Justify Why The Function With $f^{\prime}(x) = 2 \cos X + 2 \cos X \sin X$ Is Decreasing From $\frac{\pi}{2}$ To $\frac{3 \pi}{2}$.A. $f^{\prime}(x) \ \textgreater \ 0$ For $\left(\frac{\pi}{2}, \frac{3

Feb 26, 2025 by ADMIN 205 views

Justifying the Decreasing Nature of a Function

In calculus, the behavior of a function can be understood by analyzing its derivative. The derivative of a function represents the rate of change of the function with respect to its input. In this article, we will justify why the function with derivative $f^{\prime}(x) = 2 \cos x + 2 \cos x \sin x$ is decreasing from $\frac{\pi}{2}$ to $\frac{3 \pi}{2}$ .

The given derivative is $f^{\prime}(x) = 2 \cos x + 2 \cos x \sin x$ . To understand why this function is decreasing, we need to analyze the behavior of its derivative. The derivative represents the rate of change of the function, and if the derivative is negative, the function is decreasing.

To analyze the derivative, we can rewrite it as $f^{\prime}(x) = 2 \cos x (1 + \sin x)$ . This form makes it easier to understand the behavior of the derivative.

The derivative $f^{\prime}(x)$ is a product of two functions: $2 \cos x$ and $(1 + \sin x)$ . The behavior of the derivative depends on the behavior of these two functions.

Behavior of $2 \cos x$ : The function $2 \cos x$ is a cosine function with amplitude 2. The cosine function oscillates between -1 and 1, and its amplitude is 1. In this case, the amplitude is 2, so the function oscillates between -2 and 2.
Behavior of $(1 + \sin x)$ : The function $(1 + \sin x)$ is a sine function with amplitude 1, shifted upwards by 1. The sine function oscillates between -1 and 1, and its amplitude is 1. In this case, the function oscillates between 0 and 2.

Now that we have analyzed the behavior of the two functions, we can combine them to understand the behavior of the derivative.

Behavior of $2 \cos x (1 + \sin x)$ : The derivative $f^{\prime}(x)$ is a product of the two functions $2 \cos x$ and $(1 + \sin x)$ . Since both functions are oscillating, the product will also oscillate. However, the amplitude of the product will be the product of the amplitudes of the two functions, which is $2 \cdot 2 = 4$ .

Now that we have analyzed the behavior of the derivative, we can justify why the function is decreasing from $\frac{\pi}{2}$ to $\frac{3 \pi}{2}$ .

Behavior of the derivative in the given interval: In the interval $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ , the derivative $f^{\prime}(x)$ is negative. This is because the function $2 \cos x (1 + \sin x)$ is oscillating between -4 and 4, and in this interval, it is always negative.
Conclusion: Since the derivative $f^{\prime}(x)$ is negative in the interval $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ , the function is decreasing in this interval.

In conclusion, we have justified why the function with derivative $f^{\prime}(x) = 2 \cos x + 2 \cos x \sin x$ is decreasing from $\frac{\pi}{2}$ to $\frac{3 \pi}{2}$ . We analyzed the behavior of the derivative and showed that it is negative in the given interval, which means that the function is decreasing in this interval.

[1] Calculus, 3rd edition, Michael Spivak
[2] Calculus, 2nd edition, James Stewart

The following is a proof of the statement that the derivative $f^{\prime}(x)$ is negative in the interval $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ .

Let $x \in \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ . Then, we have:

f^{\prime}(x) = 2 \cos x (1 + \sin x)

Since $x \in \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ , we have:

\cos x < 0

and

\sin x > 0

Therefore, we have:

f^{\prime}(x) = 2 \cos x (1 + \sin x) < 0

This shows that the derivative $f^{\prime}(x)$ is negative in the interval $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ .

The final answer is that the function with derivative $f^{\prime}(x) = 2 \cos x + 2 \cos x \sin x$ is decreasing from $\frac{\pi}{2}$ to $\frac{3 \pi}{2}$ .
Q&A: Justifying the Decreasing Nature of a Function

In our previous article, we justified why the function with derivative $f^{\prime}(x) = 2 \cos x + 2 \cos x \sin x$ is decreasing from $\frac{\pi}{2}$ to $\frac{3 \pi}{2}$ . In this article, we will answer some frequently asked questions related to this topic.

A: The derivative of a function represents the rate of change of the function with respect to its input. It is a crucial tool in understanding the behavior of a function, including its increasing or decreasing nature.

A: To determine if a function is increasing or decreasing, you need to analyze its derivative. If the derivative is positive, the function is increasing. If the derivative is negative, the function is decreasing.

A: The derivative of a function represents the rate of change of the function. It is a measure of how fast the function is changing at a given point.

A: Yes, the function $f(x) = -x^2$ is a classic example of a decreasing function. Its derivative is $f^{\prime}(x) = -2x$ , which is negative for all $x > 0$ .

A: There are several methods to find the derivative of a function, including:

Power rule: If $f(x) = x^n$ , then $f^{\prime}(x) = nx^{n-1}$ .
Product rule: If $f(x) = u(x)v(x)$ , then $f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x)$ .
Quotient rule: If $f(x) = \frac{u(x)}{v(x)}$ , then $f^{\prime}(x) = \frac{u^{\prime}(x)v(x) - u(x)v^{\prime}(x)}{v(x)^2}$ .

A: The interval $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ is significant because it is the interval in which the function is decreasing. The function's derivative is negative in this interval, which means that the function is decreasing.

A: Yes, the following graph shows the function $f(x) = 2 \cos x + 2 \cos x \sin x$ and its derivative $f^{\prime}(x) = 2 \cos x (1 + \sin x)$ .

[Insert graph here]

In conclusion, we have answered some frequently asked questions related to the decreasing nature of a function. We hope that this article has provided a better understanding of the topic and has helped to clarify any doubts.

[1] Calculus, 3rd edition, Michael Spivak
[2] Calculus, 2nd edition, James Stewart