Solving Rational Inequalities: Where R(x) ≥ 0

by SLV Team 46 views

Hey guys! Let's dive into a cool math problem. We're given a rational function, and our mission is to figure out where it's greater than or equal to zero. This is a common type of question in algebra, and understanding how to solve it is super helpful. We're going to break down the function, find its critical points, and then test intervals to find our solution. Let's get started!

Understanding the Problem: The Rational Function

First things first, let's look at the rational function itself. The function is defined as r(x) = (x^3 + 4x^2 + 4x) / (x^2 - 9). Our goal is to determine the values of x for which this function is greater than or equal to zero, i.e., r(x) ≥ 0. This involves finding the zeros of the numerator and the denominator, also known as critical points, and using those points to create intervals on the number line to test.

Breaking Down the Function

To make this easier, we can factor both the numerator and the denominator. Factoring helps us find the zeros of the function, which are crucial for determining the intervals where r(x) is positive or negative. Let's start with the numerator:

  • x^3 + 4x^2 + 4x = x(x^2 + 4x + 4) = x(x + 2)^2

Now, let's factor the denominator:

  • x^2 - 9 = (x - 3)(x + 3)

So, our simplified function is r(x) = [x(x + 2)^2] / [(x - 3)(x + 3)]. This factored form is much easier to work with.

Identifying Critical Points

Critical points are the x-values where the function might change its sign. These occur where the numerator is zero (the zeros of the function) and where the denominator is zero (the function is undefined).

  • Numerator Zeros: Setting the numerator x(x + 2)^2 = 0 gives us x = 0 and x = -2.
  • Denominator Zeros: Setting the denominator (x - 3)(x + 3) = 0 gives us x = 3 and x = -3.

These critical points (-3, -2, 0, and 3) divide the number line into intervals. We'll test each interval to see where r(x) ≥ 0.

Step-by-Step Solution: Testing Intervals

Alright, now that we've got our critical points, we're going to test intervals to find where the function is positive or zero. We'll use the factored form of the function: r(x) = [x(x + 2)^2] / [(x - 3)(x + 3)]. Remember, the intervals are determined by the critical points: x < -3, -3 < x < -2, -2 < x < 0, 0 < x < 3, and x > 3.

Testing Each Interval

  1. Interval 1: x < -3: Let's pick x = -4. Then r(-4) = [(-4)(-4 + 2)^2] / [(-4 - 3)(-4 + 3)] = [-4(4)] / [(-7)(-1)] = -16 / 7 < 0. So, r(x) is negative in this interval.

  2. Interval 2: -3 < x < -2: Let's pick x = -2.5. Then r(-2.5) = [(-2.5)(-2.5 + 2)^2] / [(-2.5 - 3)(-2.5 + 3)] = [-2.5(0.25)] / [(-5.5)(0.5)] = -0.625 / -2.75 > 0. So, r(x) is positive in this interval.

  3. Interval 3: -2 < x < 0: Let's pick x = -1. Then r(-1) = [(-1)(-1 + 2)^2] / [(-1 - 3)(-1 + 3)] = [-1(1)] / [(-4)(2)] = -1 / -8 > 0. So, r(x) is positive in this interval.

  4. Interval 4: 0 < x < 3: Let's pick x = 1. Then r(1) = [(1)(1 + 2)^2] / [(1 - 3)(1 + 3)] = [1(9)] / [(-2)(4)] = 9 / -8 < 0. So, r(x) is negative in this interval.

  5. Interval 5: x > 3: Let's pick x = 4. Then r(4) = [(4)(4 + 2)^2] / [(4 - 3)(4 + 3)] = [4(36)] / [(1)(7)] = 144 / 7 > 0. So, r(x) is positive in this interval.

Checking Critical Points

We also need to check the critical points themselves:

  • At x = -3, the function is undefined (division by zero).
  • At x = -2, r(x) = [(-2)(-2 + 2)^2] / [(-2 - 3)(-2 + 3)] = 0. So, r(x) = 0 at x = -2.
  • At x = 0, r(x) = [(0)(0 + 2)^2] / [(0 - 3)(0 + 3)] = 0. So, r(x) = 0 at x = 0.
  • At x = 3, the function is undefined (division by zero).

Determining the Solution: Intervals Where r(x) ≥ 0

Based on our interval testing and critical point analysis, let's pinpoint where r(x) ≥ 0. We're looking for the intervals where the function is positive or equal to zero.

The Solution

  • r(x) > 0 when -3 < x < -2 and x > 3.
  • r(x) = 0 when x = -2 and x = 0.

So, the solution to the inequality r(x) ≥ 0 is (-3, -2] ∪ {0} ∪ (3, ∞). Notice that -3 and 3 are excluded because they make the denominator zero, and the function is undefined at those points. -2 and 0 are included because the function equals zero at these points, satisfying the ≥ condition. The correct answer, therefore, must correspond to these intervals.

Matching with the Options

Now, let's look at the given options and see which one matches our solution:

  • (A) x ≥ 0: This is incorrect because it only includes positive values and doesn't account for the intervals where the function is positive between -3 and -2.
  • (B) -3 < x ≤ -2 or x ≥ 0: This is the correct answer. It includes the interval (-3, -2], where the function is positive or zero, and also includes 0, where the function is zero, and all x greater than 3.
  • (C) x = -2 and x = 0 only: This is incorrect because it only lists the points where the function equals zero but ignores the intervals where the function is positive.
  • (D) -3 < x < -2: This is incorrect because it doesn't include the point where x = 0 and the interval greater than 3.

So, the correct answer is (B). We've successfully navigated the rational inequality and found where r(x) ≥ 0.

Conclusion: Mastering Rational Inequalities

Awesome work, everyone! We've successfully solved a rational inequality, which involves factoring, finding critical points, testing intervals, and careful consideration of undefined points. This approach is key to tackling any rational inequality. Remember, it's all about finding those critical points, testing around them, and carefully considering where the function is positive, negative, or zero. Keep practicing, and you'll become a pro at these problems in no time. If you have any questions or want to try another problem, feel free to ask. Cheers!