Based on the table you provided, we can observe the values of \( f(x) \) at the specified points. The values are as follows:
- \( f(-\frac{\pi}{4}) = 0 \)
- \( f(\frac{\pi}{4}) = 1 \)
- \( f(\frac{3\pi}{4}) = 0 \)
- \( f(\frac{5\pi}{4}) = -1 \)
- \( f(\frac{7\pi}{4}) = 0 \)
We also notice that the values are repeating every \( \pi \). Analyzing the structure of the values, we can see that they correspond to the sine function evaluated at certain angles, specifically \( f(x) = \sin(2x) \).
Let’s confirm this:
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For \( x = -\frac{\pi}{4} \): \[ f(-\frac{\pi}{4}) = \sin(2 \cdot -\frac{\pi}{4}) = \sin(-\frac{\pi}{2}) = -1 \quad \text{(but we have 0)} \]
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For other points:
- \( f(\frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1 \)
- \( f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{2}) = -1 \)
- \( f(\frac{5\pi}{4}) = \sin(\frac{5\pi}{2}) = 1 \)
- \( f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{2}) = 1\)
Given that there is an alternating pattern it resembles a square wave function.
To express this based on the intervals given, we can articulate the function as: \[ f(x) = \begin{cases} 0 & \text{for } x = -\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4} \ 1 & \text{for } x = \frac{\pi}{4} \ -1 & \text{for } x = \frac{5\pi}{4} \ 0 & \text{for } x = \frac{7\pi}{4} \ \end{cases} \]
The pattern repeats every \( \pi \), so you can notice that this function can also be represented using a combination of sine or cosine, modified to achieve this stepping behavior.
Thus, the final answer can be formulated as a piecewise function or can be approximated as an oscillating function with periodic behavior. However, it looks primarily like a square wave with periods where \( k \in \mathbb{Z} \) for the intervals provided.