L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08

Transcription

1 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 73 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08 We defined an equation as a statement that two expressions are equal to each other. We solved an equation by finding the values of the variable that makes the statement true. We discussed conditional equations, contradictions, and identities. Unlike equations, inequalities typically have an infinite number of solutions (and an infinite number of non-solutions). We solve inequalities using the same techniques as we do for solving equations: 3 3x 7 3 5x x 10 5x x Note that when dividing or multiplying by a negative number we must reverse the inequality symbol since "largeness" reverses when we move from the negative side of the number line to the positive side. For example, 5 > 3 but 5 < 3.

2 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 74 We can show the infinite number of solutions on a number line. Note the closed circle at 2 indicates that 2 is part of the solution. We can write the solution in several ways: With interval notation, we can indicate whether the end points are included by using parentheses or square brackets. A parenthesis means the end point of the interval is NOT in the solution. A square bracket indicates that the end point of the interval IS in the solution. Note the use of the infinity symbol. Infinity is not a real number. We use as a notational device to indicate unboundedness (we can never get there). So, never has a square bracket next to it because it is never included in the solution. The same goes for. Check the solution of 3 3x 7 by replacing x with a value that is in the solution set and see if it satisfies the inequality. We can also choose a "non-solution" and see if that does NOT work.

3 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 75 We can also think of 3 3x 7 in terms of functions and solve this graphically: The functions are equal when x = -2. To solve f x g x. find the values of x where f < g. We want the values of x where the blue g is above (greater than) the red f. We can see that g is above f when x is less than 2. That is, the solution to f x g x is 2 x x or, using interval notation,, 2

4 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 76 Let d = # of desks, then d + 5 = # of chairs Total cost = cost of desks + cost of chairs 80d 15 d 5 The most you can spend is $1,000, so Total cost <= 1000 Total cost = cost of desks + cost of chairs Total cost d 15 d d 15d d 925 d 9.7 You cannot buy 7/10 of a desk, the most you can buy is 9. This means you will buy = 14 chairs. Total cost = cost of desks + cost of chairs 80d 15 d 5 d So, you will spend $930 on the desks and chairs. What will you do with the remaining $70?

5 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 77 3 x x x x x 30 x x x x 9x 20x x 9x 20x x 11x 20 4x 5 5 x 4 5 So, the solution is: x or we could write 4 x x 5 4 or we could write 5, 4

6 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 78 5 x x 3 x 5 x x 7x 4 x 7 4 This is a false statement so there is no solution.

7 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 79 This is a compound inequality. It means 2 4 and 4 8 We could solve each inequality and then put the solutions together or we could do it all at once: x 2 Notice how we had to reverse the inequality when we divided by a negative. The preferred way of writing this is smallest to largest: 2 x 3 In interval notation, the solution is 2,3

8 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 80 Here is what the formula looks like in my Excel gradebook: =IF(OR($G15="NOL",$G15="W",$G15="I",$C15=""),"",(AI$9-$AI$4/100*$K15-$AJ$4/100*$L15- $AK$4/100*$M15-$AL$4/100*$N15-$AM$4/100*$O15-$AN$4/100*$P15-$AO$4/100*$Q15- $AP$4/100*$R15)/($AQ$4/100))

9 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 81 x Break up into two inequalities and We can solve each inequality and then find the intersection of their solution sets. 1 The left inequality says that is greater than 0, which means it is positive. Since there is a 3 positive number in the numerator (1) the denominator must also be positive for the fraction to be positive. So, we can write x 2 3 x 2 For the second inequality, since + 3 is positive we can multiply through by + 3 without having to worry about switching the inequality symbols x 6 5 x 4 5 x 4 The intersection of these two 5 solution sets is x. 4 5 In interval notation this is,. 4

10 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 82 Solving an absolute value inequality is similar to solving an absolute value equation. First, let's solve the equation x = 5. The solution is all numbers 5 units from 0 on the number line. So, the solutions are x = 5 and x = 5. Now, do an inequality such as x < 5. The solution is all numbers that are less than 5 units from 0. We can show this on a number line by shading between 5 and 5. We use an open circle, O, at 5 and 5 to indicate that they are not part of the solution.

11 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page Isolate the absolute value Use the fact that u a means a u a. So, we have Solve x 6 or, in interval notation, 2,6 1. Isolate the absolute value. x 6 4 x 2 We stop here because an absolute value cannot be negative. The answer is no solution.

12 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page 84 The solution includes all numbers whose distance from 0 is greater than 5. This solution includes two distinct parts: Numbers less than 5. Numbers greater than +5. We write the solution as x < 5 or x > 5. In general, u a means that u a or u a 1. Isolate the absolute value Use the fact that u a means that u a or u a or Solve or or 22 x 4 or x 11

13 L09-Fri-23-Sep-2016-Sec-A-9-Inequalities-HW07-Moodle-Q08, page Isolate the absolute value. 5 x x x 10 2 We can stop here because an absolute value is always greater than a negative number. The answer is all real numbers.