Finding The Absolute Maximum Of F(x) = 5x - 3x Ln(x)

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Hey there, math enthusiasts! Today, we're diving into a fun calculus problem: finding the absolute maximum value of the function f(x) = 5x - 3x ln(x) on the interval (0, ∞). This is a classic optimization problem that combines algebraic manipulation, calculus techniques, and a bit of logical reasoning. So, grab your thinking caps, and let's get started!

1. Understanding the Problem

Before we jump into the calculations, let's make sure we understand what the question is asking. We're given a function, f(x) = 5x - 3x ln(x), and an interval, (0, ∞). Our goal is to find the largest possible value that the function can take within that interval. This largest value is called the absolute maximum.

Now, why is the interval (0, ∞) important? Well, it tells us the domain, or the set of all possible input values (x-values), that we're considering. In this case, we're only looking at positive numbers since the natural logarithm, ln(x), is only defined for x > 0. The ∞ symbol means we're considering all positive numbers, without any upper limit.

2. Finding Critical Points

In calculus, the key to finding maximum and minimum values lies in identifying what we call critical points. These are the points where the function's derivative is either equal to zero or undefined. Why are they so important? Because at these points, the function's slope is either flat (derivative = 0) or has a sharp turn (derivative undefined), which are potential locations for maximum or minimum values.

So, our first step is to find the derivative of f(x). Remember the product rule for differentiation? It states that the derivative of u(x)v(x) is u'(x)v(x) + u(x)v'(x). We'll need this to differentiate the 3x ln(x) term.

Let's break it down:

  • f(x) = 5x - 3x ln(x)
  • f'(x) = d/dx (5x) - d/dx (3x ln(x))
  • f'(x) = 5 - [3 ln(x) + 3x (1/x)] (Applying the product rule)
  • f'(x) = 5 - 3 ln(x) - 3
  • f'(x) = 2 - 3 ln(x)

Great! We've found the derivative. Now, we need to find the x-values where f'(x) = 0 or is undefined.

The derivative, f'(x) = 2 - 3 ln(x), is defined for all x > 0 (the same domain as the original function). So, we just need to solve for where it equals zero:

  • 2 - 3 ln(x) = 0
  • 3 ln(x) = 2
  • ln(x) = 2/3

To get rid of the natural logarithm, we use the exponential function, e^x:

  • x = e^(2/3)

So, we have one critical point: x = e^(2/3). This is a crucial value that we'll need to investigate further.

3. Analyzing Critical Points and End Behavior

We've found one critical point, but remember, we're looking for the absolute maximum on the interval (0, ∞). This means we also need to consider what happens to the function as x approaches the endpoints of the interval, which are 0 and ∞.

Let's start by analyzing the behavior of f(x) as x approaches 0:

  • lim (x→0⁺) [5x - 3x ln(x)]

This is a bit tricky because we have a term that goes to zero (5x) and a term that's an indeterminate form (0 * -∞). To handle this, we can rewrite the second term using L'Hôpital's Rule. L'Hôpital's Rule is a powerful tool that allows us to evaluate limits of indeterminate forms by taking the derivatives of the numerator and denominator.

First, rewrite x ln(x) as a fraction:

  • x ln(x) = ln(x) / (1/x)

Now, as x approaches 0, we have ln(x) → -∞ and (1/x) → ∞, which is an indeterminate form. Applying L'Hôpital's Rule:

  • lim (x→0⁺) [ln(x) / (1/x)] = lim (x→0⁺) [(1/x) / (-1/x²)] = lim (x→0⁺) [-x] = 0

So, lim (x→0⁺) [x ln(x)] = 0. Now we can go back to our original limit:

  • lim (x→0⁺) [5x - 3x ln(x)] = 5(0) - 3(0) = 0

This tells us that as x approaches 0, the function f(x) approaches 0.

Now, let's consider the behavior of f(x) as x approaches infinity:

  • lim (x→∞) [5x - 3x ln(x)]

In this case, both 5x and 3x ln(x) approach infinity. However, the ln(x) function grows much slower than x, so the 3x ln(x) term will dominate and become negative. This means the entire expression will approach negative infinity:

  • lim (x→∞) [5x - 3x ln(x)] = -∞

So, as x approaches infinity, the function f(x) approaches negative infinity. This means there's no absolute maximum at infinity.

4. Evaluating the Function at the Critical Point

We know the function approaches 0 as x approaches 0, and it approaches negative infinity as x approaches infinity. This suggests that the absolute maximum, if it exists, must occur at our critical point, x = e^(2/3). Let's evaluate f(x) at this point:

  • f(e^(2/3)) = 5e^(2/3) - 3e^(2/3) ln(e^(2/3))
  • f(e^(2/3)) = 5e^(2/3) - 3e^(2/3) (2/3)
  • f(e^(2/3)) = 5e^(2/3) - 2e^(2/3)
  • f(e^(2/3)) = 3e^(2/3)

So, the value of the function at the critical point is 3e^(2/3). Since the function approaches 0 as x approaches 0 and negative infinity as x approaches infinity, this value must be the absolute maximum.

5. The Conclusion: The Absolute Maximum

After all our hard work, we've arrived at the answer! The absolute maximum value of the function f(x) = 5x - 3x ln(x) on the interval (0, ∞) is 3e^(2/3), and it occurs at x = e^(2/3). That’s it, guys! We successfully navigated this optimization problem using calculus principles and a little bit of clever thinking.

Final Answer

A. The absolute maximum is 3e^(2/3) at x = e^(2/3)

Are you trying to figure out how to find the absolute maximum value of a function? Let’s walk through the process together using the function f(x) = 5x - 3x ln(x) on the interval (0, ∞) as an example. This problem involves calculus, but don’t worry, we'll break it down into manageable steps.

Understanding the Problem and Why It Matters

First, let's understand what we're trying to find. The absolute maximum of a function on a given interval is the highest point the function reaches within that interval. In our case, we have the function f(x) = 5x - 3x ln(x), and we want to find its highest value on the interval from 0 to infinity. This is a common type of problem in calculus and has applications in various fields, such as physics, economics, and engineering, where optimization is crucial.

Why (0, ∞)? This interval means we're only considering positive values of x, since the natural logarithm, ln(x), is only defined for positive numbers. The infinity symbol tells us we're looking at all positive numbers without an upper bound. This interval is essential because it defines the scope of our search for the maximum value.

Calculating the Derivative

To find the maximum or minimum points of a function, we need to find its critical points. These are the points where the derivative of the function is either zero or undefined. The derivative gives us the slope of the function at any given point, and at the maximum or minimum points, the slope is typically zero (or undefined at a sharp turn).

Our first step is to find the derivative of f(x) = 5x - 3x ln(x). Remember the product rule? It's crucial here. The product rule states that the derivative of u(x)v(x) is u'(x)v(x) + u(x)v'(x). We'll apply this to the 3x ln(x) term.

Let's break it down:

  1. Original function: f(x) = 5x - 3x ln(x)
  2. Differentiate each term: f'(x) = d/dx (5x) - d/dx (3x ln(x))
  3. Apply the product rule to 3x ln(x): f'(x) = 5 - [3 ln(x) + 3x (1/x)]
  4. Simplify: f'(x) = 5 - 3 ln(x) - 3
  5. Final derivative: f'(x) = 2 - 3 ln(x)

Great job! We've found the derivative. This f'(x) tells us the slope of f(x) at any x-value. Now we need to find where this slope is zero.

Finding Critical Points by Setting the Derivative to Zero

Now that we have the derivative, f'(x) = 2 - 3 ln(x), we need to find the values of x where f'(x) = 0. These are the points where the function’s slope is flat, which are potential maximum or minimum points.

Let's set the derivative to zero and solve for x:

  1. Set the derivative to zero: 2 - 3 ln(x) = 0
  2. Rearrange the equation: 3 ln(x) = 2
  3. Divide by 3: ln(x) = 2/3

Now, to get rid of the natural logarithm (ln), we'll use the exponential function, e^x. Remember that e^ln(x) = x:

  1. Apply the exponential function: x = e^(2/3)

Fantastic! We’ve found one critical point: x = e^(2/3). This is a key value because it's a potential location for our absolute maximum. But we're not done yet; we need to consider the behavior of the function at the boundaries of our interval and determine if this critical point is indeed a maximum.

Analyzing End Behavior and Using Limits

We've found our critical point, but to be sure we've found the absolute maximum, we need to examine what happens to our function f(x) at the edges of our interval (0, ∞). This means looking at the limits as x approaches 0 from the right (since ln(x) is not defined for x ≤ 0) and as x approaches infinity.

Let's start with the limit as x approaches 0 from the right (0⁺):

  1. Write the limit: lim (x→0⁺) [5x - 3x ln(x)]

This looks tricky because we have a product that approaches 0 times negative infinity (0 * -∞). This is an indeterminate form, and we can tackle it using L'Hôpital's Rule. L'Hôpital's Rule helps us evaluate limits of indeterminate forms by taking the derivatives of the numerator and denominator of a fraction.

To use L'Hôpital's Rule, we need to rewrite our expression as a fraction. Let's rewrite x ln(x) as ln(x) / (1/x):

  1. Rewrite x ln(x): x ln(x) = ln(x) / (1/x)

Now our limit looks like this:

lim (x→0⁺) [ln(x) / (1/x)]

As x approaches 0⁺, ln(x) goes to negative infinity, and (1/x) goes to positive infinity, so we have the indeterminate form -∞/∞. Now we can apply L'Hôpital's Rule:

  1. Apply L'Hôpital's Rule: lim (x→0⁺) [(1/x) / (-1/x²)]

Now, let's simplify this expression:

  1. Simplify: lim (x→0⁺) [-x]

This limit is much easier to evaluate:

  1. Evaluate the limit: lim (x→0⁺) [-x] = 0

So, the limit as x approaches 0⁺ of x ln(x) is 0. Now we can go back to our original limit:

  1. Evaluate original limit at 0: lim (x→0⁺) [5x - 3x ln(x)] = 5(0) - 3(0) = 0

This tells us that as x gets very close to 0, the function f(x) approaches 0.

Next, we need to consider the limit as x approaches infinity:

  1. Write the limit as x approaches infinity: lim (x→∞) [5x - 3x ln(x)]

In this case, both 5x and 3x ln(x) approach infinity, but the ln(x) function grows much slower than x. This means the 3x ln(x) term will eventually dominate and make the whole expression negative:

  1. Evaluate limit at infinity: lim (x→∞) [5x - 3x ln(x)] = -∞

So, as x approaches infinity, the function f(x) approaches negative infinity. This means the function does not have an absolute maximum as x goes to infinity.

Evaluating the Function at the Critical Point

We know the function approaches 0 as x approaches 0, and it approaches negative infinity as x approaches infinity. This strongly suggests that the absolute maximum, if it exists, must occur at our critical point, x = e^(2/3). Let's find the value of f(x) at this point:

  1. Write the function at the critical point: f(e^(2/3)) = 5e^(2/3) - 3e^(2/3) ln(e^(2/3))
  2. Simplify the logarithm: f(e^(2/3)) = 5e^(2/3) - 3e^(2/3) (2/3)
  3. Multiply: f(e^(2/3)) = 5e^(2/3) - 2e^(2/3)
  4. Combine terms: f(e^(2/3)) = 3e^(2/3)

So, the value of the function at the critical point is 3e^(2/3). This is a positive value, and since we know the function approaches 0 as x approaches 0 and negative infinity as x approaches infinity, this value must be the absolute maximum.

Stating the Conclusion with Confidence

We've done it! After carefully analyzing the function, finding its derivative, identifying critical points, and examining end behavior, we've found the absolute maximum value of the function f(x) = 5x - 3x ln(x) on the interval (0, ∞). High five, guys!

The absolute maximum value is 3e^(2/3), and it occurs at x = e^(2/3). This result is significant because it tells us the highest point the function reaches within the specified interval.

Hey math lovers! Today, we're tackling an interesting problem in calculus: finding the absolute maximum value of the function f(x) = 5x - 3x ln(x) on the interval (0, ∞). This is a typical optimization problem that involves differentiation, critical points, and a good understanding of function behavior. Don't worry, we'll go through it step by step, and you'll see it's not as daunting as it might seem!

Understanding the Absolute Maximum Problem

First, let’s break down the question. When we talk about the absolute maximum of a function on an interval, we're looking for the highest value that the function attains within that interval. Think of it as the peak of a mountain range within a specific region. In this case, our function is f(x) = 5x - 3x ln(x), and our region is the interval (0, ∞), which means all positive numbers.

The interval (0, ∞) is crucial here. The ln(x) part of our function is only defined for positive values of x, so we can’t consider zero or negative numbers. Also, the ∞ tells us we need to think about what happens as x gets incredibly large. So, we're essentially exploring the function's behavior across all positive numbers to find its highest point.

Finding the Derivative to Locate Critical Points

In calculus, we use derivatives to find the maximum and minimum values of functions. The derivative of a function tells us its slope at any point. At the top of a peak (a maximum) or the bottom of a valley (a minimum), the slope is usually zero. These points where the slope is zero (or undefined) are called critical points, and they're our prime suspects for absolute maximums and minimums.

So, the first thing we need to do is find the derivative of our function, f(x) = 5x - 3x ln(x). Remember the product rule? We'll need it for the 3x ln(x) term. The product rule states that the derivative of u(x)v(x) is u'(x)v(x) + u(x)v'(x). Let's apply this:

  1. Original function: f(x) = 5x - 3x ln(x)
  2. Differentiate term by term: f'(x) = d/dx (5x) - d/dx (3x ln(x))
  3. Apply the product rule: f'(x) = 5 - [3 ln(x) + 3x (1/x)]
  4. Simplify: f'(x) = 5 - 3 ln(x) - 3
  5. Final derivative: f'(x) = 2 - 3 ln(x)

Excellent! We’ve found the derivative, f'(x) = 2 - 3 ln(x). This formula gives us the slope of the function at any x-value. Now, we need to find the x-values where this slope is zero.

Setting the Derivative to Zero and Solving for x

To find the critical points, we set the derivative equal to zero and solve for x. This will give us the x-values where the function has a horizontal tangent line, which are potential locations for maximum or minimum values.

Let’s set f'(x) = 0 and solve:

  1. Set the derivative to zero: 2 - 3 ln(x) = 0
  2. Rearrange the equation: 3 ln(x) = 2
  3. Divide by 3: ln(x) = 2/3

Now, to get x by itself, we need to get rid of the natural logarithm (ln). We do this by using the exponential function, e^x. Remember that e^ln(x) = x:

  1. Apply the exponential function: x = e^(2/3)

Great! We’ve found one critical point: x = e^(2/3). This is a crucial value that we'll need to investigate further to see if it corresponds to a maximum or a minimum.

Investigating End Behavior with Limits

Finding the critical points is just one piece of the puzzle. Since we're looking for the absolute maximum on the interval (0, ∞), we also need to consider what happens to our function as x approaches the endpoints of the interval: 0 and ∞. This is where limits come in handy.

Let's start by looking at the limit as x approaches 0 from the right (since the logarithm is not defined for x ≤ 0):

  1. Write the limit: lim (x→0⁺) [5x - 3x ln(x)]

We have a tricky situation here: as x approaches 0, 5x approaches 0, but ln(x) approaches negative infinity. So, we have a product of the form 0 * -∞, which is an indeterminate form. To handle this, we'll use L'Hôpital's Rule. L'Hôpital's Rule allows us to evaluate limits of indeterminate forms by taking the derivatives of the numerator and denominator of a fraction.

To use L'Hôpital's Rule, we need to rewrite our expression as a fraction. Let's rewrite x ln(x) as ln(x) / (1/x):

  1. Rewrite x ln(x): x ln(x) = ln(x) / (1/x)

Now our limit looks like this:

lim (x→0⁺) [ln(x) / (1/x)]

As x approaches 0⁺, ln(x) goes to negative infinity, and (1/x) goes to infinity, so we have the indeterminate form -∞/∞. Let's apply L'Hôpital's Rule:

  1. Apply L'Hôpital's Rule: lim (x→0⁺) [(1/x) / (-1/x²)]

Now, simplify this expression:

  1. Simplify: lim (x→0⁺) [-x]

This limit is much easier to evaluate:

  1. Evaluate the limit: lim (x→0⁺) [-x] = 0

So, the limit as x approaches 0⁺ of x ln(x) is 0. Now we can go back to our original limit:

  1. Evaluate original limit at 0: lim (x→0⁺) [5x - 3x ln(x)] = 5(0) - 3(0) = 0

This tells us that as x gets very close to 0, the function f(x) approaches 0.

Next, let's consider the limit as x approaches infinity:

  1. Write the limit as x approaches infinity: lim (x→∞) [5x - 3x ln(x)]

In this case, both 5x and 3x ln(x) approach infinity, but the ln(x) function grows much slower than x. This means the 3x ln(x) term will eventually dominate and make the whole expression negative:

  1. Evaluate limit at infinity: lim (x→∞) [5x - 3x ln(x)] = -∞

So, as x approaches infinity, the function f(x) approaches negative infinity. This means we won’t find an absolute maximum as x goes to infinity.

Finding the Absolute Maximum Value at the Critical Point

We've determined that our function approaches 0 as x approaches 0 and negative infinity as x approaches infinity. This strongly indicates that the absolute maximum, if it exists, must occur at our critical point, x = e^(2/3). Now, let's plug this value back into our original function to find the maximum value:

  1. Write the function at the critical point: f(e^(2/3)) = 5e^(2/3) - 3e^(2/3) ln(e^(2/3))
  2. Simplify the logarithm: f(e^(2/3)) = 5e^(2/3) - 3e^(2/3) (2/3)
  3. Multiply: f(e^(2/3)) = 5e^(2/3) - 2e^(2/3)
  4. Combine terms: f(e^(2/3)) = 3e^(2/3)

So, the value of the function at the critical point is 3e^(2/3). Since we know the function approaches 0 as x approaches 0 and negative infinity as x approaches infinity, this value must be the absolute maximum.

Conclusion The Peak Value Found!

Phew! We made it through all the steps! We've found the absolute maximum value of the function f(x) = 5x - 3x ln(x) on the interval (0, ∞). Way to go, team!

The absolute maximum value is 3e^(2/3), and it occurs at x = e^(2/3). This is the highest point our function reaches on the given interval. We used a combination of calculus techniques, including differentiation and limits, to solve this optimization problem. Keep up the fantastic work, and remember that every calculus problem is just another step toward mastering the art of mathematics!

Okay, mathletes, let's dive into a classic calculus problem: finding the absolute maximum value of the function f(x) = 5x - 3x ln x on the interval (0, ∞). This is a fun exercise that combines differentiation, critical points, and limit analysis. Let’s break it down step-by-step so you can ace similar problems in the future.

Step 1: Grasping the Problem

First things first, let's make sure we're all on the same page. What does it mean to find the absolute maximum? Simply put, it's the highest point the function reaches within the specified interval. Think of it as the peak of a rollercoaster on a particular track. In this case, our track is the interval (0, ∞), meaning we’re only considering positive values of x. Why (0, ∞), you ask? Well, the natural logarithm, ln x, is only defined for positive x-values. So, we're exploring all positive numbers to find where our function hits its highest point. This might sound tricky, but with the right tools, it's totally manageable.

Step 2: Hunting Down Critical Points via Differentiation

Now, let's roll up our sleeves and get into some calculus. The secret to finding maximum and minimum values of a function lies in the derivative. The derivative, denoted as f'(x), tells us the slope of the function at any point. At a maximum or minimum, the slope is usually zero (think of the top of the hill where it levels out momentarily). These points where the slope is zero or undefined are called critical points. So, our mission is to find the derivative of f(x) and then find where it equals zero. Buckle up; we’re about to use the product rule!

Remember the product rule? It's essential when differentiating a product of two functions. If we have u(x)v(x), its derivative is u'(x)v(x) + u(x)v'(x). Our function has a term, 3x ln x, that's a product, so we’ll need this rule. Let's break it down:

  1. The function: f(x) = 5x - 3x ln x
  2. Take the derivative term by term: f'(x) = d/dx(5x) - d/dx(3x ln x)
  3. Apply the product rule to 3x ln x: f'(x) = 5 - [3 ln x + 3x(1/x)]
  4. Simplify: f'(x) = 5 - 3 ln x - 3
  5. The derivative: f'(x) = 2 - 3 ln x

Boom! We've got our derivative. Now, we need to find the x-values where this derivative equals zero. These are our potential hotspots for maximum or minimum values.

Step 3: Setting the Derivative to Zero and Unearthing the Critical Point

With the derivative in hand, our next task is to set f'(x) to zero and solve for x. This will pinpoint the x-value(s) where the function has a horizontal tangent line, indicating a potential maximum or minimum. Let's get to it:

  1. Set the derivative to zero: 2 - 3 ln x = 0
  2. Rearrange the equation: 3 ln x = 2
  3. Divide by 3: ln x = 2/3

Now, to isolate x, we need to get rid of the natural logarithm (ln). We’ll use the exponential function, e^x, for this. Recall that e^(ln x) = x. So:

  1. Apply the exponential function: x = e^(2/3)

Hooray! We’ve unearthed our critical point: x = e^(2/3). This is a key player in our quest for the absolute maximum. But wait, we’re not done yet! We need to ensure this critical point truly represents a maximum and that there aren't other contenders lurking at the edges of our interval.

Step 4: Investigating End Behavior through the Lens of Limits

We've found a critical point, but to be absolutely sure we've nailed the absolute maximum, we need to examine the behavior of our function, f(x), at the boundaries of our interval, (0, ∞). This means we need to consider what happens as x approaches 0 from the right (0⁺) and as x approaches infinity. We’ll use limits to help us.

Let's start by analyzing the limit as x approaches 0 from the right:

  1. The limit expression: lim (x→0⁺) [5x - 3x ln x]

Uh-oh, this looks a bit tricky. As x gets closer to 0, 5x clearly approaches 0. However, the term 3x ln x is a bit more complex. As x approaches 0, ln x approaches negative infinity. So, we have a situation of 0 * (-∞), which is an indeterminate form. To crack this, we’re going to call in the big guns: L'Hôpital's Rule.

L'Hôpital's Rule is our go-to method for evaluating limits of indeterminate forms like 0/0 or ∞/∞. It involves taking the derivatives of the numerator and the denominator. To use it, we first need to rewrite our expression as a fraction. Let's transform x ln x into ln x / (1/x):

  1. Rewrite x ln x: x ln x = ln x / (1/x)

Now, our limit looks like this:

lim (x→0⁺) [ln x / (1/x)]

As x approaches 0⁺, ln x heads to negative infinity, and (1/x) zooms to positive infinity. We now have the indeterminate form -∞/∞, so we can confidently apply L'Hôpital's Rule:

  1. Apply L'Hôpital's Rule: lim (x→0⁺) [(1/x) / (-1/x²)]

Let’s simplify this expression:

  1. Simplify: lim (x→0⁺) [-x]

This is much easier to handle:

  1. Evaluate the limit: lim (x→0⁺) [-x] = 0

So, the limit as x approaches 0⁺ of x ln x is 0. Now, let's plug that back into our original limit:

  1. Evaluate the original limit at 0: lim (x→0⁺) [5x - 3x ln x] = 5(0) - 3(0) = 0

This tells us that as x gets incredibly close to 0, our function f(x) approaches 0.

Next up, let's tackle the limit as x approaches infinity:

  1. The limit as x approaches infinity: lim (x→∞) [5x - 3x ln x]

In this scenario, both 5x and 3x ln x surge towards infinity. However, the natural logarithm function, ln x, grows much more slowly than x. This means that the 3x ln x term will eventually dominate and pull the whole expression downwards, towards negative infinity:

  1. Evaluate the limit at infinity: lim (x→∞) [5x - 3x ln x] = -∞

Therefore, as x heads to infinity, our function f(x) plunges towards negative infinity. This is super helpful information because it rules out the possibility of an absolute maximum at infinity.

Step 5: Pinpointing the Absolute Maximum Value at the Critical Point

We've navigated the treacherous waters of limits and end behavior, and here’s what we’ve discovered: our function approaches 0 as x nears 0 and nosedives to negative infinity as x shoots to infinity. This strongly suggests that the absolute maximum, if it exists, must reside at our critical point, x = e^(2/3). It’s time to plug this value back into our original function to find that maximum value:

  1. The function at the critical point: f(e^(2/3)) = 5e^(2/3) - 3e^(2/3) ln(e^(2/3))
  2. Simplify the logarithm: f(e^(2/3)) = 5e^(2/3) - 3e^(2/3) (2/3)
  3. Multiply: f(e^(2/3)) = 5e^(2/3) - 2e^(2/3)
  4. Combine like terms: f(e^(2/3)) = 3e^(2/3)

So, the value of our function at the critical point is 3e^(2/3). Considering what we learned about the function’s behavior at the boundaries of our interval, this value has to be the absolute maximum!

Step 6: Celebrating the Grand Finale

We’ve conquered the problem! After a thorough analysis involving derivatives, critical points, and limits, we've triumphantly discovered the absolute maximum value of the function f(x) = 5x - 3x ln x on the interval (0, ∞). Give yourselves a round of applause, you brilliant minds!

The absolute maximum value is 3e^(2/3), and it occurs at x = e^(2/3). This marks the highest peak our function reaches within the specified interval. We successfully employed a variety of calculus tools and techniques to solve this optimization problem. You've now added another valuable skill to your mathematical toolkit, so keep that curiosity burning and those problem-solving gears turning!

In conclusion, finding the absolute maximum value of a function involves a multi-step process. We first find the critical points by calculating the derivative and setting it to zero. Then, we analyze the end behavior of the function using limits to understand what happens as x approaches the boundaries of the interval. Finally, we evaluate the function at the critical points and the endpoints (if included) to determine the absolute maximum value. For the function f(x) = 5x - 3x ln(x) on the interval (0, ∞), the absolute maximum value is 3e^(2/3), which occurs at x = e^(2/3). This process showcases the power of calculus in solving optimization problems, which are crucial in various fields like physics, economics, and computer science.

The final answer is: A. The absolute maximum is 3e^(2/3) at x = e^(2/3).