Finding Local Minima And Maxima Using The Second Derivative Test F(x) = 9ln(x) - 6x

Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of calculus to explore how the second derivative test can help us pinpoint those crucial local minimum and maximum points of a function. Our specimen for today's mathematical dissection is the function f(x) = 9ln(x) - 6x. So, buckle up, grab your thinking caps, and let's embark on this exciting journey together!

Cracking the Code: The Second Derivative Test

Before we jump into the nitty-gritty details of our function, let's take a moment to understand the second derivative test itself. It's a powerful tool in our calculus arsenal that helps us determine the nature of critical points – those special points where the function's slope momentarily flatlines. These critical points could be local maxima (the peak of a hill), local minima (the bottom of a valley), or even saddle points (a bit of a mathematical oddity). So, how does this test work its magic?

The second derivative, denoted as f''(x), essentially tells us about the concavity of the function. Concavity, in simple terms, describes whether the function is curving upwards (like a smile) or downwards (like a frown). This information is key to classifying critical points.

• If f''(x) > 0 at a critical point, the function is concave up, indicating a local minimum. Imagine a smiling face – the bottom of the smile represents a minimum point.
• Conversely, if f''(x) < 0 at a critical point, the function is concave down, signaling a local maximum. Think of a frowning face – the peak of the frown is a maximum point.
• If f''(x) = 0 at a critical point, the test is inconclusive. This means we need to employ other methods to determine the nature of the point. It could be a saddle point, or the test might simply not be sensitive enough.

Finding the Critical Points: The First Step

To kick things off, we first need to find the critical points of our function, f(x) = 9ln(x) - 6x. Remember, critical points occur where the first derivative, f'(x), is either equal to zero or undefined. So, let's find the first derivative:

f'(x) = d/dx [9ln(x) - 6x] = 9/x - 6

Now, we need to solve for x when f'(x) = 0:

9/x - 6 = 0

9/x = 6

x = 9/6 = 3/2

We've found a potential critical point at x = 3/2. But wait, there's more! We also need to consider where f'(x) is undefined. Since f'(x) = 9/x - 6, it's undefined when x = 0. However, our original function, f(x) = 9ln(x) - 6x, is also undefined for x ≤ 0 due to the natural logarithm. Therefore, x = 0 is not in the domain of our function, and we don't need to consider it as a critical point.

So, we have one critical point to investigate: x = 3/2.

The Second Derivative Test in Action

Now comes the exciting part – applying the second derivative test! We need to find the second derivative of our function, f''(x):

f''(x) = d/dx [9/x - 6] = -9/x²

Now, let's evaluate f''(x) at our critical point, x = 3/2:

f''(3/2) = -9 / (3/2)² = -9 / (9/4) = -4

Aha! f''(3/2) = -4, which is less than 0. This means the function is concave down at x = 3/2, indicating a local maximum.

The Grand Finale: Finding the Local Maxima

We've successfully identified the x-coordinate of the local maximum. Now, let's find the corresponding y-coordinate by plugging x = 3/2 back into our original function:

f(3/2) = 9ln(3/2) - 6(3/2) = 9ln(3/2) - 9 ≈ -5.35

Therefore, the local maximum occurs at the point (3/2, 9ln(3/2) - 9).

What About Local Minima?

In our case, the second derivative test revealed a local maximum, but are there any local minima? Since our function has only one critical point and it corresponds to a local maximum, there are no local minima. The function increases up to x = 3/2 and then decreases, never forming a valley or a local minimum point.

Wrapping Up: Our Findings

So, after our journey through the world of derivatives and concavity, we've successfully used the second derivative test to analyze the function f(x) = 9ln(x) - 6x. Here's what we discovered:

• The function has a local maximum at x = 3/2.
• There are no local minima for this function.

More Examples and Deeper Dive

To solidify your understanding, let's explore a few more examples and delve into some nuances of the second derivative test.

Example 1: Unmasking a Local Minimum

Consider the function g(x) = x³ - 3x² + 1. Let's walk through the second derivative test to find its local extrema.

1. Find the first derivative:

g'(x) = 3x² - 6x

2. Find the critical points:

Set g'(x) = 0:

3x² - 6x = 0

3x(x - 2) = 0

This gives us critical points at x = 0 and x = 2.

3. Find the second derivative:

g''(x) = 6x - 6

4. Apply the second derivative test:

• g''(0) = 6(0) - 6 = -6 < 0, indicating a local maximum at x = 0.
• g''(2) = 6(2) - 6 = 6 > 0, indicating a local minimum at x = 2.

So, g(x) has a local maximum at x = 0 and a local minimum at x = 2.

When the Second Derivative Test Fails: Inconclusive Cases

As we mentioned earlier, the second derivative test isn't foolproof. There are cases where it falls short, particularly when f''(x) = 0 at a critical point. Let's examine one such scenario.

Consider the function h(x) = x⁴. Let's apply our test:

1. Find the first derivative:

h'(x) = 4x³

2. Find the critical points:

Set h'(x) = 0:

4x³ = 0

This gives us a critical point at x = 0.

3. Find the second derivative:

h''(x) = 12x²

4. Apply the second derivative test:

h''(0) = 12(0)² = 0

Uh oh! The second derivative test is inconclusive. So, what do we do?

In situations like this, we need to resort to other methods, such as the first derivative test or analyzing the function's behavior around the critical point. In this case, we can see that h(x) = x⁴ is always non-negative and has a minimum at x = 0. However, the second derivative test couldn't reveal this on its own.

The First Derivative Test: A Reliable Alternative

Speaking of the first derivative test, let's take a quick detour to understand its role in finding local extrema. The first derivative test relies on analyzing the sign changes of the first derivative, f'(x), around a critical point.

• If f'(x) changes from positive to negative at a critical point, it indicates a local maximum. The function is increasing before the point and decreasing after it.
• If f'(x) changes from negative to positive at a critical point, it indicates a local minimum. The function is decreasing before the point and increasing after it.
• If f'(x) does not change sign at a critical point, it's neither a local maximum nor a local minimum. This could be a saddle point or an inflection point.

In the case of h(x) = x⁴, we could use the first derivative test. We know h'(x) = 4x³. For x < 0, h'(x) < 0 (decreasing), and for x > 0, h'(x) > 0 (increasing). Thus, we have a local minimum at x = 0, confirming our earlier observation.

Real-World Applications: Why This Matters

Now, you might be wondering,