Finding Intervals For Sin(3x)cos(3x) Growth And Decay

Hey guys, ever found yourself staring at a function like sin(3x)cos(3x) and wondering, "When is this thing going up, and when is it going down?" You're in the right place! Today, we're diving deep into how to figure out the increasing and decreasing intervals for this trigonometric beast. It might look a little intimidating at first, but trust me, with a few calculus tricks up our sleeves, we'll conquer it together. We're going to break down the process step-by-step, making sure you understand why we do each thing. So, grab your favorite study buddy, maybe a coffee, and let's get this party started. We'll cover everything from simplifying the function to finding critical points and ultimately defining those crucial intervals where our function is on the rise or taking a nosedive. Get ready to boost your understanding of function behavior and ace those calculus problems!

Understanding the Core Concept: Derivatives and Monotonicity

Alright, before we get lost in the trigonometric weeds, let's quickly chat about the fundamental idea behind finding increasing and decreasing intervals. The magic word here is derivative. Remember, the derivative of a function, f'(x), tells us about the slope of the original function, f(x), at any given point. If the derivative f'(x) > 0 on an interval, it means the slope is positive, and thus, the original function f(x) is increasing on that interval. Conversely, if f'(x) < 0, the slope is negative, and f(x) is decreasing. If f'(x) = 0, we've hit a critical point – these are the potential turning points where the function might switch from increasing to decreasing, or vice versa.

Now, when we're dealing with a function like sin(3x)cos(3x), we need to find its derivative first. But before we jump into differentiation, there's a neat little trigonometric identity that can make our lives a whole lot easier. Remember the double-angle identity 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). Look familiar? Our function sin(3x)cos(3x) fits this pattern perfectly if we let θ = 3x. So, we can rewrite our original function as:

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

See? Much simpler to work with! This transformation is key because it reduces the complexity of the function, making the differentiation process more straightforward and less prone to errors. It's like finding a shortcut on a long hike; you get to the same destination, but with less effort. This simplification is a crucial first step in analyzing the function's behavior, setting the stage for us to find where it's growing and where it's shrinking.

Differentiating the Simplified Function

Okay, guys, we've simplified our function to f(x) = (1/2)sin(6x). Now comes the calculus part: finding the derivative, f'(x). This is where we apply the chain rule. The derivative of sin(u) is cos(u), and the derivative of 6x with respect to x is 6. So, applying the chain rule:

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

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

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

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

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

So, our derivative is f'(x) = 3cos(6x). This is the tool we'll use to determine when our original function is increasing or decreasing. Remember, we're looking for intervals where f'(x) > 0 (increasing) and where f'(x) < 0 (decreasing). The points where f'(x) = 0 are our critical points, the boundaries between these intervals. It's like finding the peaks and valleys on a roller coaster track; they mark where the ride changes direction. This derivative, 3cos(6x), is our key to unlocking the secrets of sin(3x)cos(3x)'s journey up and down the graph. Keep this handy as we move on to the next crucial step!

Finding Critical Points: Where the Action Happens

Now that we have our derivative, f'(x) = 3cos(6x), the next step is to find the critical points. These are the points where the derivative is either zero or undefined. For f'(x) = 3cos(6x), it's defined for all real numbers, so we only need to find where f'(x) = 0.

Setting the derivative to zero:

3cos(6x) = 0

Dividing by 3:

cos(6x) = 0

Now, we need to recall the values of θ for which cos(θ) = 0. These occur at π/2, 3π/2, 5π/2, and so on. In general, cos(θ) = 0 when θ = π/2 + nπ, where n is any integer.

In our case, θ = 6x. So, we set 6x equal to these values:

6x = π/2 + nπ

To find x, we divide both sides by 6:

x = (π/2 + nπ) / 6

x = π/12 + nπ/6

These are our critical points! Let's list a few of them by plugging in different integer values for n:

• If n = 0, x = π/12
• If n = 1, x = π/12 + π/6 = π/12 + 2π/12 = 3π/12 = π/4
• If n = 2, x = π/12 + 2π/6 = π/12 + 4π/12 = 5π/12
• If n = 3, x = π/12 + 3π/6 = π/12 + 6π/12 = 7π/12
• If n = -1, x = π/12 - π/6 = π/12 - 2π/12 = -π/12

These critical points divide the number line into intervals. Within each interval, the sign of the derivative f'(x) will be constant. This means the original function f(x) will either be increasing or decreasing throughout that entire interval. Finding these points is like marking the important milestones on a map; they tell us where potential changes in direction occur, guiding us to the next phase of our analysis.

Determining Intervals of Increase and Decrease

With our critical points in hand – x = π/12 + nπ/6 – we can now determine the increasing and decreasing intervals. These critical points act as boundaries. We need to test the sign of the derivative, f'(x) = 3cos(6x), in the intervals between these points. Let's consider a typical interval, for example, between x = π/12 and x = π/4 (which corresponds to n=0 and n=1).

The interval is (π/12, π/4). We need to pick a test value within this interval. A good choice would be x = π/6. Let's plug this into our derivative:

f'(π/6) = 3cos(6 * π/6) = 3cos(π)

Since cos(π) = -1, we have:

f'(π/6) = 3 * (-1) = -3

Since f'(π/6) = -3, which is negative, our original function f(x) is decreasing on the interval (π/12, π/4).

Let's try another interval, say between x = π/4 and x = 5π/12 (corresponding to n=1 and n=2). We'll pick a test value, perhaps x = π/3.

f'(π/3) = 3cos(6 * π/3) = 3cos(2π)

Since cos(2π) = 1, we have:

f'(π/3) = 3 * 1 = 3

Since f'(π/3) = 3, which is positive, our original function f(x) is increasing on the interval (π/4, 5π/12).

We continue this process for all intervals defined by our critical points. The general pattern for cos(θ) is that it's positive in the first and fourth quadrants and negative in the second and third quadrants. Since our argument is 6x, the sign of cos(6x) will alternate as 6x crosses π/2, 3π/2, 5π/2, etc. This means the sign of f'(x) will alternate between intervals defined by our critical points.

To summarize the pattern based on f'(x) = 3cos(6x):

• f'(x) > 0 (Increasing) when cos(6x) > 0. This happens when 6x is in intervals like (-π/2 + 2nπ, π/2 + 2nπ). Dividing by 6, we get x in intervals like (-π/12 + nπ/3, π/12 + nπ/3).
• f'(x) < 0 (Decreasing) when cos(6x) < 0. This happens when 6x is in intervals like (π/2 + 2nπ, 3π/2 + 2nπ). Dividing by 6, we get x in intervals like (π/12 + nπ/3, π/4 + nπ/3).

By systematically testing intervals or recognizing the pattern of the cosine function, we can map out precisely where sin(3x)cos(3x) is climbing and where it's falling. This is the core of understanding function behavior!

Visualizing the Behavior: Graphing the Function

So, we've done the heavy lifting with calculus: simplified the function, found its derivative, identified critical points, and determined the intervals where sin(3x)cos(3x) is increasing and decreasing. But how does this all look visually? Graphing the function can really solidify your understanding. Remember, we simplified f(x) = sin(3x)cos(3x) 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.

Let's recall our critical points, which are the points where the derivative is zero. These points correspond to the local maxima and minima of the function. Our critical points are x = π/12 + nπ/6.

• At x = π/12 (n=0), f(π/12) = (1/2)sin(6 * π/12) = (1/2)sin(π/2) = 1/2. This is a local maximum.
• At x = π/4 (n=1), f(π/4) = (1/2)sin(6 * π/4) = (1/2)sin(3π/2) = -1/2. This is a local minimum.
• At x = 5π/12 (n=2), f(5π/12) = (1/2)sin(6 * 5π/12) = (1/2)sin(5π/2) = 1/2. This is a local maximum.

Our analysis told us:

• The function is decreasing on intervals like (π/12, π/4).
• The function is increasing on intervals like (π/4, 5π/12).

If you were to plot f(x) = (1/2)sin(6x), you'd see a smooth, wave-like curve. Between x = π/12 and x = π/4, the graph would be sloping downwards, hitting a low point at x = π/4. Then, from x = π/4 to x = 5π/12, the graph would be sloping upwards, reaching a peak at x = 5π/12. This pattern repeats due to the periodic nature of the sine function.

Visualizing this helps immensely. You can see the peaks (maxima) where the function stops increasing and starts decreasing, and the valleys (minima) where it stops decreasing and starts increasing. The points where the function crosses the x-axis (where f(x) = 0) are also interesting, but for monotonicity, we focus on the points where the slope f'(x) changes sign. The graph is your friend in confirming the results derived from calculus. It provides an intuitive feel for the function's dynamics, making the abstract calculus concepts more concrete and understandable. So, don't shy away from sketching it out!

Conclusion: Mastering Monotonicity

So there you have it, guys! We’ve successfully navigated the world of increasing and decreasing intervals for the function sin(3x)cos(3x). By employing the power of trigonometric identities to simplify the function to (1/2)sin(6x), we made the subsequent calculus steps much more manageable. We found the derivative, f'(x) = 3cos(6x), identified the critical points where f'(x) = 0, and then systematically tested intervals to determine where the function's slope was positive (increasing) or negative (decreasing).

Remember the key takeaway: the derivative is your roadmap to understanding a function's behavior. A positive derivative means the function is climbing, and a negative derivative means it's falling. The critical points are simply the places where the direction might change. Whether you're solving homework problems, preparing for an exam, or just expanding your mathematical horizons, mastering the concept of monotonicity is a fundamental skill in calculus. Keep practicing, and don't hesitate to use graphing tools to visualize these concepts. It makes all the difference! You've got this!