Unlocking Rational Inequalities: When Is $h(x) \geq 0$?

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Hey there, math explorers! Ever looked at a complex fraction involving polynomials and thought, "Ugh, where do I even begin?" Well, guess what? You're not alone! Today, we're diving deep into a super interesting problem that deals with rational inequalities. We're going to break down how to figure out when a function like $h(x) = \frac{f(x)}{g(x)}$ is greater than or equal to zero. This isn't just about getting the right answer; it's about understanding the why and how behind it, building a rock-solid foundation for future math adventures. So, grab your favorite beverage, get comfy, and let's unravel this mystery step-by-step. By the end of this, you'll feel like a pro, I promise!

This journey into rational inequalities is incredibly valuable, not just for passing your next math test, but for developing a type of logical thinking that's useful in so many aspects of life. We'll be tackling a specific problem involving $f(x)=x^3-9x$ and $g(x)=x^2-x-6$, and our goal is to find all the intervals where $h(x)=\frac{f(x)}{g(x)}$ is non-negative, meaning $h(x) \geq 0$. This might sound a bit intimidating at first glance, but trust me, by meticulously breaking it down, we'll see that it's actually quite manageable and even fun once you get the hang of it. We're going to learn about factoring polynomials, identifying critical points, understanding function domains, and using a super effective technique called sign analysis. These skills are absolutely fundamental in algebra and calculus, opening doors to understanding more complex mathematical models. So, let's gear up and get ready to become masters of rational inequalities!

Understanding the Building Blocks: $f(x)$ and $g(x)$

Before we can even think about $h(x)$, we need to get intimately familiar with its components: $f(x)$ and $g(x)$. Think of them as the individual Lego bricks we're using to build something bigger. Understanding each piece individually makes the whole construction much easier. Our main keywords here are polynomial factoring and function analysis, as these are the first crucial steps. When dealing with rational functions and inequalities, the absolute first thing you should always do is factorize your polynomials completely. Why? Because factoring reveals the roots or zeros of the polynomials, which are incredibly important points on our number line. These points are where the function might change its sign, and they'll be key to solving our inequality.

Factoring $f(x)$: The Numerator's Story

Let's start with $f(x) = x^3 - 9x$. This is a cubic polynomial, but don't let the "cubic" part scare you! The very first rule of factoring any polynomial is to look for a greatest common factor (GCF). Do you see one here? Absolutely! Both terms, $x^3$ and $-9x$, share an $x$. So, we can factor out $x$ from both terms: $f(x) = x(x^2 - 9)$. Now, take a closer look at what's inside the parentheses: $x^2 - 9$. Does that look familiar? It should! That's a classic example of a difference of squares. Remember the formula: $a^2 - b^2 = (a-b)(a+b)$. In our case, $a=x$ and $b=3$. So, $x^2 - 9$ factors into $(x-3)(x+3)$. Putting it all together, the fully factored form of $f(x)$ is $f(x) = x(x-3)(x+3)$. This factored form is gold, guys! It immediately tells us the zeros of $f(x)$ are $x=0$, $x=3$, and $x=-3$. These are the points where the numerator is zero, which means $h(x)$ itself could be zero at these points (as long as the denominator isn't also zero).

Factoring polynomials isn't just a math exercise; it's a fundamental skill that unlocks deeper understanding. By breaking $f(x)$ down into its simplest multiplicative components, we're essentially finding the DNA of the function. Each factor, $(x)$, $(x-3)$, and $(x+3)$, represents a linear term, and the zeros of these terms (0, 3, and -3 respectively) are crucial for our sign analysis later. They are the exact points where the sign of $f(x)$ could flip from positive to negative, or vice versa. If we hadn't factored, trying to determine when $x^3 - 9x$ is positive or negative would be a much more arduous task, likely involving test points and potentially graphing, which is less precise for inequalities. This systematic approach ensures we don't miss any critical turning points. Always remember, a fully factored polynomial is your best friend when tackling inequalities!

Factoring $g(x)$: The Denominator's Role

Next up, we have $g(x) = x^2 - x - 6$. This is a quadratic polynomial, and our goal is to factor it into two linear expressions. For a quadratic in the form $ax^2 + bx + c$, we're looking for two numbers that multiply to $c$ (which is -6 in our case) and add up to $b$ (which is -1). Let's list pairs of factors for -6: (1, -6), (-1, 6), (2, -3), (-2, 3). Which pair adds up to -1? Bingo! It's 2 and -3. So, $g(x)$ factors into $(x-3)(x+2)$. Just like with $f(x)$, this factored form is super important because it tells us the zeros of $g(x)$. Here, the zeros are $x=3$ and $x=-2$. Why are these particular zeros so critical? Because these are the values of $x$ for which the denominator becomes zero, and division by zero is a big no-no in mathematics! This means that $x=3$ and $x=-2$ are points where $h(x)$ is undefined. We need to keep these values firmly in mind as we move forward, as they will define the boundaries of our solution intervals.

Understanding the denominator's zeros is arguably even more critical than understanding the numerator's zeros when it comes to rational functions. While $h(x)$ can be zero when its numerator is zero, it can never exist where its denominator is zero. These points represent vertical asymptotes or holes in the graph of $h(x)$, and they must be excluded from our solution set, regardless of whether the inequality includes "equal to" (i.e., $\geq$ or $\leq$). Missing this step is a common pitfall, and it leads to incorrect intervals. So, remember: the zeros of the denominator are always excluded. We're building a complete picture here, and every piece of information, especially these exclusion points, helps us accurately sketch that picture. This careful consideration of the domain is what truly distinguishes mastering rational functions from just solving equations.

Defining Our Hero: The Rational Function $h(x)$

Alright, guys, now that we've got $f(x)$ and $g(x)$ all neatly factored, it's time to put them together to form our main event: $h(x)$. This is where the magic of rational functions really starts to unfold. A rational function, by definition, is a function that can be written as the ratio of two polynomials. In our case, $h(x) = \frac{f(x)}{g(x)}$. But simply writing it out isn't enough; we need to carefully consider its properties, especially its domain and how it might simplify.

Setting Up $h(x)$

Using our factored forms, we can write $h(x)$ as:

$h(x) = \frac{x(x-3)(x+3)}{(x-3)(x+2)}$

Look at that! It's much cleaner than the original polynomial mess, right? This factored form immediately highlights something super interesting. Do you see any common factors in the numerator and the denominator? Absolutely! Both have an $(x-3)$ term. This tells us a lot about the behavior of $h(x)$ near $x=3$. While this common factor might tempt us to immediately cancel it out, we need to be incredibly careful. Canceling it prematurely without considering the original domain would be a classic blunder that many students make. This is where understanding the concept of domain becomes paramount.

Crucial Step: Identifying the Domain (Excluding Values)

Before we do any simplification, we must determine the domain of the original function $h(x)$. The domain of any rational function is all real numbers except for those values of $x$ that make the denominator equal to zero. From our factored $g(x) = (x-3)(x+2)$, we know that $g(x)=0$ when $x=3$ or $x=-2$. Therefore, $h(x)$ is undefined at $x=3$ and $x=-2$. These points must be excluded from our final solution, no matter what the simplified form looks like or what the inequality sign is. Even if our inequality were $h(x) > 0$, these points would still be excluded because $h(x)$ simply doesn't exist there. This is a non-negotiable rule when working with rational functions, and it's a common source of error if overlooked. Ignoring these crucial points can lead to including values in your solution that make the original function invalid, which is a major no-no in mathematics. This rigorous attention to detail is what sets apart a thorough understanding from a superficial one, ensuring our solution is mathematically sound.

Simplifying $h(x)$ (and why we can do it with caveats)

Now that we've identified and mentally (or physically!) marked our exclusion points, we can simplify $h(x)$. Since $(x-3)$ is a common factor in both the numerator and the denominator, we can cancel it out, provided that $x \neq 3$. So, for all $x$ where $x \neq 3$ and $x \neq -2$, our function simplifies to:

$h(x) = \frac{x(x+3)}{x+2}$

This simplified form is much easier to work with for our sign analysis. The fact that $(x-3)$ canceled out means there's a hole in the graph of $h(x)$ at $x=3$, not a vertical asymptote. If it hadn't canceled, $x=3$ would have been a vertical asymptote. The behavior around $x=-2$ (where $(x+2)$ is still in the denominator) indicates a vertical asymptote there. Understanding the difference between a hole and a vertical asymptote is key for sketching graphs of rational functions, but for inequalities, the main takeaway is that both are points where the function is undefined. The simplification step makes the subsequent interval testing much less computationally intensive because we're working with simpler expressions. However, the caveat about the domain ($x \neq 3, x \neq -2$) is paramount. Never forget to carry those exclusions forward, even after simplification. They are permanent scars on the function's domain, defining where our hero $h(x)$ simply cannot exist. This entire process of careful domain analysis and simplification is a testament to the methodical approach required for mastering rational function problems, allowing us to accurately predict and describe the function's behavior across the entire real number line, except for those critical exclusion points.

The Core Challenge: Solving $h(x) \geq 0$

Alright, squad! This is where all our hard work on factoring and understanding the domain comes together. Our mission now is to find all values of $x$ for which $h(x) \geq 0$. This means we're looking for intervals where $h(x)$ is either positive or zero. The main tool we'll use here is sign analysis, a powerful technique that helps us systematically determine the sign of a function across different intervals. The first step in sign analysis is identifying all the critical points. These are the points on the number line where the function might potentially change its sign. For a rational function, critical points come from two sources: the zeros of the numerator and the zeros of the denominator. We've already done the legwork, so let's list them out.

Finding Critical Points (Zeros and Undefined Points)

From our factored $h(x) = \frac{x(x+3)}{x+2}$ (with the understanding that $x \neq 3$ from the original function), our critical points are:

• Zeros of the numerator: These are the values of $x$ that make $x(x+3) = 0$. So, $x=0$ and $x=-3$.
• Zeros of the denominator: These are the values of $x$ that make $x+2 = 0$. So, $x=-2$.
• Original function's undefined points: Remember $x=3$ from the original $g(x)$? Even though it simplified out, it's still a point where $h(x)$ is undefined. So, $x=3$ is also a critical point for setting up our intervals.

So, our complete set of critical points, in increasing order, is: $-3$, $-2$, $0$, $3$. These critical points divide the number line into distinct intervals. It's like drawing lines in the sand, creating separate regions where the function's sign will remain consistent. The magic of sign analysis is that within each of these intervals, the sign of $h(x)$ (either positive or negative) will not change. This means we only need to pick one test point from each interval to determine the sign for the entire interval. This saves us a ton of work compared to checking every single point!

This method of identifying critical points is absolutely vital for correctly solving any inequality, whether it's polynomial or rational. Each critical point acts as a potential 'sign-flip' location. By meticulously finding all zeros of the numerator and denominator, we ensure that we haven't missed any boundary where the function's behavior might transition from being positive to negative, or vice versa. The inclusion of the point $x=3$, even though it led to a removable discontinuity (a hole), is a testament to the importance of always referring back to the original form of the function when establishing its domain. These are the details that separate a robust, accurate solution from one that might look correct but is fundamentally flawed. Keeping these points straight ensures that our subsequent analysis on the number line is built on a solid, unwavering foundation, ready for accurate interval testing and precise conclusions about where $h(x)$ is truly non-negative.

The Magic of the Number Line: Interval Analysis

Now, let's set up our number line using these critical points: $-\infty \dots -3 \dots -2 \dots 0 \dots 3 \dots +\infty$. These points create five distinct intervals. For each interval, we'll pick a simple test value, plug it into our simplified $h(x) = \frac{x(x+3)}{x+2}$, and determine the sign. Remember, we're looking for $h(x) \geq 0$.

Testing Interval 1: $(-\infty, -3)$

Let's pick $x = -4$. This is a pretty straightforward choice and well within our interval. Now, plug it into $h(x)$:

$h(-4) = \frac{(-4)(-4+3)}{-4+2} = \frac{(-4)(-1)}{-2} = \frac{4}{-2} = -2$

Since $-2$ is negative, $h(x) < 0$ in this entire interval. So, $(-\infty, -3)$ is not part of our solution.

Testing Interval 2: $[-3, -2)$

Let's choose $x = -2.5$. This value sits comfortably between $-3$ and $-2$. Let's see what happens:

$h(-2.5) = \frac{(-2.5)(-2.5+3)}{-2.5+2} = \frac{(-2.5)(0.5)}{-0.5} = \frac{-1.25}{-0.5} = 2.5$

Since $2.5$ is positive, $h(x) > 0$ in this interval. Because our inequality is $h(x) \geq 0$, and $x=-3$ makes the numerator zero (so $h(-3)=0$), we include $-3$. However, $x=-2$ makes the denominator zero, so it must be excluded. Therefore, this interval contributes $[-3, -2)$ to our solution. This is a crucial distinction: points where the numerator is zero are included if the inequality has an "equal to" component, but points where the denominator is zero are always excluded, full stop. This small detail can make or break your final answer, so always pay close attention to the open or closed brackets!

Testing Interval 3: $(-2, 0)$

Let's pick $x = -1$. Easy peasy! Plug it into $h(x)$:

$h(-1) = \frac{(-1)(-1+3)}{-1+2} = \frac{(-1)(2)}{1} = -2$

Since $-2$ is negative, $h(x) < 0$ in this interval. Thus, $(-2, 0)$ is not part of our solution. This emphasizes that even if an interval is bordered by critical points, it doesn't automatically mean it's part of the solution; you always have to test it!

Testing Interval 4: $[0, 3)$

For this interval, let's go with $x = 1$. It's a nice, simple positive number. Let's calculate $h(1)$:

$h(1) = \frac{(1)(1+3)}{1+2} = \frac{1(4)}{3} = \frac{4}{3}$

Since $\frac{4}{3}$ is positive, $h(x) > 0$ in this interval. Since $x=0$ makes the numerator zero ($h(0)=0$), and $0 \geq 0$ is true, we include $0$. However, remember that $x=3$ was an original excluded value because it made $g(x)=0$. So, we exclude $3$. This interval therefore contributes $[0, 3)$ to our solution. This interval testing process is the heart of solving rational inequalities. Each test point acts as a representative for its entire interval, saving us from checking infinite values. The careful notation with square and round brackets is essential to precisely define the solution set. Square brackets [ or ] mean the endpoint is included, typically when the function is zero at that point and the inequality allows for equality. Round brackets ( or ) mean the endpoint is excluded, either because the function is undefined there or the inequality is strictly less than/greater than. Mastering this notation ensures your answer is not only correct but also mathematically rigorous.

Testing Interval 5: $(3, +\infty)$

Let's pick a value like $x = 4$. This is our final interval to check! Substituting $x=4$ into $h(x)$:

$h(4) = \frac{(4)(4+3)}{4+2} = \frac{4(7)}{6} = \frac{28}{6} = \frac{14}{3}$

Since $\frac{14}{3}$ is positive, $h(x) > 0$ in this interval. Since $x=3$ is an excluded value, we use a round bracket. Thus, this interval contributes $(3, +\infty)$ to our solution. This final test point confirms the behavior of our rational function at the far right of the number line. The positive result here, coupled with the exclusion of $x=3$, gives us a complete picture of where $h(x)$ is non-negative. This thorough, systematic approach, moving from factoring to critical points, then to detailed interval testing, is the gold standard for solving such problems. It minimizes errors and builds a deep understanding of the function's behavior across its entire domain, preparing you for even more complex mathematical challenges. Remember, every step is important, and none should be skipped if you want to arrive at a perfectly accurate and defensible solution. This diligent sign analysis, where each test point validates the behavior of an entire segment of the number line, is the cornerstone of mastering rational inequalities.

Putting It All Together: The Solution Intervals

Awesome work, guys! We've systematically analyzed every single interval. Now, all that's left is to combine the intervals where $h(x) \geq 0$. Looking back at our tests, the intervals where $h(x)$ was positive or zero were:

• $[-3, -2)$ (from Test Interval 2)
• $[0, 3)$ (from Test Interval 4)
• $(3, +\infty)$ (from Test Interval 5)

So, the complete solution, representing all the intervals on which $h(x) \geq 0$, is the union of these intervals:

$\boxed{[-3, -2) \cup [0, 3) \cup (3, +\infty)}$

This is precisely the answer we were aiming for! Each component of this solution carries specific meaning. The square brackets at $-3$ and $0$ indicate that these points are included because $h(x)$ equals zero at these points, satisfying the "equal to" part of our inequality $h(x) \geq 0$. Conversely, the round brackets at $-2$ and $3$ signify that these points are excluded. $x=-2$ is excluded because it makes the denominator of $h(x)$ zero, rendering the function undefined (a vertical asymptote). $x=3$ is excluded because, although it was a removable discontinuity (a hole) after simplification, it still makes the original denominator zero, meaning $h(x)$ is undefined there. The use of the union symbol ($\cup$) is crucial as it correctly combines these distinct, non-overlapping intervals into a single, coherent solution set. This final step is a powerful demonstration of how logical deduction and meticulous calculation come together to solve complex mathematical problems. It's a moment of triumph where all your hard work on factoring, domain analysis, and interval testing culminates into a precise and accurate representation of the function's behavior. This mastery of interval notation and the underlying principles is what truly solidifies your understanding of rational inequalities, making you ready for more advanced challenges in your mathematical journey. This solution is not just an answer; it's a narrative of the function's life on the number line.

Why This Matters: Real-World Connections and Beyond

"Okay, great, I solved a math problem. But why should I care?" That's a fair question, and it's one I hear a lot! The truth is, guys, understanding rational inequalities goes far beyond just getting a good grade in your algebra class. This isn't just abstract math; it's a fundamental concept that underpins a huge range of real-world applications and more advanced mathematical fields. Think about it: whenever you're dealing with ratios, rates, or relationships where one quantity depends on another in a fractional way, you're essentially working with rational functions. For instance, in economics, rational functions can model cost-benefit analyses, where increasing production might initially lower average costs (making $h(x)$ positive) but eventually lead to diminishing returns or increased marginal costs (making $h(x)$ negative or undefined at certain production levels). Businesses use this kind of analysis to find optimal production quantities to maximize profits or minimize costs.

In engineering, particularly in fields like electrical engineering or control systems, rational functions are used to describe the transfer functions of circuits and systems. Analyzing when these functions are positive, negative, or undefined can tell engineers about the stability of a system, its response to different inputs, or even potential points of failure. Imagine designing a bridge: you'd want to know under what conditions (e.g., wind speed, load weight) a certain stress ratio remains within acceptable, non-negative limits, or when it becomes critically high (undefined or negative in a structural sense). Similarly, in physics, ratios of forces, velocities, or concentrations often lead to rational functions. Understanding their inequalities allows scientists to predict behavior, like when a reaction rate is favorable or when a projectile's altitude is above a certain threshold.

Beyond these direct applications, the skills you honed today – polynomial factoring, domain analysis, critical point identification, and sign analysis – are incredibly versatile. They build your logical reasoning capabilities, teach you methodical problem-solving, and prepare you for higher-level mathematics like calculus, where these concepts are expanded upon significantly. For example, finding where a function's derivative (often a rational function itself) is positive or negative tells you where the original function is increasing or decreasing. Understanding discontinuities, like the holes and asymptotes we discussed, is crucial for graphing and limits. So, every time you tackle a problem like this, you're not just solving for $x$; you're sharpening your mind, equipping yourself with powerful analytical tools, and laying the groundwork for understanding the complex world around us. It's truly empowering stuff!

Wrapping Up: Your Journey to Mastering Rational Inequalities

Wow, what a journey we've had today! We started with what looked like a pretty daunting problem, $h(x) = \frac{x^3-9x}{x^2-x-6}$, and systematically broke it down to find all the intervals where $h(x) \geq 0$. We leveraged the power of polynomial factoring to simplify both the numerator and the denominator, transforming $f(x)$ into $x(x-3)(x+3)$ and $g(x)$ into $(x-3)(x+2)$. This crucial first step revealed the zeros of each polynomial, which are the building blocks for our subsequent analysis. We then moved on to the absolutely critical step of domain analysis, identifying the values of $x$ (specifically $x=-2$ and $x=3$) that would make the original denominator zero, thus rendering $h(x)$ undefined. Remember, these points are always excluded from the solution set, regardless of the inequality type.

After simplifying $h(x)$ to $\frac{x(x+3)}{x+2}$ (while keeping those exclusion points in mind!), we embarked on our sign analysis. This involved identifying all critical points from the simplified numerator's zeros ($x=0, x=-3$) and the denominator's zero ($x=-2$), along with the original exclusion $x=3$. These points segmented our number line into distinct intervals: $(-\infty, -3)$, $(-3, -2)$, $(-2, 0)$, $(0, 3)$, and $(3, +\infty)$. We then methodically picked a test point from each interval, plugged it into our simplified $h(x)$, and determined the sign of the function in that entire interval. This meticulous testing led us to conclude that $h(x) \geq 0$ in the intervals $[-3, -2)$, $[0, 3)$, and $(3, +\infty)$. The final step was to combine these findings using the union symbol, giving us the elegant solution: $[-3, -2) \cup [0, 3) \cup (3, +\infty)$.

This entire process isn't just about memorizing steps; it's about developing a robust understanding of how functions behave and how different mathematical concepts interconnect. You've gained invaluable skills in polynomial manipulation, domain restrictions, and interval analysis – tools that will serve you well in countless mathematical and scientific contexts. So, next time you see a tricky rational inequality, don't sweat it! Just remember the systematic approach we took today, and you'll be able to conquer it with confidence. Keep practicing, keep exploring, and remember that every problem you solve makes you a stronger, more capable mathematician. You've totally got this!