Hey guys, let's dive into the nitty-gritty of when the function f(x) = sin(3x)cos(3x) is heading upwards (increasing) and when it's going downwards (decreasing). This might sound a bit mathy, but trust me, once we break it down, it's totally manageable. We're going to be using our good old friend, calculus, to figure this out. Specifically, we'll be looking at the first derivative of the function. Remember, the sign of the first derivative tells us everything we need to know about whether a function is increasing or decreasing. If f'(x) > 0, the function is increasing, and if f'(x) < 0, it's decreasing. So, our main mission here is to find f'(x) and then analyze its sign over different intervals. It's like being a detective, and the derivative is our clue! We'll be using trigonometric identities too, because, let's be honest, they make things a whole lot simpler. Keep those eyes peeled, and let's get this done!

The Power of Trigonometric Identities

Before we even think about taking derivatives, let's make our lives easier by using a super handy trigonometric identity. You know that double angle formula for sine? It goes like this: sin(2θ) = 2sin(θ)cos(θ). Now, look closely at our function, f(x) = sin(3x)cos(3x). It looks almost like the right side of that identity, right? We just need a '2' in front! So, we can rewrite our function like this:

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

And voilà! Using the identity with θ = 3x, we get:

f(x) = (1/2)sin(2 * 3x) = (1/2)sin(6x)

See? This simplified form, f(x) = (1/2)sin(6x), is way easier to work with for differentiation and analysis. It's always a good strategy to simplify your functions before diving into calculations. It saves time and reduces the chances of making silly mistakes. This is a common trick in calculus problems involving trigonometric functions, so try to spot these opportunities whenever you can. It's like finding a shortcut on a long road trip!

Finding the First Derivative

Alright, now that we have our simplified function f(x) = (1/2)sin(6x), it's time to find its first derivative, f'(x). This is where calculus comes into play. We'll use the chain rule here. Remember the chain rule? If you have a function inside another function, you differentiate the outer function, multiply by the derivative of the inner function.

Here, our outer function is (1/2)sin(u) and our inner function is u = 6x.

The derivative of (1/2)sin(u) with respect to u is (1/2)cos(u).

The derivative of u = 6x with respect to x is 6.

So, applying the chain rule, f'(x) = (1/2)cos(6x) * 6.

And if we clean that up a bit, we get:

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

This is our crucial derivative! This expression, f'(x) = 3cos(6x), is what we'll use to determine the intervals where our original function is increasing or decreasing. Keep this result handy, as it's the key to unlocking the rest of our problem. Remember, differentiating correctly is half the battle. If you're ever unsure, it's always a good idea to double-check your derivative rules.

Determining Intervals of Increase and Decrease

Now for the main event, guys! We need to find out where f'(x) = 3cos(6x) is positive (for increasing) and where it's negative (for decreasing). To do this, we first need to find the critical points, which are the values of x where f'(x) = 0 or where f'(x) is undefined. Since 3cos(6x) is defined for all real x, we only need to solve 3cos(6x) = 0.

This simplifies to cos(6x) = 0.

We know that the cosine function equals zero at angles like π/2, 3π/2, 5π/2, and so on. In general, cos(θ) = 0 when θ = π/2 + nπ, where 'n' is any integer.

So, in our case, 6x = π/2 + nπ.

To find x, we divide both sides by 6:

x = π/12 + nπ/6

These are our critical values. They divide the number line into intervals. Now, we need to test a value of x from each interval in f'(x) = 3cos(6x) to see if the derivative is positive or negative. Let's consider the interval [0, 2π) for a complete cycle of the original function's behavior.

Here are some of the critical points within this interval (for n = 0, 1, 2, 3, 4, 5):

• n=0: x = π/12
• n=1: x = π/12 + π/6 = 3π/12 = π/4
• n=2: x = π/12 + 2π/6 = 5π/12
• n=3: x = π/12 + 3π/6 = 7π/12
• n=4: x = π/12 + 4π/6 = 9π/12 = 3π/4
• n=5: x = π/12 + 5π/6 = 11π/12
• n=6: x = π/12 + 6π/6 = 13π/12
• n=7: x = π/12 + 7π/6 = 15π/12 = 5π/4
• n=8: x = π/12 + 8π/6 = 17π/12
• n=9: x = π/12 + 9π/6 = 19π/12
• n=10: x = π/12 + 10π/6 = 21π/12 = 7π/4
• n=11: x = π/12 + 11π/6 = 23π/12

These points divide the interval [0, 2π) into subintervals. We'll pick a test value within each subinterval and plug it into f'(x) = 3cos(6x).

Let's analyze the sign of cos(6x). The function cos(θ) is positive in the first and fourth quadrants, and negative in the second and third quadrants. Since our 'θ' is '6x', we need to consider where '6x' falls within these quadrants.

Interval 1: (0, π/12)

Let's pick x = π/24. Then 6x = 6(π/24) = π/4. cos(π/4) is positive. So, f'(x) is positive. The function is increasing.

Interval 2: (π/12, π/4)

Let's pick x = π/8. Then 6x = 6(π/8) = 3π/4. cos(3π/4) is negative. So, f'(x) is negative. The function is decreasing.

Interval 3: (π/4, 5π/12)

Let's pick x = π/3. Then 6x = 6(π/3) = 2π. cos(2π) is positive. So, f'(x) is positive. The function is increasing.

Interval 4: (5π/12, 7π/12)

Let's pick x = π/2. Then 6x = 6(π/2) = 3π. cos(3π) is negative. So, f'(x) is negative. The function is decreasing.

We can see a pattern emerging here. The sign of f'(x) alternates between positive and negative as we move through the intervals determined by the critical points. This is directly related to the oscillating nature of the cosine function. The period of cos(6x) is 2π/6 = π/3. This means the pattern of increase and decrease will repeat every π/3.

To be thorough, let's list the general intervals for increase and decrease over the entire domain of real numbers. Based on x = π/12 + nπ/6 where n is an integer:

• Increasing Intervals: Where cos(6x) > 0. This happens when 6x is in the intervals (-π/2 + 2kπ, π/2 + 2kπ) for any integer k. So, -π/12 + 2kπ/6 < x < π/12 + 2kπ/6, which simplifies to -π/12 + kπ/3 < x < π/12 + kπ/3.

• Decreasing Intervals: Where cos(6x) < 0. This happens when 6x is in the intervals (π/2 + 2kπ, 3π/2 + 2kπ) for any integer k. So, π/12 + 2kπ/6 < x < 3π/12 + 2kπ/6, which simplifies to π/12 + kπ/3 < x < 5π/12 + kπ/3.

So, to summarize, the function f(x) = sin(3x)cos(3x), which we simplified to f(x) = (1/2)sin(6x), is increasing on the intervals (-π/12 + kπ/3, π/12 + kπ/3) and decreasing on the intervals (π/12 + kπ/3, 5π/12 + kπ/3), where 'k' is any integer. Pretty neat, right? We used calculus and trigonometry to map out the function's journey!

Visualizing the Behavior

To really get a grasp on this, guys, let's visualize what's happening. Remember, we simplified our original function f(x) = sin(3x)cos(3x) to f(x) = (1/2)sin(6x). This form tells us a lot. It's essentially a sine wave that's been scaled vertically by a factor of 1/2 and horizontally by a factor of 1/6. The amplitude is 1/2, meaning the peaks will reach 1/2 and the troughs will go down to -1/2. The period of this function is 2π / 6 = π/3. This means the entire wave pattern repeats every π/3 units along the x-axis.

When we found that f'(x) = 3cos(6x), we were looking at a cosine wave that's been scaled vertically by 3. The critical points we found, x = π/12 + nπ/6, are where this cosine wave crosses the x-axis (i.e., where its value is zero). These are precisely the points where the slope of our original sine wave changes from positive to negative (a peak) or negative to positive (a trough).

Let's think about the graph of y = cos(x). It starts at its maximum at x=0, crosses the x-axis at π/2, reaches its minimum at π, crosses the x-axis again at 3π/2, and completes a cycle at 2π. Now, consider y = cos(6x). The