Increasing And Decreasing Intervals Of F(x) = 3cos(2x) On [-3π/2, 3π/2]

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Introduction

In this article, we will delve into the process of identifying the intervals where the function f(x) = 3cos(2x) is increasing and decreasing within the specified domain of [-3π/2, 3π/2]. This involves utilizing the principles of calculus, particularly the first derivative test, to analyze the function's behavior. Understanding where a function increases or decreases is crucial in various mathematical and real-world applications, such as optimization problems, curve sketching, and analyzing rates of change. By the end of this exploration, you will gain a comprehensive understanding of how to determine these intervals for trigonometric functions and apply these techniques to other functions as well.

Understanding Increasing and Decreasing Functions

Before we dive into the specifics of our function, let’s solidify our understanding of what it means for a function to be increasing or decreasing. A function f(x) is said to be increasing on an interval if, for any two points x₁ and x₂ in that interval where x₁ < x₂, we have f(x₁) < f(x₂). In simpler terms, as x moves to the right, the value of f(x) goes up. Conversely, a function f(x) is said to be decreasing on an interval if, for any two points x₁ and x₂ in that interval where x₁ < x₂, we have f(x₁) > f(x₂). Here, as x moves to the right, the value of f(x) goes down.

The first derivative of a function, denoted as f'(x), provides a powerful tool for determining these intervals. The sign of the first derivative tells us whether the function is increasing or decreasing. If f'(x) > 0 on an interval, the function is increasing on that interval. If f'(x) < 0 on an interval, the function is decreasing on that interval. Points where f'(x) = 0 or is undefined are called critical points, and they often mark the transition between increasing and decreasing intervals. These critical points are essential for our analysis, as they help us partition the domain into intervals where the function's behavior is consistent.

Step 1: Find the First Derivative

To determine the intervals where the function f(x) = 3cos(2x) is increasing and decreasing, the first crucial step is to find its first derivative, f'(x). We will use the chain rule, which is a fundamental concept in calculus for differentiating composite functions. The chain rule states that if we have a function f(g(x)), then its derivative is f'(g(x)) * g'(x). In our case, we have f(x) = 3cos(2x), which can be seen as a composite function where the outer function is 3cos(u) and the inner function is u = 2x.

Applying the chain rule, we first differentiate the outer function with respect to u, which gives us -3sin(u). Then, we differentiate the inner function u = 2x with respect to x, which gives us 2. Multiplying these results together, we get:

f'(x) = -3sin(2x) * 2 = -6sin(2x)

This first derivative, f'(x) = -6sin(2x), is the key to understanding the increasing and decreasing behavior of our original function. By analyzing the sign of f'(x) over the given interval [-3π/2, 3π/2], we can pinpoint where f(x) is increasing and where it is decreasing. This process involves identifying the critical points, where f'(x) = 0 or is undefined, and then testing the sign of f'(x) in the intervals between these critical points.

Step 2: Find the Critical Points

Now that we have the first derivative, f'(x) = -6sin(2x), the next step is to find the critical points. Critical points are the values of x where the derivative is either equal to zero or undefined. These points are crucial because they often mark the transitions between intervals where the function is increasing and intervals where it is decreasing. In other words, critical points are potential turning points of the function's graph. To find these points, we set the derivative equal to zero and solve for x:

-6sin(2x) = 0

Dividing both sides by -6, we get:

sin(2x) = 0

To solve this trigonometric equation, we need to find the values of 2x for which the sine function is zero. We know that sin(θ) = 0 when θ is an integer multiple of π, i.e., θ = nπ, where n is an integer. Therefore, we have:

2x = nπ

Dividing both sides by 2, we get:

x = nπ/2

This gives us a general solution for the critical points. However, we are only interested in the critical points that lie within our specified interval of [-3π/2, 3π/2]. To find these specific critical points, we substitute integer values for n and check if the resulting x falls within our interval. Let's consider different values of n:

  • If n = -3, then x = -3π/2
  • If n = -2, then x = -π
  • If n = -1, then x = -π/2
  • If n = 0, then x = 0
  • If n = 1, then x = π/2
  • If n = 2, then x = π
  • If n = 3, then x = 3π/2

So, the critical points within the interval [-3π/2, 3π/2] are x = -3π/2, -π, -π/2, 0, π/2, π, 3π/2. These points will help us divide the interval into subintervals where we can analyze the sign of the first derivative.

Step 3: Create Intervals and Test the Sign of f'(x)

With the critical points identified, we can now divide the given interval [-3π/2, 3π/2] into subintervals. These subintervals are bounded by the critical points, and within each subinterval, the sign of the first derivative f'(x) will be consistent (either positive or negative). This consistency is crucial because it tells us whether the function f(x) is increasing or decreasing throughout that subinterval. The critical points we found were x = -3π/2, -π, -π/2, 0, π/2, π, 3π/2. This divides our interval into the following subintervals:

  1. [-3π/2, -π)
  2. (-π, -π/2)
  3. (-π/2, 0)
  4. (0, π/2)
  5. (π/2, π)
  6. (π, 3π/2]

Now, we need to test the sign of f'(x) = -6sin(2x) in each of these intervals. To do this, we choose a test value within each interval and evaluate f'(x) at that point. The sign of f'(x) at the test value will indicate the sign of f'(x) throughout the entire interval.

Let's perform the test for each interval:

  1. Interval: [-3π/2, -π)

    • Test value: x = -5π/4
    • f'(-5π/4) = -6sin(2(-5π/4)) = -6sin(-5π/2) = -6(-1) = 6 > 0
    • Conclusion: f(x) is increasing on [-3π/2, -π).

  2. Interval: (-π, -π/2)

    • Test value: x = -3π/4
    • f'(-3π/4) = -6sin(2(-3π/4)) = -6sin(-3π/2) = -6(1) = -6 < 0
    • Conclusion: f(x) is decreasing on (-π, -π/2).

  3. Interval: (-π/2, 0)

    • Test value: x = -π/4
    • f'(-π/4) = -6sin(2(-π/4)) = -6sin(-π/2) = -6(-1) = 6 > 0
    • Conclusion: f(x) is increasing on (-π/2, 0).

  4. Interval: (0, π/2)

    • Test value: x = π/4
    • f'(π/4) = -6sin(2(π/4)) = -6sin(π/2) = -6(1) = -6 < 0
    • Conclusion: f(x) is decreasing on (0, π/2).

  5. Interval: (π/2, π)

    • Test value: x = 3π/4
    • f'(3π/4) = -6sin(2(3π/4)) = -6sin(3π/2) = -6(-1) = 6 > 0
    • Conclusion: f(x) is increasing on (π/2, π).

  6. Interval: (π, 3π/2]

    • Test value: x = 5π/4
    • f'(5π/4) = -6sin(2(5π/4)) = -6sin(5π/2) = -6(1) = -6 < 0
    • Conclusion: f(x) is decreasing on (π, 3π/2].

Step 4: Summarize the Intervals

After testing the sign of the first derivative in each subinterval, we can now summarize the intervals where the function f(x) = 3cos(2x) is increasing and decreasing on the interval [-3π/2, 3π/2]. This summary provides a clear and concise understanding of the function's behavior across its domain.

Increasing Intervals:

Based on our analysis, the function f(x) = 3cos(2x) is increasing on the following intervals:

  • [-3π/2, -π)
  • (-π/2, 0)
  • (π/2, π)

On these intervals, the first derivative f'(x) = -6sin(2x) is positive, indicating that the function's value increases as x increases. This means that the graph of the function is sloping upwards in these intervals.

Decreasing Intervals:

Similarly, the function f(x) = 3cos(2x) is decreasing on the following intervals:

  • (-π, -π/2)
  • (0, π/2)
  • (π, 3π/2]

On these intervals, the first derivative f'(x) = -6sin(2x) is negative, indicating that the function's value decreases as x increases. This means that the graph of the function is sloping downwards in these intervals.

Conclusion

In conclusion, by applying the principles of calculus and the first derivative test, we have successfully identified the intervals where the function f(x) = 3cos(2x) is increasing and decreasing on the interval [-3π/2, 3π/2]. This process involved finding the first derivative, identifying critical points, dividing the domain into subintervals, and testing the sign of the derivative within each subinterval. The resulting intervals provide a comprehensive understanding of the function's behavior, allowing us to visualize its graph and predict its values across the given domain. This method can be applied to a wide range of functions, making it a fundamental tool in calculus and mathematical analysis. Understanding the increasing and decreasing behavior of functions is essential in various applications, from optimization problems to modeling real-world phenomena.

By mastering these techniques, you are well-equipped to analyze the behavior of various functions and apply this knowledge to solve practical problems.