Analyzing Cosine Function Symmetry And Intervals

Hey everyone, let's dive into analyzing the cosine function: $v=\frac{1}{4} \cos \left(-\frac{1}{4} x\right)$. Our goal is to figure out which statements about its graph are true. Sounds fun, right? We'll break it down step by step, so you can understand the properties of this function, focusing on symmetry and intervals where it increases or decreases. Let's get started! This function is a transformed cosine function, and understanding its properties requires knowing how these transformations affect the original cosine graph. The general form of a cosine function is $y = A$cos(B(x - C))}$ + D$, where * A affects the amplitude. * B affects the period. * C affects the phase shift. * D affects the vertical shift. In our given function, $v=\frac{1{4} \cos \left(-\frac{1}{4} x\right)$, we have $A = \frac{1}{4}$, $B = -\frac{1}{4}$, $C = 0$, and $D = 0$. The amplitude is $\frac{1}{4}$, which means the graph oscillates between $-\frac{1}{4}$ and $\frac{1}{4}$. The negative sign inside the cosine function is crucial for determining symmetry, and the coefficient of $x$ affects the period. Since we have a cosine function, the properties of symmetry are vital to examine. Cosine is an even function, which means $\cos(-x) = \cos(x)$. This property is essential because it affects the symmetry of the graph. The negative sign inside the cosine function doesn't change the function's value. This is because the cosine function is even, meaning it is symmetric about the y-axis. Let's dig in and check it out together, guys!

Symmetry About the y-axis

Okay, so the first statement we're looking at is whether the graph is symmetric about the y-axis. Symmetry about the y-axis means that if you fold the graph along the y-axis, the two sides will match perfectly. To check this, we can use the property of even functions. A function is even if $f(-x) = f(x)$ for all x in its domain. The cosine function is a classic example of an even function. In our case, we have $v=\frac{1}{4} \cos \left(-\frac{1}{4} x\right)$. Because of the properties of the cosine function where $\cos(-x) = \cos(x)$, the function effectively becomes $v=\frac{1}{4} \cos \left(\frac{1}{4} x\right)$. Thus, the function is indeed symmetric about the y-axis. The negative sign inside the cosine function doesn't affect the symmetry because cosine is an even function. So, the answer to the first question is yes. The function $v$ is symmetric about the y-axis. This symmetry is a direct consequence of the property that $\cos(-x) = \cos(x)$. This means that for every point $(x, y)$ on the graph, there's also a point $(-x, y)$. It's like looking in a mirror; the y-axis acts as the mirror line. This is a crucial property to understand because it helps us visualize and predict the behavior of the graph. Any function of the form $f(x) = \cos(kx)$, where k is a constant, will be symmetric about the y-axis. The transformation doesn't alter the essential characteristic of the cosine function. Therefore, the graph will always display symmetry along the y-axis. It's a fundamental property of the cosine function that we need to acknowledge.

To confirm this, let's consider some specific values. If $x = 2$, then $v = \frac{1}{4} \cos(-\frac{1}{2}) = \frac{1}{4} \cos(\frac{1}{2})$. If $x = -2$, then $v = \frac{1}{4} \cos(\frac{1}{2})$. So the values are equal at $x$ and $-x$, thus confirming the symmetry. This symmetry property simplifies the graph's analysis, enabling us to concentrate on one side of the y-axis to understand the entire function's behavior. The practical application of the symmetry property also comes to light when we analyze real-world scenarios and applications.

Function Behavior on the Interval (0, 4π)

Now, let's discuss the function's behavior on the interval $(0, 4\pi)$. Specifically, we need to figure out whether the function decreases on this interval. The period of the function $v=\frac{1}{4} \cos \left(-\frac{1}{4} x\right)$ is determined by the coefficient of x inside the cosine function. The general formula for the period of a cosine function is $\frac{2\pi}{|B|}$, where B is the coefficient of x. In our case, B = -1/4 (or 1/4, because the negative sign doesn't change the period). Therefore, the period $T = \frac{2\pi}{\frac{1}{4}} = 8\pi$. The interval $(0, 4\pi)$ is a portion of one period of the function. The standard cosine function $\cos(x)$ decreases from $x = 0$ to $x = \pi$, then increases from $x = \pi$ to $x = 2\pi$. In our transformed function, the period is $8\pi$. Therefore, the function completes one full cycle from $0$ to $8\pi$. We want to examine if the function decreases over the interval $(0, 4\pi)$. Now, let's consider the behavior of the function $v=\frac{1}{4} \cos \left(-\frac{1}{4} x\right)$ on the interval $(0, 4\pi)$. Notice that, because of the negative sign inside the cosine, the function essentially becomes $\frac{1}{4} \cos \left(\frac{1}{4} x\right)$. So, we can focus on how the regular cosine function behaves and apply the transformation. The cosine function starts at its maximum value at $x = 0$ and decreases until it reaches its minimum. The cosine function decreases from 0 to π. So we'll figure out where we get the $\pi$ point in our modified function. Because the period is $8\pi$, the half-period is $4\pi$. The function completes half a cycle over the interval $(0, 4\pi)$. Consequently, the function decreases from its maximum value at $x = 0$ to its minimum value. Therefore, on the interval $(0, 4\pi)$, the function does not simply decrease; instead, it completes half a cycle. This means the function decreases over a portion of the interval, but it doesn't decrease over the entire range. So, the function does not decrease throughout the entire interval $(0, 4\pi)$. It decreases in the first half of the interval but increases in the second half. When examining functions over an interval, it is very important to determine the complete cycle. If the interval fits in one complete cycle, we can determine whether the function is increasing or decreasing. If it doesn't, like our case, we should consider the interval's specific position on the graph.

Breaking it Down Further

Let's break down the period of the function $v = \frac{1}{4} \cos \left(-\frac{1}{4} x\right)$ further: * Period: $8\pi$ * Interval of Consideration: $(0, 4\pi)$ The interval $(0, 4\pi)$ is half of the period $(0, 8\pi)$. The cosine function decreases from 0 to π, so it decreases in a portion of the interval $(0, 4\pi)$. Specifically, it decreases from $x = 0$ to $x = 4\pi$. So the function does not decrease on the entire interval $(0, 4\pi)$. Therefore, the function decreases on the interval $(0, 4\pi)$.

Final Verdict, Guys!

So, to summarize our analysis: * The graph is symmetric about the y-axis - TRUE. * The function decreases on the interval $(0, 4\pi)$ - TRUE. Both statements are accurate. I hope this detailed explanation was helpful, and if you have any questions, don't hesitate to ask! Keep practicing, and you'll become a math pro in no time. Analyzing functions can be both fun and rewarding. Keep up the great work! And don't be afraid to break down complex concepts into manageable chunks. Analyzing functions is one of those skills that gets better with practice, so keep at it, and you'll be a pro in no time!