美国高中数学1-3 Inequalities

美国高中 数学 数学高考答题卡模板高考数学答题卡模板三年级数学混合运算测试卷数学作业设计案例新人教版八年级上数学教学计划 1-3 Inequalities Section 3-1: Linear inequalities and Absolute Value Solving a linear inequality is very similar to solving a linear equation. You can basically solve it the same way except for one situation. When you multiply/divide both sides of an inequality by a negative number, change the sense (direction) of the inequality. Sample problems Problem Problem 2 1 8 - 2x > 3x - 5 < 9 6 + x -2x > 2x - 5 < -2 9 x < 2x < 1 14 x < 7 In problem 1, it was solved exactly like an equation. The solution is all numbers less than 7 make the statement true. The graph looks like: The open dot means does not include! In problem two, notice we changed the sense in the last step because we divided both sides by -2. The graph looks like: The closed dot means it includes this point! Absolute Value Recall that absolute value means the distance from any point to zero. Study the following chart: Sentence Meaning Graph Solution The distance from x to 0 is exactly a x = a or x = |x| = a units -a The distance from x to 0 is less than a |x| < a -a < x < a units. The distance from x to 0 is greater x < -a or x > |x| > a than a units a Sample Problems 2|3x - 5| = 8 This is an equality, choice number 1 on the chart. First, islolate the absolute value to use the chart above. Divide both sides by 2. |3x - 5| = 4 Now apply the definition in the last column. 3x - 5 = 4 or 3x - 5 = -4 Solve each one individually. Add 5 and divide by 3. 3x = 9 or 3x = 1 x = 3 or x = 1/3 3|2x + 3| < 12 This is an inequality involving the less than symbol. Use choice two from the chart. Remember to isolate the absolute value first! Divide by 3 |2x + 3| < 4 Now apply rule #2 from the above chart. -4 < 2x + 3 < 4 Isolate for x in the middle by subtracting 3 and dividing by 2 -7 < 2x < 1 -7/2 < x < 1/2 The solution is all numbers between -7/2 and 1/2. |x - 4|/3 > 2 This is the last choice involving a greater than symbol. Again, isolate the absolute value before you apply the appropriate rule! Multiply by 3 |x - 4| > 6 Now apply the last rule from the chart. x - 4 > 6 or x - 4 < -6 Add 4 to both sides. x > 10 or x < -2 All numbers greater than or equal to 10 work, and all numbers less than or equal to -2 work! A Slightly tougher problem!! 1 < |x| < 5 Method 1. Split into two inequalities 1 < |x| and |x| < 5 Solve individually |x| > 1 and |x| < 5 The first inequality is rule 3 and the second is rule 2. Apply these rules: x > 1 or x < -1 and -5 < x < 5 To find the solution, find the area which is common to both. Look at these to graphs and what points do they have in common? That's right, the numbers between -1 and -5 and the numbers between 1 and 5!! The solution looks like: Let's head on to the next section: Section 3-2: Polynomial Inequalities, One Variable We have two methods to solve polynomial inequalities: 1) Find the zeros and use a sign analysis of P(x) 2) Use the graphing calculator and analyze the graph of P(x) Method One! 21) x + x - 6 > 0 First, factor the polynomial. (x + 3)(x - 2) > 0 Identify the zeros! x = -3, x = 2 Graph the zeros, use open dots (strickly greater than) Now do a sign analysis by picking a point in each of the three regions divided by the zeros of the function. Chosen x value Sign of the function -4 (-4 + 3)(-4 - 2), + 0 (0 + 3)(0 - 2), - 3 (3 + 3)(3 - 2), + The sections we want are the positive sections because the function is greater than zero! Therefore, the answer is: x < -3 or x > 2 222) (x - 4)(x - 1) < 0 Factor the polynomial to find the zeros 2(x - 2)(x + 2)(x - 1) < 0 The zeros are: x = 2, x = -2, x = 1 (double root) Do a sign analysis for the four sections divided by the three zeros Chosen x Sign of function value (-3 - 2)(-3 + 2)(-3 - -3 21), + (0 - 2)(0 + 2)(0 - 0 21), - (1.5 - 2)(1.5 + 2)(1.5 - 1.5 21), - (3 - 2)(3 + 2)(3 - 3 21), + We want the sections that are negative because the polynomial is less than 0. Therefore, the answer is: -2 < x < 2 3) (x - 3)(x + 4)/(x - 5) > 0 This is already factored. Find the zeros of the numerator and denominator but keep in mind the bottom can never equal zero, so the zeros of the denominator will always have open dots. The zeros are: x = 3, x = -4 and x = 5 Now do your sign analysis for the four sections seperated by the three zeros. Chosen x sign function value (-5 - 3)(-5 + 4)/(-5 - -5 5), - (0 - 3)(0 + 4)/(0 - 0 5), + (4 - 3)(4 + 4)/(4 - 4 5), - (6 - 3)(6 + 4)/(6 - 6 5), + The sections that we want are positive because the polynomial is greater than zero. Therefore, the solutions are: -4 < x < 3 or x > 5 Method two! This method is accomplished by using your graphing calculator!! Graph the function and identify the zeros. Then, any part above the x-axis is the positive area (> 0) and any part below the x-axis represents the negative area(< 0). 321) 2x + x - 5x - 2 < 0 Use your graphing feature on the TI-82 to graph this function. Only type in upto the < symbol. This is the graph: Use the calc section choice 2 to find each of the zeros. The zeros (rounded to tenths place) are: x = -1.6, x = -.4, and x = 1.5 We want the area that is less than zero, or the area below the x axis. Thus, the solution is: x < -1.6 or -.4 < x < 1.5 This is a good method to use when you can't factor the polynomial!! Section 3-3: Polynomial Inequalities in Two Variables Demo: Systems of Linear Inequalities Slope-Int Form(Exploremath - requires Shockwave) Demo: Systems of Inequalities, Standard form (Exploremath - requires Shockwave) The graphs of inequalities in two variables consist of points in the x-y graphing plane. Two find the graph, first graph the equation. This is the boundary line between what makes the inequality greater than or less than. The line can be solid (points on the line make the statement true) or dotted (does not include the points on the line). You must decide which side of the line to shade. The easiest way to tell is to pick a point on one side of the line. If it makes the statement true, you have the correct side. Go ahead and shade it. If it makes the statement false, you have the wrong side. Shade the other side. Sample Problems 1) Graph the inequality y < 2x + 1 The boundary is a line. The y-intercept is (0,1), with slope of 2. Graph the line! Notice the line is dotted because the line is not included ( no or equal to in the statement). Now pick a point not on the line. Use (0,0). Is 0 < 2(0) + 1? Sure is, so shade this side of the line! 2) Graph the solution of y > x + 3 and y < 9 2- x The first has a boundary line. The second boundary is a parabola. The line graphs with slope 1 and y-intercept (0, 3). The parabola has a vertex point at (0, 9) and crosses the x-axis at 3 and -3. Look back to graphing parabolas if you forgot about that!! Now pick a point. Try (0, 0). Try the line. 0 > 0 + 3. This is false. Shade the other direction, above the line. Now try (0, 0) in the parabola. 0 < 9 - 0. This is true. Shade inside the parabola. Viola!! You get: The solution is the cross-hatched area!! 3) Find the solution for |y| > 1 and |x| < 3 in two variables. Both are absolute value graphs The first can be written as y > 1 or y < -1. The second can be written as -3 < x < 3. The first absolute value graph means we want values above one and below -1 including the equality. The second absolute value means x must be between -3 and 3 including these points. Add that graph!! The solution is the crossed-hatched area!! 4) Find the solution for y < x + 1, y > -x + 4 and y > 2 Each of these has a boundary which is a line. Two are slanted and the other is horizontal. The first has slope 1 and y-intercept 1. The second has slope -1 and y-intercept 4. The last is a horizontal line at 2. You can try (0,0) for all three lines. In the first line 0 < 0 + 1 is true. Shade below the blue line. In the second line, 0 > 0 + 4 is false. Shade above the green line. In the third line, 0 > 2 is false. Shade above the purple line. The result is: On to linear programming: Back it up dude!! I'm lost!! Section 3-4: Linear Programming Demo: Linear Programming (Exploremath - requires Shockwave) Try the quiz at the bottom of the page! go to quiz Linear programming is a method used to identify optimal maximum or minimum values. It is used in business for practical planning, decision-making problems, and many other problems that can be done using a computer. Each different resource can be written as a linear inequality called a constraint. These constraints can be resources like the number of workers, amount of time on a given shift, number of machines, availability of these machines, etc., etc. By using what we call the corner point theorem, we can find an optimal solution(s) for our problem. When we graph these constraints, we will get a feasible region that contains our solutions. The corner point theorem says that if a maximum or minimum value exists, it will occur at a corner point of this feasible region. Sample problem 1) Find and graph the feasible region for the following constraints: x + y < 5 2x + y > 4 x > 0, y > 0 First step is to write each of these inequalities isolating for y y < -x + 5 y > -2x + 4 x > 0, y > 0 Now graph all 4 of the constraints. The first inequality has a slope of -1 and y-intercept of 5. The second has a slope of -2 and y-intercept of 4. The third is a vertical line (y-axis) and the fourth is a horizontal line (x-axis). The area that we want is below the blue line, above the green line, to the right of the purple line and above the red line. The feasible region looks like: The feasible region is outlined in black!! Now, let's find the maximum value if the function we want to maximize is: P = 3x + 5y To find the maximum value, we need to use the corner point theorem. The graph above is easy to find the corner points. They are: (0, 4), (0, 5), (2, 0) and (5, 0). Plug these four values in the function and take the largest value. Point function Value (0, 4) 0 + 5(4) 20 (0, 5) 0 + 5(5) 25 (2, 0) 3(2) + 0 6 (5, 0) 3(5) + 0 15 The maximum value from this chart would be at (0, 5), with a maximum of 25! 2) Minimize the function C = x + 4y under the following constraints: x + y < 10 5x + 2y > 20 -x + 2y > 0 x > 0, y > 0 First, write all constraints isolated for y. y < -x + 10 y > -2.5 x + 10 y > .5x x > 0, y > 0 The first line has slope -1 and y-intercept 10. The second has slope -2.5 and y-intercept 10. The third has slope .5 and y-intercept 0. The next two are vertical and horizontal lines. We want the area below the blue line, to the right of the green line and above the purple line. The area forms a triangle like the next picture. Now to find the corner points, you need to find the intersection of two lines solving simultaneously. x + y = 10 -2x - 2y = -20 5x + 2y = 20 3x = 0 x = 0 Intersects at (0, 10) x + y = 10 -x + 2y = 0 3y = 10 y = 10/3 x = 30/3 - 10/3 = 20/3 Intersects at (20/3, 10/3) -x + 2y = 0 x - 2y = 0 5x + 2y = 20 6x = 20 x = 20/6 = 10/3 10/3 -2y = 0 -2y = -10/3 y = 10/6 = 5/3 Intersects at (10/3, 5/3) Now plug each of these into the original function C = x + 4y Point Function Value (0, 10) 0 + 4(10) 40 (20/3, 10/3) 20/3 + 4(10/3) 20 (10/3, 5/3) 10/3 + 4(5/3) 10 The minimum value of 10 happens at (10/3, 5/3) 3) A studio sells photographs and prints. It cost $20 to purchase each photograph and it takes 2 hours to frame it. It costs $25 to purchase each print and it takes 5 hours to frame it. The store has at most $400 to spend and at most 60 hours to frame. It makes $30 profit on each photograph and $50 profit on each print. Find the number of each that the studio should purchase to maximize profits. The first thing to do is to organize the information into a chart. Let x = the number of photographs and y = the number of prints Photos Prints Total Cost 20x 25y < 400 Time 2x 5y < 60 Profit 30x 50y Write the constraints for the problem from the above chart: 20x + 25y < 400 2x + 5y < 60 x > 0, y > 0 Write the profit function from the chart above: P = 30x + 50y Now graph the constraints like we did in the first two examples. Notice that the last two constraints put the graph in the first quadrant. The corner points are (0, 0), (0, 12), (20, 0). The last corner point is found by the intersection of the two lines. 20x + 25y = 400 2x + 5y = 60 -10x - 25y = -300 10x = 100 x = 10 The other corner point is (10, 8) Now set up a chart to find the maximum value for the profit function. Point function value (0, 0) 30(0) + 50(0) 0 (0, 12) 30(0) + 50(12) 600 (20, 0) 30(20) + 50(0) 600 (10, 8) 30(10) + 50(8) 700 The maximum value happens at (10, 8) for a profit of $700. This means to purchase 10 photos and 8 prints. Bring on the practice test!! No, I better go back and study more!!! <> Current quizaroo # 3 1) Solve the inequality | 2x - 3| < 5 a) x < -1 or x > 4 b) x > 4 c) -1 < x < 4 d) x < 4 e) x > 0 2) Solve the inequality | 4x - 5| > 8 a) x > 13/4 or x < -3/4 b) x > 13/4 c) -3/4 < x < 13/4 d) x > 0 e) x < 0 3) Solve the inequality (x - 5)(x + 3) > 0 a) -3 < x < 5 b) x < -3 or x > 5 c) x < -3 d) x > 5 e) empty set 24) Solve the inequality (x - 3) /(x - 4) < 0 a) 3 < x < 4 b) x > 4 c) x < 3 d) x < 4 e) x < 4, but not including 3 5) The inequalities of the above graph are y < x + 1 and y > -x + 3. Identify the region that represents the intersection. a) section 1 b) section 2 c) section 3 d) section 4 e) empty set click here for answers!! Sample test 1) Solve each equation or inequality. Graph the solution on a number line. If there is no solution, write "no solution." a) 7x - 5 > 19 b) (2 - x)/3 < (3 - 2x)/4 c) |x| > 5 d) 3|x - 5| < 21 e) |5 - 2x| < 23 f) 3 < |x - 2| < 7 2) Solve each inequality by the indicated method. Graph on a number line. Do by the factoring method. 2 a) x - 5x + 4 > 4 - 1 < 0 b) x 0 c) (x - 4)(x + 3)/(x - 5) < 0 Use your graphing calculator to do the following. 32 d) 2x + x - 12x + 8 < 320 e) 2x + 3x - 1 > 0 3) Sketch the graph of each of the given inequality in the x-y plane. a) 3x - 4y > 2 + 3x + 12 b) y > x 2 c) 2 < | x - 3| < 5 4) Graph the solution of the given system of inequalities in the x-y plane. a) x > 2 b) y 2 - 3 c) |x| < 4 < x x + 2y < 2 y > x |y| > 1 5) Maximize P = 2x + 3y under the following constraints: x + 2y < 4 x + y < 3 x > 0, y > 0 6) A popular cereal combines wheat and corn. At least 27 tons of the cereal is to be made. For the best flavor, the amount of corn should be at least twice the amount of wheat. Corn costs $250 per ton and wheat costs $350 per ton. How much of each should be used to minimize the cost? This page hosted by Get your own Free Home Page Show me the answer key!! I know I got 100% I made a few mistakes. I better head back and study!! Answers to Sample Test! 1) a) 7x > 24 x > 24/7 o-------------> 24/7 b) Multiply both sides by 12 to get 8 - 4x < 9 - 6x 2x + 8 < 9 2x < 1 x < 1//2 <-----------] closed dot! 1/2 c) x < -5 or x > 5 <----------] [-------------> -5 5 d) -7 < x - 5 < 7 -2 < x < 12 o---------------o -2 12 e) -23 < 5 - 2x < 23 -28 < -2x < 18 14 > x > -9 [-------------------------] -9 14 f) 3 < x - 2 < 7 or -3 > x - 2 > -7 5 < x < 9 or -1 > x > -5 [---------] [-------------] -5 -1 5 9 2) a) (x - 1)(x - 4) > 0 - - + (x - 4) - + + (x - 1) <-----------0-----------0-------------> 1 4 We want positive. Answer: <-----------0 0------------> 1 4 22 b) (x - 1)(x + 1) < 0 2 (x - 1)(x + 1)(x + 1) < 0 2 + + + (x + 1) - - + (x - 1) - + + (x + 1) <---------0------------0--------------------------> -1 1 We want negative. The answer is: 0--------------0 -1 1 c) - + + + (x + 3) - - + + (x - 4) - - - + (x - 5) <---------------[]-------------[]-----------0--------------- > -3 4 5 We want negative: Answer <----------------] [-----------0 -3 4 5 d) The zeros are at -2.6, .8 and 1.7 We want the area to be < 0. We need all areas of the above graph that are beneath the x-axis. Answer: <---------------0 0-----0 -2.6 .8 1.7 e) The zeros are -1, and .5 We want the area above the x-axis 0--------------------> .5 3) a) b) c) The area is from -2 to 1 and 5 to 8 exclusive. The lines are dotted! 4) a) b) c) 5) The corner points are: (0, 0), (0, 2), (2, 1) and (3, 0) P = 2x + 3y (0, 0) yields 0 (0, 2) yields 6 (2, 1) yields 7 (3, 0) yields 6 The maximum value is 7 and it happens at (2, 1) 6) x = amount of corn y = amount of wheat constraints x > 0 y > 0 x + y > 27 x > 2y objective function: C = 250 x + 350y The corner points are (18, 9) and (27, 0) (18, 9) yields 7650 (27, 0) yields 6750 To minimize, take (27, 0) This page hosted by Get your own Free Home Page