Demystifying F(x) = X(√x - 2)²: Derivative & Monotonicity

Hey there, math explorers! Ever looked at a function like f(x) = x(√x - 2)² and felt a tiny shiver of what on earth is this? Well, you're in the right place, because today, we're going to totally demystify it. Forget those boring, textbook explanations; we're breaking this down for real people, just like us, who want to understand the magic behind calculus. This isn't just about finding answers; it's about building your mathematical intuition and making these complex problems feel like a fun puzzle. We'll dive deep into function analysis, tackling everything from calculating its derivative—which, trust me, is super important for understanding a function's behavior—to figuring out where it's strictly increasing or decreasing. We'll specifically focus on proving that f'(x) = 2(√x - 2)(√x - 1) for x ∈ ℝ* and demonstrating why f is strictly increasing on the interval [0 ; 1]. And just for kicks, we’ll even compare f(x) to a simple x to see what kind of relationship they share. So, grab your favorite beverage, get comfy, and let's turn this seemingly complex function into something crystal clear. Ready to transform your math skills? Let's go!

Unpacking the Function: What is f(x) = x(√x - 2)² Anyway?

Alright, guys, before we jump into the heavy-duty calculus, let's take a moment to really understand the star of our show: f(x) = x(√x - 2)². At first glance, it might look a bit intimidating with that square root and the whole thing squared, but let's break it down piece by piece. First off, notice that √x part? That immediately tells us something crucial about our function's domain. Since you can't take the square root of a negative number in the real number system, our x values must be greater than or equal to zero. So, we're primarily concerned with x ∈ [0, +∞). This isn't just a minor detail; it's fundamental for every step we're about to take. Think of it like setting the rules for a game before you start playing!

Now, let's look at the structure: it's a product of two main components: x and (√x - 2)². This (√x - 2)² part is a binomial squared, which means (√x - 2) * (√x - 2). You could expand this out if you wanted to, to get (x - 4√x + 4), and then multiply by x to get f(x) = x² - 4x√x + 4x. Both forms are equivalent and might be useful depending on the specific task. For instance, f(x) = x² - 4x^(3/2) + 4x is a great form for simple power rule differentiation, but for this specific problem, the factored form x(√x - 2)² is actually a blessing in disguise because it leads directly to the derivative form we're asked to prove. This is where good mathematical strategy comes in—sometimes, simplifying isn't always expanding; sometimes, it's about keeping things in a form that makes the next step easier. We're looking at a function that combines polynomial-like behavior (the x terms) with radical behavior (the √x terms). Understanding these individual pieces helps us anticipate how the function will behave. When x is small, near zero, √x is also small. When x gets really big, √x also gets big, but slower than x. The (√x - 2) term will be negative for x < 4 and positive for x > 4. Squaring it always makes it non-negative. So, f(x) will always be non-negative for x ≥ 0! These initial observations, before even touching calculus, give us a powerful mental map of f(x). It's like checking the map before you start your road trip! This foundational understanding is key to not just solving the problem, but to truly mastering function analysis. We need to respect the domain, understand the components, and then, only then, unleash our calculus tools.

The Core Challenge: Proving the Derivative, f'(x) = 2(√x - 2)(√x - 1)

Alright, my friends, it's time to flex those calculus muscles! The first big task presented to us is to prove that for ∀ x ∈ ℝ* (meaning all positive real numbers), the derivative of our function f(x) is f'(x) = 2(√x - 2)(√x - 1). This isn't just about finding the derivative; it's about showing that our derivation matches a specific target. This is a common exercise in higher-level math, demanding precision and careful application of derivative rules. Remember, x ∈ ℝ* specifically excludes x = 0, which is important because √x has an undefined derivative at x = 0. So, when tackling derivatives involving square roots, always be mindful of that edge case.

Now, looking at f(x) = x(√x - 2)², what's the first rule that springs to mind? If you said the product rule, give yourself a pat on the back! We have u = x and v = (√x - 2)². The product rule states that (uv)' = u'v + uv'. But wait, there's a chain rule lurking inside that v term, too! That (√x - 2)² is a composite function, g(h(x)), where g(y) = y² and h(x) = √x - 2. So, v' = 2(√x - 2) * (derivative of √x - 2). And what's the derivative of √x? It's 1/(2√x). The derivative of -2 is simply 0. So, v' = 2(√x - 2) * (1/(2√x)). See how these rules nest together? It's like Russian dolls, each rule contained within another. This is where many students trip up, so paying close attention to these nested structures is crucial for success. We're not just mindlessly applying formulas; we're strategically breaking down the function into manageable parts.

Step-by-Step Differentiation of f(x)

Let's meticulously go through the differentiation process for f(x) = x(√x - 2)². This is where we put our product and chain rule knowledge into action, step by careful step, to ensure we arrive at the exact derivative we need to prove. It's like following a recipe—each ingredient and step matters!

First, identify our components for the product rule: Let u = x Then u' = 1 (that's the easy part, right?)

Now for v = (√x - 2)². This is where the chain rule comes into play. Let h(x) = √x - 2. The derivative of h(x), h'(x), is 1/(2√x) - 0 = 1/(2√x). Let g(y) = y². The derivative of g(y), g'(y), is 2y.

So, using the chain rule, v' = g'(h(x)) * h'(x) = 2(√x - 2) * (1/(2√x)). We can simplify v' a bit: v' = (√x - 2) / √x.

Now, let's plug these into the product rule formula: f'(x) = u'v + uv'. f'(x) = (1) * (√x - 2)² + (x) * ((√x - 2) / √x)

Let's simplify this expression carefully. We have (√x - 2)² in the first term, and x / √x simplifies to √x in the second term. So, f'(x) = (√x - 2)² + √x * (√x - 2)

Look at that! We have a common factor: (√x - 2). Let's factor it out, guys! This is a super smart move because it directly pushes us towards the target form. f'(x) = (√x - 2) * [ (√x - 2) + √x ]

Now, just simplify the terms inside the square brackets: f'(x) = (√x - 2) * [ √x - 2 + √x ] f'(x) = (√x - 2) * [ 2√x - 2 ]

We're almost there! Notice that 2√x - 2 has a common factor of 2. Let's pull that out! f'(x) = (√x - 2) * 2 * (√x - 1)

And voilà! Rearranging the terms to match the required format: f'(x) = 2(√x - 2)(√x - 1)

Boom! We did it! This proof wasn't just about mechanical differentiation; it was about strategic simplification and recognizing common factors. Each step builds on the previous one, and understanding why we apply each rule and when to simplify is what truly makes you a math wizard. This confirms our derivative for x ∈ ℝ*, exactly as the problem asked. This detailed breakdown ensures that every single person, regardless of their current math level, can follow along and build confidence in handling such problems. Remember, practice makes perfect, and understanding each piece of the puzzle is the real victory here. This rigorous process is fundamental for anyone looking to master calculus and understand the behavior of functions from their derivatives.

Journey to Monotonicity: Is f(x) Increasing or Decreasing?

Alright, team, we've successfully conquered the derivative! Now, let's use that powerful tool, f'(x) = 2(√x - 2)(√x - 1), to unravel another fascinating aspect of our function f(x): its monotonicity. In plain English, monotonicity tells us whether a function is generally going up (increasing) or going down (decreasing) over a specific interval. This is one of the coolest applications of derivatives, because it gives us direct insight into the function's shape without having to plot a single point! The golden rule here is simple:

• If f'(x) > 0 on an interval, then f(x) is strictly increasing on that interval.
• If f'(x) < 0 on an interval, then f(x) is strictly decreasing on that interval.
• If f'(x) = 0 at a point, we might have a local maximum, minimum, or a saddle point.

Our next mission, should we choose to accept it (and we do!), is to show that f(x) is strictly increasing on the interval [0 ; 1]. This means we need to prove that f'(x) > 0 for all x within that interval. Remember, the derivative f'(x) is only defined for x ∈ ℝ*, so for x = 0, we'll typically analyze the limit of the derivative or consider the behavior at 0 from the right. However, for "strictly increasing on [0, 1]", it generally means f'(x) > 0 for x ∈ (0, 1) and then checking the endpoint 0 separately or by continuity of f. Let's focus on x ∈ (0, 1). To do this, we need to analyze the signs of the factors in f'(x) = 2(√x - 2)(√x - 1). The constant 2 is positive, so we just need to worry about (√x - 2) and (√x - 1). This systematic approach to sign analysis is critical; it’s like being a detective, looking for clues to build a solid case. We're not guessing here; we're deducing based on mathematical facts. So, let's roll up our sleeves and break down these factors over our target interval. Understanding how each piece of the derivative contributes to the overall sign is what gives us the power to predict the function's behavior. This kind of detailed analysis is a hallmark of strong mathematical reasoning and moves us beyond just getting the right answer to truly understanding why the answer is right. It's a skill that pays dividends in all areas of quantitative analysis.

Analyzing f'(x) for Signs on [0, 1]

Okay, folks, let's zoom in on the interval [0 ; 1] and dissect our derivative: f'(x) = 2(√x - 2)(√x - 1). Our goal is to prove that f(x) is strictly increasing on this interval, which means we need f'(x) > 0 for x ∈ (0, 1). We'll also implicitly check the endpoint behaviors.

First, let's consider the individual factors for any x in the interval (0, 1).

Factor 1: (√x - 2)

If x is in the interval (0, 1), what can we say about √x? Well, if 0 < x < 1, then 0 < √x < 1. Think about it: √0 = 0, √1 = 1, and for any number between 0 and 1 (like 0.25), its square root (0.5) is also between 0 and 1. Now, let's subtract 2 from √x: 0 < √x < 1 Subtract 2 from all parts of the inequality: 0 - 2 < √x - 2 < 1 - 2 -2 < √x - 2 < -1

So, for x ∈ (0, 1), the term (√x - 2) is always negative! It's actually quite strongly negative, falling between -2 and -1. That's a crucial piece of information.

Factor 2: (√x - 1)

Let's do the same for the second factor, (√x - 1). Again, for x ∈ (0, 1), we know 0 < √x < 1. Now, subtract 1 from √x: 0 - 1 < √x - 1 < 1 - 1 -1 < √x - 1 < 0

Aha! For x ∈ (0, 1), the term (√x - 1) is also always negative! It's a small negative number, but negative nonetheless.

Putting it all together for f'(x):

We have f'(x) = 2 * (negative number) * (negative number). The 2 is a positive constant. A negative number multiplied by another negative number results in a positive number. So, f'(x) = (positive constant) * (positive result) = a positive number!

Therefore, for all x ∈ (0, 1), we have f'(x) > 0. This is exactly what we needed to show!

What about x = 0? At x = 0, f(x) = 0(√0 - 2)² = 0. The derivative f'(x) as we calculated it is not strictly defined at x = 0 because of 1/(2√x). However, the function f(x) is continuous at x=0. To assess strict monotonicity at the endpoint, we examine the behavior as x approaches 0 from the right. Since f'(x) > 0 for x ∈ (0, 1), and f(x) is continuous on [0, 1], we can confidently conclude that f(x) is indeed strictly increasing on the entire interval [0 ; 1].

So, guys, we've demonstrated that f'(x) is positive throughout (0, 1). This means that as x moves from 0 to 1, the value of f(x) is consistently going up. How cool is that? We've painted a clear picture of the function's behavior in this critical interval, all thanks to our derivative analysis. This deep dive into sign analysis is a cornerstone of understanding function behavior, not just for calculus exams, but for real-world applications where predicting trends is paramount. Mastering this technique truly empowers you to "see" how a function behaves without even needing a graph! This methodical approach not only solves the problem but also fortifies your understanding of core calculus principles, making you more confident in tackling future challenges.

Beyond Monotonicity: Comparing f(x) with x

Alright, my awesome math enthusiasts, we've differentiated f(x) and analyzed its monotonicity on [0, 1]. Now, let's tackle the third and final part of our problem: proving the identity f(x) - x = x(√x - 1)(√x). This isn't just a random algebraic manipulation; it's a fantastic way to compare our original function f(x) with the simpler function g(x) = x. Understanding the difference f(x) - x can tell us a lot about whether f(x) is greater than, less than, or equal to x over different intervals. This kind of comparison is super useful in various mathematical contexts, from inequality proofs to understanding how one function deviates from a simpler baseline. It's like asking: "How much 'extra' stuff does f(x) have compared to just x?" The beauty of this specific problem is that it gives us a target factored form for f(x) - x, which is a huge hint. Instead of just expanding everything and hoping for the best, we can strategically work towards that factored expression. This requires a bit of algebraic finesse and an eye for common factors, which are skills that truly elevate your mathematical game.

Remember our initial function: f(x) = x(√x - 2)². Our goal is to show that f(x) - x can be written as x(√x - 1)(√x). The easiest path here is often to start with f(x) - x and perform algebraic manipulations until we reach the desired factored form. Sometimes, it might be tempting to start from the right-hand side and expand, but working from f(x) - x usually feels more direct when f(x) itself is already somewhat structured. Let's make sure we're careful with our square roots and binomial expansions. This step reinforces your algebraic foundations, proving that calculus isn't just about derivatives and integrals; it's built upon a solid understanding of fundamental algebra. Being comfortable with these manipulations will make your calculus journey much smoother, allowing you to focus on the conceptual aspects without getting bogged down by the algebra. It's truly about seeing the forest for the trees, understanding how each part connects to the whole, and simplifying expressions strategically to reveal hidden relationships.

Proving the Identity: f(x) - x = x(√x - 1)(√x) (and Addressing a Curious Twist!)

Alright, math heroes, for our grand finale, we're challenged to prove the identity f(x) - x = x(√x - 1)(√x). This kind of proof is super valuable because it helps us understand the relationship between our complex function f(x) and the simpler linear function x. Is f(x) generally bigger than x, smaller than x, or do they cross paths? This identity, if proven, would give us direct insight.

Let's begin by taking our function f(x) = x(√x - 2)² and subtracting x from it. Our goal is to manipulate this expression algebraically until it matches the right-hand side, x(√x - 1)(√x).

Start with the left-hand side: f(x) - x = x(√x - 2)² - x

The very first thing you should spot, guys, is that x is a common factor in both terms! Let's pull it out; this is a power move that often simplifies things dramatically. f(x) - x = x [ (√x - 2)² - 1 ]

Now, look closely at the expression inside the square brackets: (√x - 2)² - 1. Does it ring a bell? It should! This is a perfect example of the difference of squares formula: a² - b² = (a - b)(a + b). In our case, a = (√x - 2) and b = 1 (because 1 is 1²). Recognizing these algebraic patterns is key to efficient problem-solving.

Applying the difference of squares: a - b = (√x - 2) - 1 = √x - 3 a + b = (√x - 2) + 1 = √x - 1

So, substituting these back into our expression, we get: f(x) - x = x (√x - 3)(√x - 1)

Now, hold on a second! This is where things get interesting, and a little bit tricky. The problem asked us to prove f(x) - x = x(√x - 1)(√x). But our meticulous, step-by-step derivation, using correct algebraic principles, has led us to x(√x - 1)(√x - 3). Notice the difference? The factor (√x - 3) is what we derived, but the problem statement's final factor is (√x).

This is a super important moment in math. It means that, based on our exact function f(x) = x(√x - 2)², the identity as stated in the problem, f(x) - x = x(√x - 1)(√x), is not mathematically true. The correct identity for f(x) - x is actually x(√x - 1)(√x - 3).

What does this teach us?

1. Always verify: Even in a problem statement, there can be typos or slight inaccuracies. Your job as a rigorous mathematician (or aspiring one!) is to follow the logic and verify.
2. Trust your process: If you've applied all the rules correctly, step by step, and your result doesn't match, it's more likely that the target is incorrect than your method.
3. Value of detailed steps: By showing every single step, we can pinpoint exactly where a discrepancy might lie. If we had just tried to force (√x) into our derivation, we would have made an algebraic error.

So, while we've proven an identity for f(x) - x, it's not the one precisely stated by the problem in its fragmented form. We've shown that f(x) - x = x(√x - 1)(√x - 3). This accurate derivation provides immensely more value than a forced, incorrect proof. It shows critical thinking and respect for mathematical truth. This deep dive not only fulfills the request to analyze f(x) - x but also offers a powerful lesson in mathematical integrity and the importance of verifying every detail, making you a more astute problem-solver!

Wrapping It Up: Key Takeaways and Your Math Superpowers

And there you have it, fellow math enthusiasts! We’ve journeyed through the intricate world of function analysis, starting with a seemingly complex f(x) = x(√x - 2)², and systematically broken it down into manageable, understandable pieces. We didn't just solve problems; we unlocked the secrets hidden within this function, transforming it from a intimidating expression into a clear example of calculus in action.

What have we accomplished?

• We mastered the art of differentiation by successfully proving that f'(x) = 2(√x - 2)(√x - 1) for all x ∈ ℝ*. This wasn't just about applying rules; it was about strategically combining the product rule and the chain rule, and then using clever algebraic factorization to reach our target form. This skill is foundational, allowing you to understand the rate of change for virtually any function you encounter.
• We delved into the function's behavior by determining its monotonicity. By meticulously analyzing the signs of the factors in f'(x), we definitively showed that f(x) is strictly increasing on the interval [0 ; 1]. This ability to predict whether a function is rising or falling without even seeing its graph is a true mathematical superpower, incredibly useful in fields like physics, economics, and engineering.
• We tackled a challenge that tested not just our algebra, but our critical thinking. While working on f(x) - x, we uncovered a subtle but important discrepancy in the problem's stated identity. Instead of forcing an incorrect answer, we diligently followed our algebraic steps and correctly derived f(x) - x = x(√x - 1)(√x - 3). This taught us a valuable lesson: always trust your process and verify every assumption. This level of mathematical integrity is what truly sets apart a diligent student from someone just looking for quick answers. It's about building a robust understanding and developing an analytical mindset.

So, guys, what's next? The best way to solidify these newfound superpowers is to practice, practice, practice! Find similar functions, try to differentiate them, analyze their monotonicity, and play around with their algebraic properties. Each problem you tackle is a chance to sharpen your skills and deepen your intuition. Remember, math isn't just about memorizing formulas; it's about understanding concepts, developing logical reasoning, and learning to approach complex problems with confidence and a systematic mindset. You've got this! Keep exploring, keep questioning, and keep growing your amazing mathematical abilities. The world of functions is vast and full of wonders, and you're now better equipped than ever to explore it. Happy calculating!