Solving H(x) ≥ 0: A Step-by-Step Guide

Alright guys, let's dive into solving this inequality: $h(x) = \frac{5x}{(x-1)(x+2)} \geq 0$. It looks a bit intimidating at first, but don't worry, we'll break it down step-by-step. Inequalities involving rational functions (that's what we have here) are best tackled by finding critical points and testing intervals. So grab your thinking caps, and let's get started!

1. Finding the Critical Points

The critical points are the values of x that make either the numerator or the denominator of our rational function equal to zero. These points are crucial because they're where the function can change its sign (from positive to negative or vice versa).

• Numerator: The numerator is $5x$. Setting it to zero, we get $5x = 0$, which gives us $x = 0$. So, $x = 0$ is one of our critical points.
• Denominator: The denominator is $(x-1)(x+2)$. Setting it to zero, we get $(x-1)(x+2) = 0$. This gives us two critical points: $x = 1$ and $x = -2$.

Therefore, our critical points are $x = -2$, $x = 0$, and $x = 1$. These are the points we'll use to divide the number line into intervals.

Understanding critical points is super important. Think of them as the potential turning points for our function. They are the values where the function might cross the x-axis (numerator equals zero) or where the function is undefined (denominator equals zero). Because the function can only change signs at these points, we know that within each interval defined by these points, the function's sign will remain constant. This allows us to test just one value within each interval to determine the sign of the entire interval. This approach greatly simplifies solving the inequality. So, remember always start by identifying these crucial critical points.

2. Creating the Intervals

Now that we have our critical points, $x = -2$, $x = 0$, and $x = 1$, we can divide the number line into intervals. These points split the number line into four distinct intervals:

• $(-\infty, -2)$
• $(-2, 0)$
• $(0, 1)$
• $(1, \infty)$

Each of these intervals represents a range of x values where the function h(x) will have a consistent sign (either positive or negative). The endpoints of these intervals are our critical points, which, as we discussed, are the only places where h(x) can potentially change its sign. So, by determining the sign of h(x) within each interval, we can identify the intervals where h(x) ≥ 0, which is what we are trying to solve.

Visualizing this on a number line can be incredibly helpful. Draw a line and mark the points -2, 0, and 1. You'll clearly see the four intervals we've created. This visual representation makes it easier to understand how the critical points divide the number line and how we'll test each region to find our solution. Remember, these intervals are the key to unlocking the solution to our inequality!

3. Testing Each Interval

Next, we need to determine the sign of $h(x)$ in each of these intervals. We do this by picking a test value within each interval and plugging it into $h(x)$.

• Interval $(-\infty, -2)$: Let's pick $x = -3$. Then $h(-3) = \frac{5(-3)}{(-3-1)(-3+2)} = \frac{-15}{(-4)(-1)} = \frac{-15}{4} < 0$. So, $h(x)$ is negative in this interval.
• Interval $(-2, 0)$: Let's pick $x = -1$. Then $h(-1) = \frac{5(-1)}{(-1-1)(-1+2)} = \frac{-5}{(-2)(1)} = \frac{-5}{-2} = \frac{5}{2} > 0$. So, $h(x)$ is positive in this interval.
• Interval $(0, 1)$: Let's pick $x = 0.5$. Then $h(0.5) = \frac{5(0.5)}{(0.5-1)(0.5+2)} = \frac{2.5}{(-0.5)(2.5)} = \frac{2.5}{-1.25} = -2 < 0$. So, $h(x)$ is negative in this interval.
• Interval $(1, \infty)$: Let's pick $x = 2$. Then $h(2) = \frac{5(2)}{(2-1)(2+2)} = \frac{10}{(1)(4)} = \frac{10}{4} = \frac{5}{2} > 0$. So, $h(x)$ is positive in this interval.

Important Note: When choosing test values, make sure they fall strictly within the interval. Avoid using the critical points themselves, as these are the points where the function equals zero or is undefined, and we want to know the sign of the function within the interval, not at its boundary. This systematic approach ensures that we accurately determine the sign of h(x) across all the intervals.

4. Determining the Solution

We want to find where $h(x) \geq 0$. Based on our interval testing:

• $h(x) > 0$ on the intervals $(-2, 0)$ and $(1, \infty)$.
• $h(x) = 0$ at $x = 0$.

Therefore, the solution to $h(x) \geq 0$ is the union of these intervals, including the point where $h(x) = 0$.

The solution is $(-2, 0] \cup (1, \infty)$. Note the square bracket at 0, indicating that 0 is included in the solution because the inequality is non-strict (i.e., $\geq$ instead of $>$). The parentheses around -2 and 1 indicate that these values are not included because they make the denominator zero, and thus, h(x) is undefined at these points. Always remember to consider whether to include or exclude the critical points based on the inequality sign and whether the points make the function undefined. This careful consideration ensures you get the correct solution set for the inequality.

5. Final Answer

So, the solution to the inequality $h(x) = \frac{5x}{(x-1)(x+2)} \geq 0$ is:

$x \in (-2, 0] \cup (1, \infty)$

And that's it! We've successfully solved the inequality. Remember the key steps: find the critical points, create intervals, test each interval, and then determine the solution based on the sign of the function in each interval. Keep practicing, and you'll become a pro at solving these types of problems. Good job, guys!

Understanding how to solve inequalities, especially those involving rational functions, is crucial for various areas of mathematics and its applications. These skills are frequently used in calculus, optimization problems, and even in modeling real-world scenarios. By mastering this technique, you are not just solving a specific problem, but also building a strong foundation for tackling more complex mathematical challenges in the future. So, keep honing your skills and remember the systematic approach we've discussed, and you'll be well-equipped to handle any inequality that comes your way! Keep up the great work, and don't hesitate to revisit these steps whenever you encounter similar problems.