Hey there, math enthusiasts! Ever found yourself staring at a trigonometric function like sin(3x)cos(3x) and wondering, "Is this thing going up or down?" You're not alone, guys! Figuring out the increasing and decreasing behavior of functions is a super crucial skill in calculus, and today, we're going to dive deep into how to tackle this for our friend, sin(3x)cos(3x). Get ready to unravel the mysteries of its slope!

Understanding the Basics: What Does Increasing and Decreasing Mean?

Before we get our hands dirty with sin(3x)cos(3x), let's quickly recap what it means for a function to be increasing or decreasing. Imagine you're walking along the graph of a function. If you're generally moving upwards as you move from left to right, the function is increasing. If you're moving downwards, it's decreasing. Mathematically, this is all about the derivative of the function. The derivative, often denoted as f'(x) or dy/dx, tells us the instantaneous rate of change, or the slope, of the function at any given point.

So, the golden rule is: if the derivative f'(x) is positive over an interval, the function f(x) is increasing on that interval. Conversely, if the derivative f'(x) is negative over an interval, the function f(x) is decreasing on that interval. If the derivative is zero, we've hit a critical point – a potential peak or valley, where the function might change its direction. These critical points are super important for finding the intervals of increase and decrease.

Now, you might be thinking, "Okay, cool, but what does this have to do with sin(3x)cos(3x)?" Well, everything! To analyze the increasing and decreasing nature of sin(3x)cos(3x), we need to find its derivative and then determine where that derivative is positive or negative. It sounds straightforward, but with trig functions, especially with that '3x' inside, things can get a little more involved. We'll need to use our trusty differentiation rules and some clever trigonometric identities to make our lives easier. So, buckle up, because we're about to go on a calculus adventure!

Simplifying sin(3x)cos(3x) with Trigonometric Identities

Alright, guys, before we jump into differentiation, let's see if we can make our function, sin(3x)cos(3x), a bit more manageable. Dealing with a product of sine and cosine can be a bit fiddly. Luckily, there's a fantastic trigonometric identity that can help us simplify this. Remember the double angle formula for sine? It states that sin(2θ) = 2sin(θ)cos(θ). If we rearrange this, we get sin(θ)cos(θ) = (1/2)sin(2θ).

Now, let's apply this to our function. If we let θ = 3x, then our function sin(3x)cos(3x) becomes (1/2)sin(2 * 3x), which simplifies beautifully to (1/2)sin(6x). See? Much cleaner! This simplified form, f(x) = (1/2)sin(6x), will make finding the derivative and analyzing its sign way easier. It's like finding a shortcut on a long road trip – definitely worth it!

Why is this simplification so powerful? Because now, instead of dealing with the product rule for differentiation (which can be a bit more work), we're dealing with a simple constant multiplied by a sine function. The core behavior of increasing and decreasing is determined by the sine part, and the (1/2) factor just scales the amplitude. The '6x' inside the sine function tells us about the period and frequency of the oscillations, which will directly impact the intervals where our function increases and decreases. So, keep this simplified form in mind; it's going to be our best friend for the rest of this analysis.

Thinking about this simplification also gives us a hint about the overall behavior of the function. We know that the sine function oscillates between -1 and 1. Therefore, (1/2)sin(6x) will oscillate between -1/2 and 1/2. This means our original function sin(3x)cos(3x) will also have a range of [-1/2, 1/2]. The '6x' term compresses the graph horizontally, meaning it will complete its cycles faster than a regular sin(x) function. Understanding these basic properties helps us anticipate the kind of increasing and decreasing patterns we might find. It's all about building a solid foundation before we start digging into the calculus!

Finding the Derivative of sin(3x)cos(3x)

Okay, fam, we've simplified our function to f(x) = (1/2)sin(6x). Now comes the moment of truth: finding the derivative. This is where the magic of calculus really shines. We'll use the chain rule, which is essential when you have a function inside another function – exactly what we have with sin(6x).

The chain rule states that the derivative of a composite function f(g(x)) is f'(g(x)) * g'(x). In our case, the outer function is (1/2)sin(u) and the inner function is u = 6x. The derivative of (1/2)sin(u) with respect to u is (1/2)cos(u). The derivative of the inner function u = 6x with respect to x is simply 6.

Putting it all together using the chain rule:

f'(x) = d/dx [(1/2)sin(6x)]

f'(x) = (1/2) * [d/dx sin(6x)]

Now, apply the chain rule to sin(6x):

d/dx sin(6x) = cos(6x) * d/dx(6x)

d/dx sin(6x) = cos(6x) * 6

So, substituting this back:

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

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

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

And there you have it! The derivative of sin(3x)cos(3x), which we simplified to (1/2)sin(6x), is f'(x) = 3cos(6x). This derivative is the key to unlocking the secrets of our function's increasing and decreasing intervals. Remember, we're looking for where this derivative is positive (function increasing) and where it's negative (function decreasing). So, our next big mission is to analyze the sign of 3cos(6x).

It's important to double-check our work here. If we had tried to differentiate the original sin(3x)cos(3x) using the product rule, we would have gotten:

d/dx [sin(3x)cos(3x)] = [d/dx sin(3x)]cos(3x) + sin(3x)[d/dx cos(3x)]

Using the chain rule for each part:

d/dx sin(3x) = cos(3x) * 3 = 3cos(3x)

d/dx cos(3x) = -sin(3x) * 3 = -3sin(3x)

So, the derivative would be:

[3cos(3x)]cos(3x) + sin(3x)[-3sin(3x)] = 3cos²(3x) - 3sin²(3x) = 3(cos²(3x) - sin²(3x))

Now, recall another trigonometric identity: cos(2θ) = cos²(θ) - sin²(θ). If we let θ = 3x, then cos(2 * 3x) = cos²(3x) - sin²(3x), which means cos(6x) = cos²(3x) - sin²(3x).

Substituting this back, we get:

3(cos²(3x) - sin²(3x)) = 3cos(6x).

Wow! We got the exact same derivative, 3cos(6x), using both methods. This gives us a lot of confidence in our answer and reinforces the power of using trigonometric identities to simplify problems. So, the derivative is definitely 3cos(6x)!

Determining Intervals of Increase and Decrease

Alright, mathletes, we've got our derivative: f'(x) = 3cos(6x). Now, the real detective work begins: finding where our original function sin(3x)cos(3x) (or (1/2)sin(6x)) is increasing or decreasing. Remember the rule: the function is increasing when f'(x) > 0, and decreasing when f'(x) < 0.

So, we need to solve the inequality 3cos(6x) > 0 for increasing intervals and 3cos(6x) < 0 for decreasing intervals. The critical points occur when f'(x) = 0, meaning 3cos(6x) = 0. This simplifies to cos(6x) = 0.

Let's think about the cosine function. Cosine is zero at odd multiples of π/2. So, we need:

6x = π/2 + nπ, where 'n' is any integer (..., -2, -1, 0, 1, 2, ...).

To find the values of x, we divide by 6:

x = (π/12) + (nπ/6)

These values of x are our critical points. They divide the number line into intervals where the derivative will either be consistently positive or consistently negative. Let's list a few of these critical points to get a feel for them:

• For n = 0: x = π/12
• For n = 1: x = π/12 + π/6 = π/12 + 2π/12 = 3π/12 = π/4
• For n = 2: x = π/12 + 2π/6 = π/12 + 4π/12 = 5π/12
• For n = 3: x = π/12 + 3π/6 = π/12 + 6π/12 = 7π/12
• For n = 4: x = π/12 + 4π/6 = π/12 + 8π/12 = 9π/12 = 3π/4

And so on. These points are where the slope of our function is zero. Now we need to check the sign of f'(x) = 3cos(6x) in the intervals between these critical points.

Consider the behavior of cos(u). It's positive in quadrants I and IV, and negative in quadrants II and III. Our argument is u = 6x. The period of cos(u) is 2π. The period of cos(6x) is 2π/6 = π/3. This means the pattern of increase and decrease will repeat every π/3 units.

Let's analyze the sign of cos(6x) over one period of cos(u), say from u = 0 to u = 2π. This corresponds to 6x from 0 to 2π, so x from 0 to π/3.

• Interval 1: Let's pick a test value between x = 0 and x = π/12. Let x = π/24. Then 6x = 6(π/24) = π/4. cos(π/4) = √2/2, which is positive. So, f'(x) = 3 * (positive) = positive. Thus, f(x) is increasing on (0, π/12).

• Interval 2: Let's pick a test value between x = π/12 and x = π/4. Let x = π/8. Then 6x = 6(π/8) = 3π/4. cos(3π/4) = -√2/2, which is negative. So, f'(x) = 3 * (negative) = negative. Thus, f(x) is decreasing on (π/12, π/4).

• Interval 3: Let's pick a test value between x = π/4 and x = 5π/12. Let x = 3π/16. Then 6x = 6(3π/16) = 9π/8. cos(9π/8) is negative (it's in the third quadrant). So, f'(x) = 3 * (negative) = negative. Wait, this doesn't seem right. Let's re-evaluate the angles where cos(6x)=0. They are π/12, 3π/12, 5π/12, 7π/12, 9π/12...

Let's reconsider the general solution for cos(u) = 0: u = π/2, 3π/2, 5π/2, 7π/2, ...

So, 6x = π/2, 3π/2, 5π/2, 7π/2, ...

Which gives:

x = π/12, 3π/12 (π/4), 5π/12, 7π/12, 9π/12 (3π/4), 11π/12, ...

Let's test intervals within one period of cos(6x), which is π/3. For example, from x=0 to x=π/3.

• Interval 1: (0, π/12). Let x = π/24. 6x = π/4. cos(π/4) > 0. f'(x) > 0. Increasing.
• Interval 2: (π/12, π/4). Let x = π/8. 6x = 3π/4. cos(3π/4) < 0. f'(x) < 0. Decreasing.
• Interval 3: (π/4, 5π/12). Let x = 3π/16. 6x = 9π/8. cos(9π/8) < 0. Hmm, still negative. Let's check the critical points again. The points where cos(6x) = 0 are indeed π/12, 3π/12, 5π/12, etc.

Okay, let's think about the graph of cos(u). It starts positive, goes negative, then positive again. So cos(6x) will be positive, then negative, then positive. We need to find the boundaries of these intervals.

cos(6x) > 0 when 6x is in the intervals (-π/2 + 2nπ, π/2 + 2nπ). Divide by 6:

x is in (-π/12 + nπ/3, π/12 + nπ/3).

cos(6x) < 0 when 6x is in the intervals (π/2 + 2nπ, 3π/2 + 2nπ). Divide by 6:

x is in (π/12 + nπ/3, 3π/12 + nπ/3) = (π/12 + nπ/3, π/4 + nπ/3).

Let's test for n=0:

• Increasing interval: (-π/12, π/12).
• Decreasing interval: (π/12, π/4).

Now let's test for n=1:

• Increasing interval: (-π/12 + π/3, π/12 + π/3) = (-π/12 + 4π/12, π/12 + 4π/12) = (3π/12, 5π/12) = (π/4, 5π/12).
• Decreasing interval: (π/12 + π/3, π/4 + π/3) = (5π/12, 7π/12).

This pattern continues. So, the function f(x) = (1/2)sin(6x) is:

Increasing on the intervals (-π/12 + nπ/3, π/12 + nπ/3) for any integer n. Decreasing on the intervals (π/12 + nπ/3, 5π/12 + nπ/3) for any integer n.

Let's verify this with our earlier test points. For n=0, we have increasing on (-π/12, π/12) and decreasing on (π/12, π/4). For n=1, we have increasing on (π/4, 5π/12) and decreasing on (5π/12, 7π/12). This matches the behavior of the cosine wave repeating its positive and negative segments.

Visualizing the Behavior

To really nail this down, guys, let's visualize what's happening. Our original function is sin(3x)cos(3x), which we simplified to f(x) = (1/2)sin(6x). This is a sine wave with an amplitude of 1/2 and a period of 2π/6 = π/3. The '6' inside the sine function means it oscillates much faster than a standard sin(x) function.

We found that the derivative, f'(x) = 3cos(6x), is positive when cos(6x) is positive. The cosine function is positive in its