# INEQUALITY SOLVING MATH 152, SECTION 55 (VIPUL NAIK)

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1 INEQUALITY SOLVING MATH 152, SECTION 55 (VIPUL NAIK) Difficulty level: Moderate for people who have been inequalities before, high for people who haven t. Corresponding material in the book: Section 1.3. However, this section does not cover inequalities involving rational functions, and only does the polynomial case. The rational functions case is only a mild generalization of the polynomial case, but it is essential that you understand the rational functions case from the notes. Things that students should definitely get: Set notation for writing intervals, how to manipulate and solve inequalities, inequalities for rational functions, inequalities for absolute value. Words... Executive summary (1) When solving inequalities, we can add the same thing to both sides. We can subtract the same thing from both sides. We can add two inequalities with the same direction of inequality. (2) We can multiply a positive number to both sides and preserve the direction of inequality. If we multiply by a negative number, we reverse the direction of inequality. If we multiply by a number that we know is nonnegative (But we re not sure if it is positive or zero) then the direction of inequality is preserved but an equality case gets introduced. e.g., if a > b and x 0, we get ax ab. Actions (think back to examples where you ve dealt with these issues)... (1) If x takes values in a certain set A, then the set of possible values for x is the union of A and the negatives of numbers in A. (2) Generally, when solving inequalities involving absolute value, consider the case that the thing inside the absolute value is negative and the case that it is nonnegative. Solve both cases and then take the union of the solutions. (3) When solving inequalities involving rational functions, bring everything to one side. So it reduces to trying to find when a given rational function is positive, negative, zero, and undefined. (4) A rational function is undefined wherever the denominator is 0. It is 0 where the numerator is 0 but the denominator isn t. For positive and negative, start from the far positive end and remember that every time you cross over a linear factor of the numerator or the denominator, the sign changes... (5)... except that the sign changes only if the total multiplicity of that factor is odd. If the total multiplicity is even, then the sign doesn t change. (What do I mean? Review the examples we did and try figuring out)... (6)... and what if there are quadratic factors in the numerator or denominator, such as x 2 + 1, that do not factorize further? These are anyway always positive or always negative, so they don t affect the sign of things... (7)... and what if there are quadratic factors that you don t know how to factorize? Well, use the quadratic formula. 1. Equality versus inequality In this lecture, we cover inequalities. This is almost like solving equations, except that instead of an equality sign, there is an inequality sign. The main difference is that because we have an inequality sign, the solution sets are usually much bigger they re not just a few isolated points, they are unions of intervals. Further, the solution to the corresponding equation is usually on the boundary of the solution set to the inequality. 1

3 To complete the discussion, we need to talk about. Now, is a big and mind-boggling concept and we re not really going to discuss it here. For our purposes, (positive infinity) is a placeholder for no upper limit and is a placeholder for no lower limit. So the interval (a, ) is the set {x R : a < x}. and [a, ) is the set {x R : a x}. And similarly, (, a) is the set {x R : x < a} and (, a] is the set {x R : x a}. And on the number line, you just use an arrow to indicate that it ll go on forever. Notice that the parentheses around are always the round ones, meaning that is never included. And that makes sense because isn t real. Nor is. Whether these things exist and what they mean is beyond the scope of our discussion. I just want you to think of them as placeholders As a union of intervals. Okay, here s a quick test. Express ( 1, 1) \ {0} as a union of intervals? What is it? [Draw diagram]. You see the picture. It is ( 1, 0) (0, 1). Or, what about ( 2, 2) \ ( 1, 1)? [Draw diagram]. That s ( 2, 1] [1, 2). Note the way the circles fill and unfill. 3. Inequality solving 3.1. Some basic rules. Let s discuss some very basic rules of inequality-solving. The way you solve inequalities is very similar to the way you solve equalities, except this: when you multiply both sides by a negative number, you change the direction of the inequality. And the one thing you shouldn t do is multiply both sides by zero. Okay, so consider the inequality: Moving all stuff to one side gives: x + 4 5(x 1) 4x Now, you multiply both sides by 1, so change the sign of the inequality, and get: 4x 9 0 And then simplify to x 9/4. Which is the interval [9/4, ) on the number line. Inequality-solving for polynomial functions. Let s now consider some polynomial functions. The goal is to consider a polynomial function f, and try to determine where it is zero, where it is positive, and where it is negative. For now, we ll focus on polynomial functions that split completely into linear factors. For instance, consider the polynomial function: f(x) = x(x + 1)(x 1) This polynomial is a product of three linear factors. How do we figure out the sign of this polynomial function at a point? Its sign is the product of the signs of the three factors. Now, if you have a linear factor x a, where is it positive, where is it zero, and where is it negative? Answer: it is positive for x > a, zero for x = a, and negative for x < a. So, we see that each linear factor switches sign at the corresponding zero (i.e., the linear factor x a switches sign at a). If only one linear factor switches sign, and the signs of the other factors remain the same, then the product also switches sign. So, for the above polynomial f, the sign changes could potentially occur at 1, 0, 1. Let s view this on a number line. To the right of 1, x is pretty large, so all the three factors are positive. A product of three positives is a positive, so for x > 1, f(x) is positive. At 1, the function becomes zero. Immediately to the left of 1, the factor x 1 becomes negative, but the other two factors are still positive. So, the overall product is negative. And so the story goes till we hit x = 0. At x = 0, the function again takes the value 0. Then, to the immediate left of 0, both x and x 1 are negative, but x + 1 is positive. So, the function is positive, and remains so till we reach x = 1. There it becomes 0, and to the left of 1, all factors are negative, hence so is the product. 3

5 x and get a new function that is defined at 0. But as written, the function is undefined at 0. [In the limit/continuity jargon, the function has a removable discontinuity at 0.] Anyway, away from these three points, the function simplifies to: (x + 1)(x + 2) (x 1)(x 2) So, the points where we could have a sign change are 2, 1, 1, 2. Let s start from the right. On the far right, everything is positive, because x is greater than all four numbers. So the expression is positive. When you cross 2, x 2 becomes negative, the others remain positive. So, negative. Then, when you cross 1, both x 1 and x 2 become negative, the others remain positive, so positive. And then after 1 it becomes negative and after 2 again positive. So the function is positive at (, 2) ( 1, 1) (2, ). But wait. The original function wasn t defined at 0, so we need to exclude that. So the original function is positive on (, 2) ( 1, 0) (0, 1) (2, ). The function is negative at ( 2, 1) (1, 2). The function is zero on { 2, 1}. It is undefined at {0, 1, 2}. By the way, there s an important difference between the point 0, where it is undefined only because we wrote the function stupidly, and {1, 2}, where it is undefined for more fundamental reasons. Namely, at 0, the discontinuity is removable, but at 1 and 2, it cannot be removed. So we have a pretty good feel. But there s an important thing that we didn t take into account: higher powers. For instance, consider: x(x + 1) 3 (x 1) 2 So here, the function is not defined at 1. It s positive to the right of 1. What happens as we cross 1? The (x 1) term changes sign, but it is squared, so the sign change doesn t affect the sign of the overall expression. And indeed, there s no sign change. So the expression continues to be positive. Then, at 0, there is a sign change, and the expression becomes negative. At 1, there is another sign change, and the expression becomes positive again. The reason? The exponent in this case is odd. So, the function is positive on (, 1) (0, 1) (1, ), negative on ( 1, 0), zero on { 1, 0}, and undefined at 1. To view a consolidated summary on how to handle inequalities involving rational functions, go to the executive summary at the beginning of the document. 4. Inequalities and absolute value 4.1. Absolute value. We ve talked about the absolute value function, so I ll just say a bit about solving inequalities involving the absolute value function. The main thing to remember is: if the absolute value of a number is in a set A of nonnegative numbers, then that number itself is in the set A A, where A is the set of negatives of elements of A. So, for instance: is equivalent to: x 2 < 3 3 < x 2 < 3 So an inequality involving absolute values is really an inequality that involves both an upper bound and a lower bound. So it is two inequalities rolled into one. Now, when you have two inequalities rolled into one, you can solve them separately and intersect the solutions. But in this case, it s easy to sort of solve them both together. Just add 2 to both sides: 1 < x < 5 so the solution set is the open interval ( 1, 5). I suggest you look at more examples in the book involving absolute values, rational functions and inequalities. There are a lot of tricks. These ideas are very important. They ll come up again and again, particularly in the context of ɛ δ proofs. 5

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