How To Maximize Y=(x²+1)/x On [-11; -0.5]?

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How to Maximize y=(x²+1)/x on [-11; -0.5]?

Hey everyone! Ever stared at a math problem and thought, "Ugh, where do I even begin?" Well, you're in the right place, because today we're tackling a classic calculus challenge: finding the greatest value of a function, specifically y = (x² + 1) / x, but with a crucial twist – it's only on a specific, closed interval of x values: [-11; -0.5]. This isn't just some abstract exercise, guys; understanding how to find maxima and minima is super important in so many real-world scenarios, from optimizing product output in a factory to minimizing costs in logistics, or even determining the peak performance of an engine. We're going to break this down step-by-step, making sure it’s crystal clear and easy to follow. Get ready to flex those brain muscles, because by the end of this, you'll be a pro at this kind of optimization problem! We'll use our trusty calculus tools to pinpoint exactly where this function hits its highest point within our given boundaries. So, let’s dive in and unlock the secrets to maximizing this function, making sure we cover all our bases and leave no stone unturned. This isn’t just about getting the right answer, it’s about understanding the journey to get there, and gaining some valuable problem-solving skills along the way. We'll be focusing on the logic and the 'why' behind each mathematical operation, ensuring you grasp the core concepts rather than just memorizing a formula. Ready for an adventure in optimization? Let's do this!

Cracking the Code: Understanding Our Optimization Challenge

Alright, let’s kick things off by really understanding what this problem is asking us to do. We're on a mission to find the absolute maximum value of the function y = (x² + 1) / x. What does that even mean, you ask? Simply put, we want to find the highest possible 'y' value that this function can spit out, but there's a catch! We're not looking at the entire universe of possible x values. Nope, our focus is narrowed down to a very specific playing field, an interval from x = -11 all the way up to x = -0.5. This is super important, like being told to find the tallest person in just one specific room instead of the entire building. The function itself, y = (x² + 1) / x, is what we call a rational function because it’s a ratio of two polynomials. You might notice right away that x can't be zero in the denominator – if x were zero, we'd have a mathematical meltdown (division by zero is a no-go!). Luckily, our interval [-11; -0.5] neatly avoids x = 0, so we don't have to worry about any nasty holes or vertical asymptotes within our relevant range. This function, often referred to as a hyperbola, has some pretty interesting behavior, but we're only concerned with its performance in this particular segment. The interval [-11; -0.5] is a closed and bounded interval, which is fantastic news for us. Why? Because it means we can confidently apply the Extreme Value Theorem, a superstar in calculus that guarantees that a continuous function on a closed interval must have both an absolute maximum and an absolute minimum somewhere within that interval. These absolute extrema will occur either at what we call "critical points" (where the function's slope is flat or undefined) or at the endpoints of our interval itself. So, our job is to meticulously check all these potential spots. Think of it like this: if you're searching for the highest point on a rollercoaster track, you'd look at the obvious peaks, but you also need to check the very start and very end of the segment of track you're interested in, just in case the highest point is right at the beginning or end of your ride section! This isn't just for textbooks, either. Imagine you're an engineer designing a circuit and need to find the peak voltage output within a specific operating range, or an economist trying to find the maximum profit for a company given certain production constraints. This exact mathematical process is what you'd use. By narrowing our search to [-11; -0.5], we're making our task manageable and precise, ensuring we find the true highest point for this function under the given conditions. So, with a clear understanding of our target and our boundaries, we’re ready to roll up our sleeves and bring out the big guns: calculus!

Unleashing Calculus: The Magic of Derivatives for Maxima

Now, for the really fun part, guys – bringing out our calculus toolkit! When we're looking for the maximum (or minimum) value of a function, our best friend is the derivative. Think of the derivative as a magical tool that tells us the slope of our function at any given point. If our function is like a hilly landscape, the derivative helps us find all the places where the ground is perfectly flat – these are our potential peaks (maxima) or valleys (minima). These flat spots are what we call critical points, and they are absolutely crucial for solving optimization problems like this one. They're like the prime suspects in our search for the highest y value! To find these critical points for our function y = (x² + 1) / x, we first need to calculate its derivative, which we denote as y'. Since our function is a quotient (one function divided by another), we'll need to use the famous Quotient Rule. Don't worry, it's not as scary as it sounds, and we'll walk through it step-by-step. The Quotient Rule states that if you have a function h(x) = u(x) / v(x), then its derivative h'(x) is given by the formula: (u'v - uv') / v². Let's break down our y = (x² + 1) / x into u and v:

  • Let u(x) = x² + 1. Its derivative, u'(x), is simply 2x. (Remember, the derivative of x² is 2x, and the derivative of a constant like 1 is 0).
  • Let v(x) = x. Its derivative, v'(x), is just 1. (The derivative of x with respect to x is 1).

Now, let's plug these into our Quotient Rule formula:

  • y' = (u'v - uv') / v²
  • y' = ((2x) * (x) - (x² + 1) * (1)) / (x)²

Time to simplify this bad boy!

  • y' = (2x² - (x² + 1)) / x²
  • y' = (2x² - x² - 1) / x²
  • y' = (x² - 1) / x²

There it is! Our derivative, y' = (x² - 1) / x². This equation now holds the key to finding our critical points. See how we meticulously followed each step of the Quotient Rule? That precision is super important in calculus, because one tiny slip can throw off your entire solution. This derivative tells us how the slope of our original function changes at different x values. Our next move will be to take this derivative and set it equal to zero, or find where it's undefined, because those are the spots where our function might be hitting a local peak or valley. This is where the detective work truly begins, as we narrow down our search for the elusive maximum within our specific interval. The entire process hinges on this critical step of finding the derivative correctly. Without it, we'd be blindly searching for our maximum without the powerful guidance of calculus. So, give yourselves a pat on the back for getting through the derivative calculation – you've mastered a significant hurdle!

Hunting for Critical Points: Where Peaks and Valleys Hide

Okay, guys, we've got our magnificent derivative: y' = (x² - 1) / x². Now it's time to put it to work and find those critical points. Remember, critical points are the x values where the derivative is either zero or undefined. These are the potential locations where our function y might reach a local maximum or minimum. Let's tackle each case:

Case 1: Where y' equals zero

We set our derivative to zero:

  • (x² - 1) / x² = 0

For a fraction to be zero, its numerator must be zero (as long as the denominator isn't zero simultaneously, which we'll address in Case 2). So, we set the numerator equal to zero:

  • x² - 1 = 0
  • x² = 1

Solving for x, we get two values:

  • x = 1
  • x = -1

Now, here's the crucial part: we need to check if these critical points actually fall within our specified interval [-11; -0.5]. This is like finding potential treasure spots, but then realizing some of them are outside the map you're given!

  • For x = 1: Is 1 within [-11; -0.5]? Nope, it's not. The interval stops at -0.5, so x = 1 is outside our region of interest. This means we can discard x = 1 for this specific problem. It might be a critical point for the function overall, but not for the part we care about.
  • For x = -1: Is -1 within [-11; -0.5]? Absolutely! -1 is comfortably nestled between -11 and -0.5. This is a valid critical point we need to consider! Mark this one down; it's a strong candidate for our maximum.

Case 2: Where y' is undefined

Our derivative, y' = (x² - 1) / x², becomes undefined when its denominator is zero. So, we set the denominator to zero:

  • x² = 0
  • x = 0

Again, we ask: Is x = 0 within our interval [-11; -0.5]? No, it's not. Our interval specifically ranges from -11 to -0.5, excluding 0. Also, recall from the start that x = 0 is not even in the domain of our original function y = (x² + 1) / x because it would lead to division by zero. So, this point is irrelevant to our search for a maximum within [-11; -0.5]. We can safely discard x = 0.

So, after all that detective work, we've found only one relevant critical point within our interval: x = -1. This means our journey to find the absolute maximum value just got a whole lot clearer! We've narrowed down the potential turning points to just one spot. The next step is to evaluate our original function y at this critical point and, equally important, at the endpoints of our interval. These values will be the contenders for our ultimate maximum. Pretty neat, right? It's all about being systematic and precise, making sure we don't miss any vital candidates while also discarding the ones that don't fit our specific conditions. This careful filtering is what makes calculus so powerful in solving real-world optimization challenges.

The Bookends of Our Search: Evaluating at the Interval Endpoints

Alright, folks, we've successfully identified our lone critical point of interest, x = -1, which lies squarely within our interval [-11; -0.5]. But here’s a super important rule from the world of calculus that we cannot forget: the Extreme Value Theorem. This theorem basically states that for a continuous function on a closed interval (which y = (x² + 1) / x is, within our [-11; -0.5] range, since x=0 is excluded), the absolute maximum and minimum values must occur either at one of the critical points inside the interval, or at the endpoints of the interval itself. It's like checking the highest points inside a room, but also remembering that the highest point might be right on the wall at the very edge of the room. So, even if our critical point looks promising, we absolutely have to evaluate our original function y at the very start and very end of our interval. These endpoints are non-negotiable contenders for the maximum value! Let's get calculating:

1. Evaluating at the left endpoint: x = -11

We plug x = -11 into our original function y = (x² + 1) / x:

  • y(-11) = ((-11)² + 1) / (-11)
  • y(-11) = (121 + 1) / (-11)
  • y(-11) = 122 / (-11)

Now, let's simplify this fraction. 122 divided by 11 is approximately 11.09, and since we're dividing a positive by a negative, our result will be negative:

  • y(-11) ≈ -11.0909...

Keep this value in your pocket; it’s one of our candidates for the maximum.

2. Evaluating at the right endpoint: x = -0.5

Next, we plug x = -0.5 into y = (x² + 1) / x. Remember that -0.5 is the same as -1/2, which can sometimes make calculations a bit cleaner, but we'll stick with decimals for clarity:

  • y(-0.5) = ((-0.5)² + 1) / (-0.5)
  • y(-0.5) = (0.25 + 1) / (-0.5)
  • y(-0.5) = 1.25 / (-0.5)

Now, let's do the division. 1.25 divided by 0.5 is 2.5. Again, positive divided by negative gives a negative result:

  • y(-0.5) = -2.5

Alright, another contender! By carefully evaluating the function at these two critical points, we're gathering all the necessary pieces of the puzzle. This meticulous approach ensures we don't accidentally miss the true maximum just because it happens to be right at the boundary of our search area. The Extreme Value Theorem is our guardian angel here, assuring us that the maximum has to be among these candidates. Without checking the endpoints, our solution would be incomplete and potentially incorrect. This step is as vital as finding the critical points themselves, making sure our search is exhaustive and accurate. Now that we have values from both our relevant critical point and our endpoints, we're ready for the grand comparison!

Bringing It All Together: Finding Our Absolute Maximum!

Alright, guys, this is it – the moment of truth! We've done all the hard work: we understood the problem, wielded the power of derivatives to find critical points, and diligently evaluated our function at both the relevant critical point and the interval endpoints. Now, all that's left is to gather these candidate values and compare them to find the absolute greatest value of y = (x² + 1) / x on our given interval [-11; -0.5]. It's like having a lineup of contenders and picking the champion!

Let's list our y values:

  1. From the critical point x = -1 (which we found earlier to be within our interval):

    • We need to calculate y(-1) using the original function y = (x² + 1) / x.
    • y(-1) = ((-1)² + 1) / (-1)
    • y(-1) = (1 + 1) / (-1)
    • y(-1) = 2 / (-1)
    • y(-1) = -2

  2. From the left endpoint x = -11 (which we calculated in the previous section):

    • y(-11) ≈ -11.0909...

  3. From the right endpoint x = -0.5 (also calculated previously):

    • y(-0.5) = -2.5

Now, let’s line up these three values: -2, -11.09, and -2.5. We're looking for the largest among them. Remember, with negative numbers, the value closest to zero (or positive infinity) is the largest.

  • Is -2 greater than -11.09? Yes! (-2 is much closer to zero).
  • Is -2 greater than -2.5? Yes! (-2 is closer to zero than -2.5).

Clearly, -2 is the biggest value among our candidates. This means our function hits its absolute maximum at x = -1 within the interval [-11; -0.5].

The absolute maximum value of the function y = (x² + 1) / x on the interval [-11; -0.5] is -2.

And there you have it! We've successfully navigated through this optimization problem. Let's quickly recap the entire process, guys:

  • First, we understood the function and the interval, recognizing the importance of the closed interval for applying the Extreme Value Theorem.
  • Next, we calculated the derivative using the Quotient Rule to find where the function's slope is flat.
  • Then, we identified critical points by setting the derivative to zero and checking where it was undefined, and most importantly, we filtered these to include only those that fell within our specified interval.
  • Crucially, we evaluated the original function at the interval's endpoints, as the maximum could reside there.
  • Finally, we compared all the candidate y values (from relevant critical points and endpoints) to pinpoint the absolute maximum.

This methodical approach is the gold standard for solving optimization problems, whether they're in a math class or a real-world application. Mastering this process equips you with a powerful tool for finding peak performance, optimal designs, or most efficient outcomes in countless fields. You've just tackled a pretty significant calculus problem, so give yourselves a huge round of applause! Keep practicing, and you'll be an optimization wizard in no time. The beauty of mathematics lies not just in solving problems, but in understanding the elegant steps and logical frameworks that guide us to the solution. Hope this guide was super helpful and made complex concepts feel a little more friendly and understandable. Now go forth and optimize everything!

Beyond the Math: Real-World Applications of Optimization

Now that we've totally crushed this math problem, finding the maximum value of our function on a specific interval, let's take a moment to appreciate why this stuff actually matters outside of a textbook. Seriously, guys, optimization problems, which is what we just solved, are everywhere in the real world. This isn't just about finding x and y; it's about making things better, faster, cheaper, or more efficient. Think about it like this: every time a company wants to maximize profit or minimize cost, they're essentially solving an optimization problem. Our little function y = (x² + 1) / x on the interval [-11; -0.5] might not seem like a profit model, but the methodology we used is precisely what engineers, economists, data scientists, and even athletes employ daily.

For instance, let's consider a manufacturing company. They want to produce a certain product. There are costs associated with raw materials, labor, electricity, and transportation. There are also selling prices and market demand. A manager might use optimization techniques, much like what we did, to find the optimal production level that maximizes their profit for a given period. The x in their function might represent the number of units produced, and the y could be the total profit. They'd also have constraints, just like our [-11; -0.5] interval, such as the maximum capacity of their factory or the minimum number of units required to break even. Finding the derivative and critical points would help them identify the sweet spot for production.

In engineering, optimization is absolutely fundamental. When designing a bridge, engineers need to find the minimum amount of material required to ensure structural integrity, thereby minimizing cost while maximizing safety. When building an airplane wing, they might optimize its shape to maximize lift and minimize drag, leading to more fuel-efficient flights. The functions they're dealing with are far more complex than ours, but the underlying calculus principles are identical: calculate derivatives, find critical points, check boundaries, and compare values. Even in designing computer chips, the placement of components is optimized to minimize signal delay and maximize processing speed.

Even in our daily lives, we encounter optimization. Think about navigation apps like Google Maps. They are constantly optimizing routes to minimize travel time or minimize distance, taking into account real-time traffic conditions, which act as dynamic constraints. Or consider a financial analyst trying to build an investment portfolio that maximizes returns while minimizing risk. They use sophisticated mathematical models that rely heavily on optimization theory, including calculus.

In environmental science, optimization helps in resource management, such as finding the optimal harvest rate for a renewable resource to maximize long-term yield without depleting the resource. In medicine, drug dosages are optimized to maximize therapeutic effect while minimizing side effects. Even in sports, athletes and coaches use data to optimize training regimens to maximize performance at specific times (like before a major competition) and minimize the risk of injury.

So, you see, the skills we just honed – understanding functions, applying derivatives, identifying critical points, and evaluating boundary conditions – are not just academic exercises. They are powerful tools that drive innovation, efficiency, and progress across virtually every industry and facet of modern life. When you solve a problem like "How to Maximize y=(x²+1)/x on [-11; -0.5]?", you're not just solving for a number; you're mastering a universal problem-solving framework that has incredible practical utility. Pretty cool, right? This journey through calculus isn't just about math; it's about gaining a superpower to make better decisions and build a better world. Keep learning, keep optimizing, and keep being awesome!