Finding Critical Points Of G(x) = Sin(3x)

Hey Plastik Magazine readers! Let's dive into some calculus fun, specifically finding the critical points of a trigonometric function. We're looking at g(x) = sin(3x), but only for values of x between 0 and π (inclusive). This means we're focusing on a specific part of the sine wave. Critical points are super important because they often pinpoint where a function might have a maximum, a minimum, or a spot where it changes direction. Ready to break it down?

Understanding Critical Points and Derivatives

First things first, what exactly are critical points? Well, they're the points where the derivative of the function is either equal to zero or doesn't exist. The derivative, which we denote as g'(x) in this case, represents the instantaneous rate of change of the function. Think of it as the slope of the tangent line at any given point on the curve. If the slope is zero, the tangent line is horizontal – a classic sign of a potential peak or valley (a maximum or minimum). If the derivative doesn't exist, it usually means there's a sharp corner, a cusp, or a vertical tangent at that point. Since we're dealing with the sine function (and a scaled version of it), we can be pretty sure that our derivative will exist everywhere within our interval [0, π].

To find these critical points, we need to get our hands dirty with some differentiation. The derivative of sin(x) is cos(x), but we have sin(3x). This calls for the chain rule, a fundamental concept in calculus. The chain rule states that the derivative of a composite function (a function within a function) is the derivative of the outer function, evaluated at the inner function, multiplied by the derivative of the inner function. In our case, the 'outer' function is the sine, and the 'inner' function is 3x. So, let's get down to it and find the derivative of our function.

The derivative of g(x) = sin(3x) is g'(x) = 3cos(3x). This is because the derivative of sin(u) is cos(u), and the derivative of 3x is 3. We then multiply the two results. Now, to locate our critical points, we set the derivative equal to zero and solve for x: 3cos(3x) = 0. Solving this equation is the next step to finding where the critical points exist. This is the heart of the problem, where we apply our understanding of derivatives and trigonometry.

Now, how do we solve 3cos(3x) = 0? We need to find the values of x that make the cosine function equal to zero. Remember the unit circle? Cosine corresponds to the x-coordinate. Cosine is zero at π/2 and 3π/2. However, we're dealing with 3x inside the cosine function, and our interval for x is [0, π]. So we need to figure out what values of 3x give us π/2 and 3π/2 (and any other values within the appropriate range). Let's start with 3x = π/2. Solving for x gives us x = π/6. Does this value of x fall within our interval [0, π]? Absolutely! This means that x = π/6 is one of our critical points. This is where g'(x) = 0.

Then, let's consider 3x = 3π/2. This gives us x = π/2, which is also within our interval. So, x = π/2 is another critical point. Now, we need to think about whether there are any other solutions to the equation 3cos(3x) = 0 within our given interval [0, π]. Remember, the cosine function repeats itself. Cosine is also zero at 5π/2, 7π/2, and so on. Let's see what happens if 3x = 5π/2. This gives us x = 5π/6, which is also within our interval. So, x = 5π/6 is also a critical point. If we go further with 3x = 7π/2, we'll get x = 7π/6, which is greater than π and outside our interval. Therefore, x = 7π/6 is not a solution.

So, we now have all our potential critical points. The next step is to examine each option presented and check if they match our findings. The critical points are, therefore, those values of x that make the derivative g'(x) equal to zero or undefined. In this case, g'(x) is always defined; therefore, we only need to consider the places where g'(x) = 0.

Evaluating the Answer Choices

Okay, let's see which of the provided answer choices are correct. We found that the critical points occur when 3cos(3x) = 0. We found these three critical points within the interval [0, π]: x = π/6, x = π/2, and x = 5π/6. Now, let's examine the options provided in the prompt.

A. x = π/6: We already determined that this is a critical point. So, this answer choice is correct.

B. x = π/2: This is also one of our critical points! So, this answer choice is correct.

C. x = 5π/6: Yes, this is another critical point we found! This answer choice is also correct.

D. g has no critical points: This is definitely incorrect. We've found multiple critical points.

Therefore, the correct answers are A, B, and C. Easy peasy!

Conclusion: Critical Points Unveiled!

So there you have it, guys! We've successfully navigated the world of derivatives and identified the critical points of g(x) = sin(3x) within the given interval. Remember, finding critical points is a crucial step in understanding the behavior of a function. These points help us locate potential maximums, minimums, and points of inflection. Keep practicing, and you'll become a critical point pro in no time! This journey underscores how essential calculus is for exploring the intricacies of functions. Understanding critical points offers a window into the behavior of the function, revealing its increasing, decreasing, or stationary nature. Keep exploring, keep learning, and keep the mathematical spirit alive! You got this!