Find Intervals Of Increasing And Decreasing F(x) = -2cos(x) - X

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In calculus, determining the intervals where a function is increasing or decreasing is a fundamental concept with wide-ranging applications. Understanding this behavior helps us sketch accurate graphs, optimize functions, and solve real-world problems. In this comprehensive guide, we will delve into the process of finding intervals of increase and decrease for a specific function, $f(x) = -2 \cos(x) - x$, defined on the interval $[0, \pi]$. This example will provide a clear roadmap for tackling similar problems, reinforcing your understanding of calculus principles.

Understanding Increasing and Decreasing Functions

Before we dive into the specifics, 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 its values increase as $x$ increases within that interval. Conversely, it's decreasing if its values decrease as $x$ increases. The key to identifying these intervals lies in the function's derivative, $f'(x)$. The derivative tells us the instantaneous rate of change of the function at any given point. A positive derivative indicates an increasing function, a negative derivative indicates a decreasing function, and a zero derivative suggests a critical point where the function might change direction.

The Role of the Derivative

The derivative, denoted as $f'(x)$, is the cornerstone of determining intervals of increasing and decreasing behavior. Here's a breakdown of how it works:

  • If $f'(x) > 0$ on an interval, then $f(x)$ is increasing on that interval.
  • If $f'(x) < 0$ on an interval, then $f(x)$ is decreasing on that interval.
  • If $f'(x) = 0$ or is undefined at a point, that point is a critical point and a potential location where the function changes from increasing to decreasing or vice versa.

Critical Points: The Turning Points

Critical points are the values of $x$ where the derivative, $f'(x)$, is either zero or undefined. These points are crucial because they often mark the transition between increasing and decreasing intervals. To find critical points, we first calculate the derivative of the function, set it equal to zero, and solve for $x$. We also need to identify any points where the derivative is undefined, as these can also be critical points. Once we've found the critical points, we'll use them to divide the domain of the function into intervals, which we will then test to determine whether the function is increasing or decreasing within each interval.

Step-by-Step Solution for $f(x) = -2cos(x) - x$ on $[0, π]$

Now, let's apply these concepts to our specific function, $f(x) = -2 \cos(x) - x$, defined on the interval $[0, \pi]$. We'll follow a systematic approach to find the intervals where the function is increasing and decreasing.

Step 1: Find the Derivative

The first step is to find the derivative of the function, $f'(x)$. Recall the derivative of $\cos(x)$ is $-\sin(x)$. Applying the rules of differentiation, we get:

f′(x)=ddx(−2cos⁡(x)−x)=−2(−sin⁡(x))−1=2sin⁡(x)−1f'(x) = \frac{d}{dx}(-2 \cos(x) - x) = -2(-\sin(x)) - 1 = 2\sin(x) - 1

Step 2: Find the Critical Points

Next, we need to find the critical points by setting the derivative equal to zero and solving for $x$:

2sin⁡(x)−1=02\sin(x) - 1 = 0

2sin⁡(x)=12\sin(x) = 1

sin⁡(x)=12\sin(x) = \frac{1}{2}

On the interval $[0, \pi]$, the solutions to this equation are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. These are our critical points.

Step 3: Create Intervals and Test Points

Now, we use the critical points to divide the interval $[0, \pi]$ into subintervals. Our critical points are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$, so we have three intervals to consider:

  • (0,π6)(0, \frac{\pi}{6})

  • (π6,5π6)(\frac{\pi}{6}, \frac{5\pi}{6})

  • (5π6,π)(\frac{5\pi}{6}, \pi)

To determine whether $f(x)$ is increasing or decreasing on each interval, we'll choose a test point within each interval and evaluate $f'(x)$ at that point. The sign of $f'(x)$ will tell us whether the function is increasing or decreasing.

Interval 1: $(0, \frac{\pi}{6})$

Let's choose a test point $x = \frac{\pi}{12}$. Then,

f′(π12)=2sin⁡(π12)−1≈2(0.2588)−1=−0.4824f'(\frac{\pi}{12}) = 2\sin(\frac{\pi}{12}) - 1 \approx 2(0.2588) - 1 = -0.4824

Since $f'(\frac{\pi}{12}) < 0$, the function $f(x)$ is decreasing on the interval $(0, \frac{\pi}{6})$.

Interval 2: $(\frac{\pi}{6}, \frac{5\pi}{6})$

Let's choose a test point $x = \frac{\pi}{2}$. Then,

f′(π2)=2sin⁡(π2)−1=2(1)−1=1f'(\frac{\pi}{2}) = 2\sin(\frac{\pi}{2}) - 1 = 2(1) - 1 = 1

Since $f'(\frac{\pi}{2}) > 0$, the function $f(x)$ is increasing on the interval $(\frac{\pi}{6}, \frac{5\pi}{6})$.

Interval 3: $(\frac{5\pi}{6}, \pi)$

Let's choose a test point $x = \frac{11\pi}{12}$. Then,

f′(11π12)=2sin⁡(11π12)−1≈2(0.2588)−1=−0.4824f'(\frac{11\pi}{12}) = 2\sin(\frac{11\pi}{12}) - 1 \approx 2(0.2588) - 1 = -0.4824

Since $f'(\frac{11\pi}{12}) < 0$, the function $f(x)$ is decreasing on the interval $(\frac{5\pi}{6}, \pi)$.

Step 4: Summarize the Results

Based on our analysis, we can conclude:

  • f(x)$ is decreasing on the interval $(0, \frac{\pi}{6})$.

  • f(x)$ is increasing on the interval $(\frac{\pi}{6}, \frac{5\pi}{6})$.

  • f(x)$ is decreasing on the interval $(\frac{5\pi}{6}, \pi)$.

Visualizing the Results

To further solidify our understanding, let's visualize these results. The graph of $f(x) = -2 \cos(x) - x$ on the interval $[0, \pi]$ would show a curve that decreases from $x = 0$ to $x = \frac{\pi}{6}$, then increases from $x = \frac{\pi}{6}$ to $x = \frac{5\pi}{6}$, and finally decreases from $x = \frac{5\pi}{6}$ to $x = \pi$. This graphical representation provides a clear visual confirmation of our calculations.

Practical Applications and Importance

Understanding intervals of increasing and decreasing functions has numerous practical applications. In optimization problems, we can use this knowledge to find the maximum and minimum values of a function, which is crucial in fields like engineering, economics, and computer science. In curve sketching, identifying these intervals helps us accurately represent the behavior of a function, providing valuable insights into its properties. Moreover, this concept forms the foundation for more advanced topics in calculus, such as concavity and inflection points.

Conclusion

In this comprehensive guide, we've explored the process of finding intervals of increasing and decreasing functions, using the example of $f(x) = -2 \cos(x) - x$ on the interval $[0, \pi]$. By calculating the derivative, finding critical points, and testing intervals, we successfully determined where the function is increasing and decreasing. This step-by-step approach provides a solid framework for analyzing the behavior of various functions. Remember, the derivative is your primary tool for unlocking the secrets of a function's behavior, and mastering this concept is essential for success in calculus and its applications. Understanding these intervals is not just a theoretical exercise; it's a practical skill that empowers you to solve real-world problems and gain a deeper appreciation for the power of calculus.