Hey guys! Let's dive into understanding when the function sin(3x)cos(3x) is increasing or decreasing. This involves a bit of calculus, so buckle up! We'll break it down step by step to make it super clear.

Understanding the Basics

Before we jump into the specifics of sin(3x)cos(3x), let's quickly recap some fundamental concepts. Knowing these will make the whole process much smoother. First, remember what it means for a function to be increasing or decreasing. A function is increasing on an interval if its values go up as you move from left to right along the x-axis. Mathematically, this means that if x₁ < x₂, then f(x₁) < f(x₂). Conversely, a function is decreasing if its values go down as you move from left to right, meaning that if x₁ < x₂, then f(x₁) > f(x₂). Calculus gives us a powerful tool to determine where a function is increasing or decreasing: the derivative. The derivative of a function, often denoted as f'(x), tells us the slope of the tangent line at any point on the function's graph. If f'(x) > 0 on an interval, the function is increasing on that interval. If f'(x) < 0, the function is decreasing. If f'(x) = 0, the function has a horizontal tangent, which could be a local maximum, a local minimum, or a saddle point. To find the intervals where a function is increasing or decreasing, we first find the derivative, then find the critical points (where the derivative is zero or undefined), and finally test the intervals between these critical points to determine the sign of the derivative. Also, let's not forget the trigonometric identities that often come in handy when dealing with sine and cosine functions. One identity that's particularly useful for simplifying expressions involving sin(x)cos(x) is the double angle formula: sin(2x) = 2sin(x)cos(x). We'll see how this plays a role shortly. Derivatives of trigonometric functions are also essential. The derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x). When dealing with functions like sin(3x) and cos(3x), we need to apply the chain rule, which states that the derivative of f(g(x)) is f'(g(x)) * g'(x). For example, the derivative of sin(3x) is cos(3x) * 3 = 3cos(3x), and the derivative of cos(3x) is -sin(3x) * 3 = -3sin(3x). Keeping these basics in mind will help us tackle the problem of finding where sin(3x)cos(3x) is increasing or decreasing with greater confidence and clarity. Now, let's get to the heart of the matter!

Step 1: Simplify the Function

Okay, so we have the function f(x) = sin(3x)cos(3x). The first thing we can do to make our lives easier is to use a trig identity. Remember that sin(2θ) = 2sin(θ)cos(θ)? We can rewrite our function using this identity. Notice that sin(3x)cos(3x) looks a lot like the right side of our identity if we let θ = 3x. To get it into the exact form, we need a 2 in front. So, let’s multiply and divide by 2:

f(x) = (1/2) * 2 * sin(3x)cos(3x)

Now we can apply the identity:

f(x) = (1/2) * sin(2 * 3x)

f(x) = (1/2)sin(6x)

This simplified form is much easier to work with! Simplifying the function sin(3x)cos(3x) into f(x) = (1/2)sin(6x) makes subsequent steps much more manageable. By leveraging the trigonometric identity sin(2θ) = 2sin(θ)cos(θ), we transform the original product of sine and cosine functions into a single sine function with a scaled argument. This simplification is crucial because it reduces the complexity of the derivative, which we need to find in the next step. Without this simplification, we would have to deal with the product rule when finding the derivative, which can be more prone to errors and require more algebraic manipulation. For instance, the derivative of sin(3x)cos(3x) using the product rule would involve differentiating both sin(3x) and cos(3x) and then combining them, whereas the derivative of (1/2)sin(6x) only requires applying the chain rule once. Moreover, the simplified form allows us to easily identify the critical points of the function, which are essential for determining the intervals where the function is increasing or decreasing. The critical points occur where the derivative is equal to zero or undefined. By simplifying to (1/2)sin(6x), finding these points becomes a straightforward algebraic exercise. Thus, simplifying the function not only reduces computational complexity but also enhances our ability to analyze the behavior of the function effectively.

Step 2: Find the Derivative

Now that we have f(x) = (1/2)sin(6x), let's find its derivative, f'(x). Remember the chain rule! The derivative of sin(u) is cos(u) * du/dx.

f'(x) = (1/2) * cos(6x) * 6

f'(x) = 3cos(6x)

So, the derivative is f'(x) = 3cos(6x). Finding the derivative of the simplified function f(x) = (1/2)sin(6x) is a crucial step in determining the intervals where the function is increasing or decreasing. The derivative, denoted as f'(x), represents the rate of change of the function at any given point x. In this case, we found that f'(x) = 3cos(6x). The process of finding this derivative involves applying the chain rule, which is a fundamental concept in calculus. The chain rule states that if you have a composite function, like f(g(x)), its derivative is f'(g(x)) * g'(x). In our scenario, we have the function (1/2)sin(6x), where the outer function is (1/2)sin(u) and the inner function is u = 6x. The derivative of the outer function with respect to u is (1/2)cos(u), and the derivative of the inner function 6x with respect to x is 6. Multiplying these together gives us (1/2)cos(6x) * 6, which simplifies to 3cos(6x). This derivative, 3cos(6x), is essential because it tells us the slope of the tangent line at any point on the graph of f(x). When f'(x) > 0, the function is increasing, and when f'(x) < 0, the function is decreasing. The points where f'(x) = 0 are critical points, which can indicate local maxima, local minima, or saddle points. Understanding and accurately calculating the derivative is therefore vital for analyzing the behavior of the function and determining its increasing and decreasing intervals.

Step 3: Find Critical Points

Critical points occur where f'(x) = 0 or is undefined. In our case, f'(x) = 3cos(6x). Cosine is never undefined, so we just need to find where 3cos(6x) = 0.

3cos(6x) = 0

cos(6x) = 0

Now, we need to find the values of 6x for which the cosine is zero. We know that cos(θ) = 0 when θ = π/2 + nπ, where n is an integer.

So, 6x = π/2 + nπ

x = (π/2 + nπ) / 6

x = π/12 + nπ/6

These are our critical points. Finding the critical points of the derivative f'(x) = 3cos(6x) is a pivotal step in identifying the intervals where the original function f(x) = (1/2)sin(6x) is either increasing or decreasing. Critical points are the x-values where the derivative is either equal to zero or undefined. These points are significant because they mark potential turning points of the function, where it transitions from increasing to decreasing or vice versa. In our case, the derivative f'(x) = 3cos(6x) is never undefined, so we only need to find where it equals zero. Setting 3cos(6x) = 0 leads to cos(6x) = 0. The cosine function equals zero at angles of the form π/2 + nπ, where n is an integer. Therefore, we set 6x = π/2 + nπ and solve for x, obtaining x = (π/2 + nπ) / 6 or x = π/12 + nπ/6. These x-values represent the critical points of the function. It's important to note that since the cosine function is periodic, there are infinitely many critical points. We typically focus on a specific interval, such as [0, 2π], to analyze the behavior of the function within that range. Once we have identified the critical points, we use them to divide the interval into subintervals. We then test a value within each subinterval to determine the sign of the derivative in that interval. This allows us to determine whether the function is increasing or decreasing in each subinterval, providing a comprehensive understanding of the function's behavior. Accurate identification of critical points is thus essential for correctly analyzing the function's intervals of increase and decrease.

Step 4: Test Intervals

Now we need to test the intervals between our critical points to see where f'(x) is positive (increasing) or negative (decreasing). Let's consider the interval [0, 2π] for simplicity. Our critical points are x = π/12 + nπ/6. We need to find the critical points within [0, 2π].

For n = 0: x = π/12 For n = 1: x = π/12 + π/6 = 3π/12 = π/4 For n = 2: x = π/12 + 2π/6 = 5π/12 For n = 3: x = π/12 + 3π/6 = 7π/12 For n = 4: x = π/12 + 4π/6 = 9π/12 = 3π/4 For n = 5: x = π/12 + 5π/6 = 11π/12 For n = 6: x = π/12 + 6π/6 = 13π/12 For n = 7: x = π/12 + 7π/6 = 15π/12 = 5π/4 For n = 8: x = π/12 + 8π/6 = 17π/12 For n = 9: x = π/12 + 9π/6 = 19π/12 For n = 10: x = π/12 + 10π/6 = 21π/12 = 7π/4 For n = 11: x = π/12 + 11π/6 = 23π/12 For n = 12: x = π/12 + 12π/6 = 25π/12 > 2π (so we stop here)

Now we have our critical points: π/12, π/4, 5π/12, 7π/12, 3π/4, 11π/12, 13π/12, 5π/4, 17π/12, 19π/12, 7π/4, 23π/12.

Let's test the intervals:

1. (0, π/12): Choose x = π/24. f'(π/24) = 3cos(6 * π/24) = 3cos(π/4) = 3 * (√2/2) > 0. Increasing.
2. (π/12, π/4): Choose x = π/6. f'(π/6) = 3cos(6 * π/6) = 3cos(π) = -3 < 0. Decreasing.
3. (π/4, 5π/12): Choose x = π/3. f'(π/3) = 3cos(6 * π/3) = 3cos(2π) = 3 > 0. Increasing.
4. (5π/12, 7π/12): Choose x = π/2. f'(π/2) = 3cos(6 * π/2) = 3cos(3π) = -3 < 0. Decreasing.
5. (7π/12, 3π/4): Choose x = 2π/3. f'(2π/3) = 3cos(6 * 2π/3) = 3cos(4π) = 3 > 0. Increasing.
6. (3π/4, 11π/12): Choose x = 5π/6. f'(5π/6) = 3cos(6 * 5π/6) = 3cos(5π) = -3 < 0. Decreasing.
7. (11π/12, 13π/12): Choose x = π. f'(π) = 3cos(6π) = 3 > 0. Increasing.
8. (13π/12, 5π/4): Choose x = 7π/6. f'(7π/6) = 3cos(6 * 7π/6) = 3cos(7π) = -3 < 0. Decreasing.
9. (5π/4, 17π/12): Choose x = 4π/3. f'(4π/3) = 3cos(6 * 4π/3) = 3cos(8π) = 3 > 0. Increasing.
10. (17π/12, 19π/12): Choose x = 3π/2. f'(3π/2) = 3cos(6 * 3π/2) = 3cos(9π) = -3 < 0. Decreasing.
11. (19π/12, 7π/4): Choose x = 5π/3. f'(5π/3) = 3cos(6 * 5π/3) = 3cos(10π) = 3 > 0. Increasing.
12. (7π/4, 23π/12): Choose x = 11π/6. f'(11π/6) = 3cos(6 * 11π/6) = 3cos(11π) = -3 < 0. Decreasing.
13. (23π/12, 2π): Choose x = 47π/24. f'(47π/24) = 3cos(6 * 47π/24) = 3cos(47π/4) = 3 * (√2/2) > 0. Increasing.

Step 5: Conclusion

So, in the interval [0, 2π], the function f(x) = sin(3x)cos(3x) is increasing on the intervals:

(0, π/12), (π/4, 5π/12), (7π/12, 3π/4), (11π/12, 13π/12), (5π/4, 17π/12), (19π/12, 7π/4), (23π/12, 2π)

And decreasing on the intervals:

(π/12, π/4), (5π/12, 7π/12), (3π/4, 11π/12), (13π/12, 5π/4), (17π/12, 19π/12), (7π/4, 23π/12)

That's it! You've successfully determined the intervals where sin(3x)cos(3x) is increasing and decreasing. Remember, the key steps are simplifying the function, finding the derivative, finding critical points, and testing the intervals. Keep practicing, and you'll get the hang of it! Determining the intervals where f(x) = sin(3x)cos(3x) is increasing and decreasing involves a systematic approach using calculus. First, we simplified the function to f(x) = (1/2)sin(6x) using the trigonometric identity sin(2θ) = 2sin(θ)cos(θ). Next, we found the derivative f'(x) = 3cos(6x) using the chain rule. Then, we identified the critical points by setting f'(x) = 0, which gave us x = π/12 + nπ/6, where n is an integer. To analyze the function's behavior within the interval [0, 2π], we found all the critical points within this interval: π/12, π/4, 5π/12, 7π/12, 3π/4, 11π/12, 13π/12, 5π/4, 17π/12, 19π/12, 7π/4, 23π/12. These critical points divide the interval into subintervals. For each subinterval, we chose a test value and evaluated the sign of f'(x) at that value. If f'(x) > 0, the function is increasing in that interval, and if f'(x) < 0, the function is decreasing. By testing each subinterval, we determined the intervals of increase and decrease: Increasing on (0, π/12), (π/4, 5π/12), (7π/12, 3π/4), (11π/12, 13π/12), (5π/4, 17π/12), (19π/12, 7π/4), (23π/12, 2π) and decreasing on (π/12, π/4), (5π/12, 7π/12), (3π/4, 11π/12), (13π/12, 5π/4), (17π/12, 19π/12), (7π/4, 23π/12). This methodical process ensures an accurate understanding of the function's increasing and decreasing behavior over the specified interval.