Hey guys! Let's dive into figuring out where the function f(x) = sin(3x)cos(3x) is increasing and decreasing. This involves a bit of calculus, specifically finding the derivative and analyzing its sign. So, buckle up, and let's get started!

Understanding the Function

Before we jump into the calculus, it's helpful to understand what our function looks like. The function f(x) = sin(3x)cos(3x) is a product of two trigonometric functions, each with a period that's been compressed by a factor of 3 compared to the standard sin(x) and cos(x). This means our function will oscillate more rapidly. To properly analyze where it's increasing or decreasing, we'll need to use the product rule and chain rule when taking the derivative.

Remember, a function is increasing where its derivative is positive and decreasing where its derivative is negative. The points where the derivative is zero or undefined are critical points, which could be local maxima, local minima, or saddle points. We'll find these critical points and then test intervals around them to determine the function's behavior.

First, we will find the derivative of the function f(x) = sin(3x)cos(3x). Then, we will need to use trigonometric identities to simplify the derivative to make it easier to analyze. After simplification, we'll set the derivative equal to zero and solve for x to find the critical points. Once we have the critical points, we will create a number line and test values in each interval to determine where the derivative is positive (increasing) and where it is negative (decreasing).

Lastly, it is very important to clearly state the intervals where the function is increasing and decreasing, and also identify any local maxima or minima that occur at the critical points. So, let's roll up our sleeves and begin this exciting calculus journey! Get ready to sharpen those pencils, because we are about to make math magic happen!

Calculating the Derivative

Okay, so the first step to figuring out where sin(3x)cos(3x) is increasing or decreasing is to find its derivative. We'll use the product rule, which states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). In our case, u(x) = sin(3x) and v(x) = cos(3x).

The derivative of sin(3x) is 3cos(3x) (using the chain rule), and the derivative of cos(3x) is -3sin(3x) (again, using the chain rule). Now, we plug these into the product rule formula:

f'(x) = (3cos(3x))cos(3x) + sin(3x)(-3sin(3x)) f'(x) = 3cos²(3x) - 3sin²(3x)

Now, we can factor out the 3:

f'(x) = 3(cos²(3x) - sin²(3x))

Recognize that cos²(3x) - sin²(3x) is the same as cos(2 * 3x) which simplifies to cos(6x). This is a super handy trig identity that makes our lives much easier!

So, our derivative simplifies to:

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

Isn't that neat? We started with a seemingly complex expression, applied the product rule and a trigonometric identity, and ended up with something much simpler. This derivative, 3cos(6x), will be much easier to analyze. Now we know that finding when f(x) is increasing or decreasing is equivalent to finding when 3cos(6x) is positive or negative.

Finding Critical Points

Alright, now that we have the derivative f'(x) = 3cos(6x), the next step is to find the critical points. Critical points occur where the derivative is either equal to zero or undefined. In this case, 3cos(6x) is never undefined, so we only need to find where it equals zero.

So, we set 3cos(6x) = 0, which simplifies to cos(6x) = 0. Now, we need to think about where cosine is zero. Cosine is zero at π/2 and 3π/2, and then every π radians after that. So, we have:

6x = π/2 + nπ, where n is an integer.

To solve for x, we divide everything by 6:

x = π/12 + nπ/6, where n is an integer.

This gives us a series of critical points. We can list out a few of them to get a sense of what's happening:

• 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

And so on. These critical points divide the number line into intervals, which we'll test to see where the function is increasing and decreasing. It's like setting up dominoes; once we know the behavior in one interval, we can predict the behavior in the adjacent intervals, given the periodic nature of the cosine function.

Determining Intervals of Increase and Decrease

Now comes the crucial part: figuring out where our function is increasing and decreasing. We do this by testing the sign of the derivative f'(x) = 3cos(6x) in the intervals between our critical points.

Let's pick some test values in each interval and plug them into f'(x):

• Interval 1: (0, π/12)
  • Test value: x = π/24
  • f'(π/24) = 3cos(6 * π/24) = 3cos(π/4) = 3 * (√2/2) > 0. So, the function is increasing in this interval.
• Interval 2: (π/12, π/4)
  • Test value: x = π/6
  • f'(π/6) = 3cos(6 * π/6) = 3cos(π) = 3 * (-1) < 0. So, the function is decreasing in this interval.
• Interval 3: (π/4, 5π/12)
  • Test value: x = π/3
  • f'(π/3) = 3cos(6 * π/3) = 3cos(2π) = 3 * (1) > 0. So, the function is increasing in this interval.
• Interval 4: (5π/12, 7π/12)
  • Test value: x = π/2
  • f'(π/2) = 3cos(6 * π/2) = 3cos(3π) = 3 * (-1) < 0. So, the function is decreasing in this interval.

We can see a pattern emerging. The function alternates between increasing and decreasing in each interval. This makes sense because the cosine function oscillates between positive and negative values.

Generalizing the Intervals:

The function f(x) = sin(3x)cos(3x) is increasing when 3cos(6x) > 0, which means cos(6x) > 0. This occurs when:

• 6x is in the intervals (-π/2 + 2nπ, π/2 + 2nπ), where n is an integer.
  • Dividing by 6, we get: x is in the intervals (-π/12 + nπ/3, π/12 + nπ/3)

The function is decreasing when 3cos(6x) < 0, which means cos(6x) < 0. This occurs when:

• 6x is in the intervals (π/2 + 2nπ, 3π/2 + 2nπ), where n is an integer.
  • Dividing by 6, we get: x is in the intervals (π/12 + nπ/3, π/4 + nπ/3)

So, to summarize:

• Increasing: x ∈ (-π/12 + nπ/3, π/12 + nπ/3)
• Decreasing: x ∈ (π/12 + nπ/3, π/4 + nπ/3)

Final Answer

Okay, after all that calculus and trigonometry, we've arrived at the final answer. The function f(x) = sin(3x)cos(3x) is:

• Increasing on the intervals (-π/12 + nπ/3, π/12 + nπ/3)
• Decreasing on the intervals (π/12 + nπ/3, π/4 + nπ/3)

Where n is any integer. This means the function oscillates between increasing and decreasing behavior in a periodic manner. The critical points we found earlier are the locations of the local maxima and minima. At x = π/12 + nπ/3, we have local maxima, and at x = π/4 + nπ/3, we have local minima.

In Conclusion:

We successfully found the intervals where sin(3x)cos(3x) is increasing and decreasing by using the product rule to find the derivative, simplifying the derivative with trigonometric identities, finding the critical points by setting the derivative equal to zero, and testing intervals around the critical points to determine the sign of the derivative. Great job, guys! You nailed it!