertical lines are parallel if and only if they have the same slope. Let s try a word problem. Ty runs 40 yards in 25 6 seconds. He starts x = 20 yards at time t = 0 and runs in the positive-x direction 2. Assume that he runs at a constant pace. What is the line relating Ty s position x to the time t? The slope is 40 25/6 = 48 yards =.6 5 seconds. At time t = 0, he s at x = 0. Using the slope-point form of the line, we get x =.6t + 20. We should probably note that x is in yards and t is in seconds. When does Ty make it to x = 00? We just solve the equation The answer is t = 25 3 = 8 3 seconds. 00 =.6t + 20. Ty s friend Rodney is running behind him, keeping the same distance between he and Ty. At time t = 4, Rodney s at x = 50. What s the line relating time and Rodney s position? The slope is the same, but the point is now (4, 50). Rodney s line is x =.6(t 4) + 50 =.6t 38.4 + 50 =.6t +.6. Two lines y = m x + b y = m 2 x + b 2 are perpendicular if m 2 = m. Suppose that I stand at the origin of a coordinate plane and swing a 3-foot chain with a weight on the end. At (5, 2), the weight snaps off. The weight travels perpendicular to the line from me to where it snaps off. What is the equation of the line the weight travels on? The line the chain is on when the weight snaps off goes through (0, 0) and (5, 2). Using rise over run, this line is y = 2 5 x. So, the weight will travel on a line with slope going through (5, 2). m = 2/5 = 5 2, 2 He s at the 30, the 40 HE COULD GO ALL THE WAYYYYY!
Using the slope-point form, y = 5 (x 5) + 2 2 y = 5 2 x + 6 2 If I wanted the general form, I would move everything to the left side and get or 5 2 x + y 6 2 = 0 5x + 2y 06 = 0. Instead of doing it algebraically, we can use a graphing calculator to find the point where two lines intersect. For example, suppose we have the two lines y = 40 x + 40 and 4.3 y = 40 5.5 x + 0. After some trial and error, I obtained the viewing rectangle [0, 8] [0, 00]. The two lines meet at about (2.8, 67.76). 2