Factoring Inequalities: A Step-by-Step Guide
Hey guys! Today, we're going to break down how to factor inequalities, step by step. We'll use a specific example to make it super clear, and I'll explain the why behind each move we make. So, let's dive right in!
Understanding the Problem
Before we get into the nitty-gritty, let's make sure we understand what we're dealing with. We've got an inequality that looks something like this:
(x² — 1) / (x² + 1) ≤ 0
Our goal is to find all the values of 'x' that make this statement true. In other words, we want to know when this fraction is less than or equal to zero. Sounds fun, right? Let's break it down.
Key Concepts
Before we jump into solving, let's refresh some key concepts. Factoring is like reverse multiplication – we're trying to break down an expression into smaller parts that multiply together. When dealing with inequalities, we're not just looking for a single answer, but a range of values that satisfy the condition. Remember, understanding the basics is crucial for tackling more complex problems.
Why Factoring Matters
Why is factoring so important in solving inequalities? Well, it helps us identify the critical points where the expression might change its sign (from positive to negative or vice versa). These critical points are the roots of the numerator and denominator, and they divide the number line into intervals. By testing values within these intervals, we can determine the solution set of the inequality. Factoring makes this process much more manageable, especially for polynomial inequalities.
Visualizing the Problem
It can be helpful to visualize what's going on. Imagine a number line. We're trying to find the sections of this line where our expression is either negative or zero. Factoring helps us pinpoint the key locations on this line where things might switch signs. Think of it like finding the turning points on a rollercoaster – these are the spots we really need to pay attention to!
Step 1: Factor the Numerator and Denominator
Okay, first things first, let's factor both the numerator and the denominator of our inequality. The numerator, x² — 1, is a classic example of the difference of squares. Remember that formula? a² — b² = (a — b)(a + b). So, we can factor x² — 1 as (x — 1)(x + 1). Easy peasy!
The denominator, x² + 1, is a bit different. Can we factor it using real numbers? Nope! This expression is always positive for any real number 'x' because squaring a number always results in a non-negative value, and adding 1 makes it strictly positive. This is a crucial observation, and we'll see why it matters later.
Detailed Explanation of Factoring the Numerator
Let's dive a bit deeper into why x² — 1 factors into (x — 1)(x + 1). This is a direct application of the difference of squares identity. Imagine you have a square with side length 'x', so its area is x². Now, you remove a smaller square with side length 1, so its area is 1. The remaining area is x² — 1. You can rearrange the remaining shape into a rectangle with sides (x — 1) and (x + 1). This geometric interpretation makes the factoring more intuitive.
Why the Denominator Doesn't Factor (in Real Numbers)
The denominator, x² + 1, cannot be factored using real numbers. Why? Because there are no real numbers that, when squared and added to 1, result in zero. The graph of y = x² + 1 is a parabola that opens upwards and never touches the x-axis. This means it has no real roots. This is a super important point to remember. In the context of inequalities, this means that the denominator will always be positive, which simplifies our analysis.
Step 2: Identify Critical Points
Now that we've factored the numerator, we need to find the critical points. These are the values of 'x' that make either the numerator or the denominator equal to zero. Why are these points so critical? Because they're the spots where our expression can potentially change its sign.
For the numerator, (x — 1)(x + 1), the critical points are x = 1 and x = -1. These are the values that make the numerator zero. For the denominator, x² + 1, we already determined that it's never zero for any real number. So, we only have two critical points to worry about: 1 and -1.
The Significance of Critical Points
Critical points are the backbone of solving inequalities. They divide the number line into intervals, and within each interval, the expression will maintain a consistent sign (either positive or negative). This is because the sign can only change at the points where the expression is either zero or undefined. Think of it like this: if you're driving along a road, you can only change direction at a fork or a turn. Critical points are like those forks and turns for our inequality.
Finding Critical Points: A Closer Look
To find the critical points, we set each factor in the numerator and denominator equal to zero and solve for 'x'. For the numerator (x — 1)(x + 1), we have:
- x — 1 = 0 => x = 1
- x + 1 = 0 => x = -1
Since the denominator x² + 1 is never zero, it doesn't contribute any critical points. This simplifies our task significantly. We now know that the only points where the sign of our expression can change are x = 1 and x = -1.
Step 3: Create a Sign Chart
Okay, here comes the fun part! We're going to create a sign chart. A sign chart is a visual tool that helps us see how the sign of our expression changes across different intervals. It's like a map that guides us to the solution.
First, draw a number line and mark our critical points, -1 and 1. These points divide the number line into three intervals: (-∞, -1), (-1, 1), and (1, ∞). Now, we'll pick a test value from each interval and plug it into our factored expression to see if it's positive or negative.
Building the Sign Chart: Step-by-Step
- Draw the Number Line: Start by drawing a horizontal line representing the real number line. Mark the critical points (-1 and 1) on this line. These points divide the line into intervals.
- Choose Test Values: Select a test value within each interval. For example:
- In the interval (-∞, -1), we can choose x = -2.
- In the interval (-1, 1), we can choose x = 0.
- In the interval (1, ∞), we can choose x = 2.
- Evaluate the Expression: Plug each test value into the factored expression (x — 1)(x + 1) / (x² + 1) and determine the sign of the result. Remember, we only care about the sign (positive or negative), not the exact value.
- Fill in the Sign Chart: Write the sign (+ or -) above each interval in the sign chart, indicating the sign of the expression in that interval.
Why Sign Charts Work
Sign charts work because the sign of a continuous expression can only change at its zeros (critical points) or at points of discontinuity. By testing a value within each interval, we can determine the sign of the expression throughout that entire interval. This is a powerful technique that simplifies solving inequalities.
Step 4: Test Values in Each Interval
Let's put our sign chart skills to the test! We'll pick a value from each interval and plug it into our factored expression, (x — 1)(x + 1) / (x² + 1). Remember, we only care about the sign, not the exact number.
- Interval (-∞, -1): Let's choose x = -2.
- ((-2) — 1)((-2) + 1) / ((-2)² + 1) = (-3)(-1) / (5) = 3/5. This is positive.
- Interval (-1, 1): Let's choose x = 0.
- (0 — 1)(0 + 1) / (0² + 1) = (-1)(1) / (1) = -1. This is negative.
- Interval (1, ∞): Let's choose x = 2.
- (2 — 1)(2 + 1) / (2² + 1) = (1)(3) / (5) = 3/5. This is positive.
So, our expression is positive in the intervals (-∞, -1) and (1, ∞), and negative in the interval (-1, 1). This is exactly what we needed to know!
Detailed Breakdown of Testing Values
Let's break down the process of testing values even further. When we plug in a test value, we're essentially evaluating the expression at a specific point. The sign of the result tells us whether the expression is positive or negative at that point. Since the expression can only change signs at the critical points, the sign we find at the test value will be the same throughout the entire interval.
For example, when we tested x = -2 in the interval (-∞, -1), we found that the expression was positive. This means that the expression is positive for every value of x in the interval (-∞, -1). We don't need to test any other values in that interval – we already know the sign.
Step 5: Determine the Solution Set
We're in the home stretch now! Remember, our original inequality was (x² — 1) / (x² + 1) ≤ 0. We want to find the intervals where our expression is less than or equal to zero. Looking at our sign chart, we see that this occurs in the interval (-1, 1). But wait, there's a slight twist!
Since we have a less than or equal to sign, we also need to include the critical points where the expression is equal to zero. These are the points where the numerator is zero, which we found to be x = -1 and x = 1. So, our solution set includes the endpoints of the interval.
Therefore, the solution to our inequality is [-1, 1]. This means that any value of 'x' between -1 and 1 (including -1 and 1) will satisfy the inequality.
Expressing the Solution Set: Interval Notation vs. Set-Builder Notation
The solution set can be expressed in different notations. We've already used interval notation, which is a concise way to represent a range of values using brackets and parentheses. In interval notation, square brackets [ ] indicate that the endpoint is included in the solution set, while parentheses ( ) indicate that the endpoint is excluded.
Another way to express the solution set is using set-builder notation. In this notation, we write the solution set as {x | condition}, where 'condition' describes the values of x that satisfy the inequality. For our example, the solution set in set-builder notation would be {x | -1 ≤ x ≤ 1}.
Checking the Solution Set
It's always a good idea to check your solution set to make sure it's correct. You can do this by plugging in values from the solution set and values outside the solution set into the original inequality and verifying that the inequality holds true for the values in the solution set and false for the values outside the solution set.
For example, let's check x = 0 (which is in our solution set [-1, 1]):
(0² — 1) / (0² + 1) = -1 / 1 = -1 ≤ 0. This is true.
Now, let's check x = 2 (which is outside our solution set):
(2² — 1) / (2² + 1) = 3 / 5 > 0. This is false.
These checks confirm that our solution set is correct.
Conclusion
And there you have it! We've successfully factored an inequality and found its solution set. Remember, the key steps are: factor the expressions, identify critical points, create a sign chart, test values in each interval, and determine the solution set. With practice, you'll be a pro at this in no time! Keep practicing, and you'll conquer any inequality that comes your way. You got this!
I hope this step-by-step explanation was helpful and easy to understand. If you have any more questions, feel free to ask! Happy factoring, guys!