Unraveling Sequences: Function F(x) Analysis & Interval Proof

Hey There, Math Enthusiasts! Understanding Our Adventure

Alright, guys, let's dive headfirst into some really cool mathematics! Today, we're not just looking at any math problem; we're going to break down a fascinating scenario involving a numerical sequence and an intriguing function. If you've ever wondered how these abstract concepts connect to real-world patterns or just love the thrill of a good mathematical puzzle, you're in the right place. We're talking about a sequence, let's call it U_n, that starts at U_0 = 0, and then each subsequent term is defined by a recursive relationship: U{n+1} = f(U_n)_. The star of our show, the function f(x), is defined as f(x) = (1 + x) / √(3 + x²). Sounds a bit intimidating with all those symbols, right? But trust me, by the end of this, you'll see just how elegant and logical it all is. Our mission, should we choose to accept it, involves two main objectives: first, we'll construct the variation table for f(x), which is like creating a detailed map of how the function behaves. This will tell us where it goes up, where it goes down, and any crucial turning points. Second, we'll prove something super neat: that if you take any number in the interval [0, 1] and plug it into f(x), the result will still be within [0, 1]. This concept of an invariant interval is incredibly powerful, especially when we start thinking about the long-term behavior of our sequence U_n. Understanding the full behavior of such a function, especially its variations and how it maps specific intervals, is absolutely fundamental in higher-level mathematics and has direct implications in fields like mathematical modeling, numerical analysis, and even computer science algorithms. So, buckle up, because we're about to embark on a journey of function analysis that’s not just about getting the right answer, but truly understanding the why and how behind it all. This entire exploration will give us a robust foundation for predicting the long-term behavior of the sequence U_n, a critical aspect of dynamic systems.

Diving Deep into f(x) = (1 + x) / √(3 + x²): A Full Breakdown

Alright, team, let's get down to business with our function, f(x) = (1 + x) / √(3 + x²). Before we jump into its variations, we need to lay some groundwork. Think of it like preparing to build a house: you can't just start with the roof! You need a solid foundation. Here, our foundation involves understanding the function's domain, checking its continuity, and peeking at its behavior at the extremes (what we call limits at infinity). This isn't just busywork; these initial steps are crucial for making sure our subsequent analysis, especially the derivative calculations, makes sense and is applied correctly. For instance, knowing the domain ensures we don't try to plug in values where the function isn't even defined. Continuity tells us if the function's graph is a smooth line or if it has any sudden jumps or breaks. And limits at infinity give us a sense of the function's global trend, telling us where it's heading as x gets really, really big (or really, really small). These preliminary insights are absolutely vital for drawing an accurate variation table later on. They provide the boundaries and overall shape within which our more detailed analysis will fit. Without this groundwork, any conclusions we draw about the function's increasing or decreasing nature might be incomplete or, worse, entirely incorrect. So, let's roll up our sleeves and tackle these foundational aspects with the care they deserve, ensuring we have a robust understanding before we move on to the more complex calculations. This comprehensive approach is what truly differentiates a superficial understanding from a deep, insightful mastery of function analysis, making our journey into the specifics of f(x) both thorough and reliable. Moreover, for any mathematical proof or analytical task, presenting this initial context demonstrates a complete understanding of the function's inherent properties and constraints.

First Steps: Domain, Continuity, and Limits

Okay, guys, let's start with the absolute basics for f(x) = (1 + x) / √(3 + x²). The very first thing we always want to figure out is the domain of the function. This basically means: what values of x can we plug into this function and get a valid output? When you look at f(x), you see two main potential troublemakers: a fraction (where the denominator can't be zero) and a square root (where the term inside can't be negative). Let's tackle the square root first: we have √(3 + x²). For this to be defined in real numbers, the expression inside the square root, (3 + x²), must be greater than or equal to zero. Well, x² is always greater than or equal to zero for any real number x. Add 3 to it, and (3 + x²) will always be strictly greater than 3 (since x² ≥ 0 implies 3 + x² ≥ 3). So, (3 + x²) is always positive, meaning √(3 + x²) is always a positive real number. Great! No issues there. Now, for the denominator not being zero: since √(3 + x²) is always strictly positive (it's always at least √3), it will never be zero. This is awesome because it means we don't have to exclude any values of x from our domain. Therefore, the domain of f(x) is all real numbers, denoted as ℝ or (-∞, +∞). This is fantastic news because it simplifies our analysis; we don't have to worry about holes or breaks due to an undefined denominator.

Next up, continuity. Since f(x) is a composition, sum, and quotient of elementary functions (polynomials, square roots), and its denominator is never zero, f(x) is continuous over its entire domain, which is ℝ. This means its graph is a smooth, unbroken curve, which is super helpful for visualizing its behavior. You won't find any sudden jumps or gaps, making our derivative calculations much more reliable.

Finally, let's check the limits at infinity. What happens to f(x) as x gets super, super large (approaches +∞) or super, super small (approaches -∞)?

• As x → +∞: f(x) = (1 + x) / √(3 + x²). For large x, the +1 in the numerator and the +3 in the denominator become negligible. So, f(x) behaves roughly like x / √(x²). Since x → +∞, √(x²) = x. So, lim (x→+∞) f(x) = lim (x→+∞) x / x = 1.

• As x → -∞: f(x) = (1 + x) / √(3 + x²). Again, for very negative x, it behaves like x / √(x²). However, when x → -∞, √(x²) = |x| = -x (because x is negative). So, we have a sign flip! lim (x→-∞) f(x) = lim (x→-∞) x / (-x) = -1.

So, as x goes to positive infinity, f(x) approaches 1 from below, and as x goes to negative infinity, f(x) approaches -1 from above. These horizontal asymptotes at y = 1 and y = -1 give us crucial boundary information for our function's overall shape. This fundamental analysis prepares us perfectly for the next step: dissecting the function's rate of change through its derivative. Knowing the function lives between -1 and 1 for extreme values of x, and is always continuous, gives us an excellent mental picture before we even calculate the derivative, making the upcoming variation table much easier to interpret and verify. We're effectively setting up all the necessary conditions and understanding the global behavior of f(x) before we zoom in on its local nuances, a practice that's invaluable in all forms of mathematical problem-solving.

Decoding f(x)'s Behavior: The Variation Table (Question 1)

Now for the really exciting part, guys: figuring out exactly how our function f(x) changes – whether it's going up, going down, or taking a momentary pause. This is where the derivative comes into play. The derivative, f'(x), is essentially a speedometer for our function, telling us its rate of change at any given point. If f'(x) > 0, the function is increasing; if f'(x) < 0, it's decreasing; and if f'(x) = 0, we've found a potential turning point, a critical point. This is the heart of constructing a variation table, which will be our ultimate map of f(x)'s ups and downs.

Let's recall our function: f(x) = (1 + x) / √(3 + x²). This is a quotient of two functions, so we'll need the quotient rule for differentiation: (u/v)' = (u'v - uv') / v².

Let's define our u and v:

• u = 1 + x => Its derivative, u' = 1.
• v = √(3 + x²) = (3 + x²)^(1/2). To find v', we use the chain rule: (d/dx) [g(h(x))] = g'(h(x)) * h'(x). Here, g(y) = y^(1/2) and h(x) = 3 + x². g'(y) = (1/2)y^(-1/2). h'(x) = 2x. So, v' = (1/2)(3 + x²)^(-1/2) * (2x) = x / √(3 + x²).

Now, let's plug these into the quotient rule: f'(x) = [ (1) * √(3 + x²) - (1 + x) * (x / √(3 + x²)) ] / (√(3 + x²))² f'(x) = [ √(3 + x²) - (x(1 + x) / √(3 + x²)) ] / (3 + x²)

To simplify the numerator, we need a common denominator: Numerator = [ (√(3 + x²)) * (√(3 + x²)) - x(1 + x) ] / √(3 + x²) Numerator = [ (3 + x²) - (x + x²) ] / √(3 + x²) Numerator = [ 3 + x² - x - x² ] / √(3 + x²) Numerator = [ 3 - x ] / √(3 + x²)

So, putting it all back together, we get: f'(x) = [ (3 - x) / √(3 + x²) ] / (3 + x²) f'(x) = (3 - x) / [ (3 + x²) * √(3 + x²) ] f'(x) = (3 - x) / (3 + x²)^(3/2)

Phew! That was a bit of work, but we got there! Now that we have f'(x) = (3 - x) / (3 + x²)^(3/2), we need to analyze its sign to determine where f(x) is increasing or decreasing. Look at the denominator: (3 + x²)^(3/2). Since 3 + x² is always positive, raising it to the power of 3/2 will also always result in a positive number. So, the sign of f'(x) is determined solely by the numerator, (3 - x).

• If 3 - x > 0 => x < 3, then f'(x) > 0, meaning f(x) is increasing.
• If 3 - x < 0 => x > 3, then f'(x) < 0, meaning f(x) is decreasing.
• If 3 - x = 0 => x = 3, then f'(x) = 0. This is our critical point.

Let's find the value of f(x) at this critical point, x = 3: f(3) = (1 + 3) / √(3 + 3²) = 4 / √(3 + 9) = 4 / √12 = 4 / (2√3) = 2/√3 = 2√3 / 3.

Now we can build our variation table:

From the table, we can see that f(x) increases from its limit of -1 (as x → -∞) all the way up to a local maximum at x = 3, where its value is 2√3 / 3 (approximately 1.15). After this point, f(x) starts decreasing and approaches its limit of 1 (as x → +∞). This detailed analysis of the derivative and the resulting variation table gives us a complete picture of the function's behavior across its entire domain, which is a fundamental skill in calculus and mathematical problem-solving. This comprehensive understanding is not only essential for solving this problem but also provides a powerful tool for analyzing a vast array of functions encountered in various scientific and engineering disciplines, truly highlighting the value of function analysis in understanding complex systems.

The Magical Interval [0, 1]: Why It Matters for Our Sequence (Question 2)

Alright, squad, let's tackle the second, equally crucial part of our mission: proving that if we start with a number x in the interval [0, 1], applying our function f(x) to it will still yield a result that stays within [0, 1]. This property is known as having an invariant interval. Why is this a big deal, you ask? Well, for a recursive sequence like our U_n where U{n+1} = f(U_n)_, if we know the starting term U_0 is within this invariant interval, then every subsequent term U_1, U_2, U_3, ... will also remain trapped inside this interval. It's like having a magical box where if you put something in, it can never escape. This is incredibly powerful because it immediately gives us a lot of information about the long-term behavior and convergence of the sequence. For instance, if a sequence is bounded (which being in an invariant interval implies) and monotonic (either always increasing or always decreasing), the famous Monotone Convergence Theorem tells us it must converge to a limit. So, proving f([0, 1]) ⊂ [0, 1] is a huge step in understanding our sequence U_n without even calculating many terms!

To prove that f([0, 1]) ⊂ [0, 1], we need to show two things for any x ∈ [0, 1]:

1. f(x) ≥ 0
2. f(x) ≤ 1

Let's use our knowledge from the variation table we just painstakingly constructed. We know that f(x) is defined and continuous on all of ℝ, and therefore certainly on [0, 1]. More importantly, our variation table showed that f(x) is increasing for all x < 3. Since the interval [0, 1] is entirely within x < 3, this means that f(x) is strictly increasing on [0, 1]. This is a critical piece of information because for an increasing function on a closed interval, its minimum value occurs at the left endpoint and its maximum value occurs at the right endpoint.

So, to find the range of f(x) on [0, 1], we just need to evaluate f(x) at its endpoints:

• Let's calculate f(0): f(0) = (1 + 0) / √(3 + 0²) = 1 / √3 = 1/√3 = √3 / 3. √3 / 3 is approximately 1.732 / 3 ≈ 0.577. This value is clearly within [0, 1].

• Next, let's calculate f(1): f(1) = (1 + 1) / √(3 + 1²) = 2 / √(3 + 1) = 2 / √4 = 2 / 2 = 1. This value is exactly 1, which is also clearly within [0, 1].

Since f(x) is strictly increasing on [0, 1], its minimum value on this interval is f(0) = √3 / 3 and its maximum value is f(1) = 1. This means that for any x ∈ [0, 1], we have: √3 / 3 ≤ f(x) ≤ 1

Since √3 / 3 ≈ 0.577, we can definitively say that 0 < √3 / 3 ≤ f(x) ≤ 1 for all x ∈ [0, 1]. Therefore, every output f(x) for an input x from [0, 1] will fall between approximately 0.577 and 1, inclusive. This range [√3 / 3, 1] is entirely contained within the larger interval [0, 1]. Voila! We have successfully proven that f([0, 1]) ⊂ [0, 1]. This means that if we start our sequence U_n with U_0 = 0 (which is definitely in [0, 1]), then U_1 = f(U_0) will be in [0, 1], U_2 = f(U_1) will be in [0, 1], and so on. Every single term of our sequence will live happily ever after within this little mathematical haven. This elegant proof not only answers the question but also unlocks a deeper understanding of the sequence's inherent stability and provides a strong foundation for further analysis of its convergence criteria. This kind of mathematical proof is essential for understanding iterative processes in various fields, reinforcing the practical power of invariant interval analysis in predicting long-term system behaviors.

Bringing It All Together: What Does This Mean for U_n?

Okay, guys, we've done some serious heavy lifting by thoroughly analyzing our function f(x) = (1 + x) / √(3 + x²). We've mapped its behavior with a variation table and, crucially, we've shown that the interval [0, 1] is invariant under f. Now, let's connect all these insights back to our original numerical sequence, U_n, which is defined by U_0 = 0 and U{n+1} = f(U_n)_. This is where all our hard work pays off, as we start to unravel the mysteries of U_n's sequence behavior.

Since we know U_0 = 0 and 0 ∈ [0, 1], and we've proven that f([0, 1]) ⊂ [0, 1], this has immediate and profound implications for every single term in our sequence. It means that U_1 = f(U_0) will be in [0, 1]. Then U_2 = f(U_1) will also be in [0, 1], and so on. By mathematical induction, every term U_n of the sequence will be contained within the interval _[0, 1]*. This tells us that the sequence U_n is bounded. It will never