QUESTION IMAGE
Question
graph in radian measure
nding period in radia
5
multiple choice 1 point
find the period of the graph shown below.
2π/1/2 = 2π·2/1 = 4π
2/3π
4π
1/2π
2π
Step1: Understand the concept of period
The period of a periodic function is the horizontal length over which the function repeats its pattern.
Step2: Count the number of cycles in a known interval
Looking at the x - axis, from \(x = 0\) to \(x = 2\pi\), we can see that there are 3 full cycles.
Let the period be \(T\). If \(n\) is the number of cycles in an interval \(L\), then \(nT=L\). Here \(n = 3\) and \(L=2\pi\). But wait, if we consider the general formula for a sinusoidal function \(y = A\sin(Bx + C)+D\) or \(y=A\cos(Bx + C)+D\), the period \(T=\frac{2\pi}{|B|}\). Another way: visually, if we assume the standard form and count the distance between two consecutive identical points (e.g., two consecutive peaks or troughs).
Looking at the graph, if we consider the fact that for a function \(y = f(x)\), if \(f(x + T)=f(x)\) for all \(x\) in the domain. From the graph, if we start at a point (say a peak) and move to the next peak, we can observe that the distance between two consecutive peaks (one full cycle) is \(\frac{2\pi}{3}\). Wait, no, let's re - examine.
Wait, actually, if we consider the formula for the period of a sinusoidal - like function. Let's use the visual approach: the period is the length of one full cycle. If we assume the function is of the form \(y = A\sin(Bx)\) (ignoring phase shift and vertical shift for simplicity). The general formula for the period \(T=\frac{2\pi}{B}\).
Looking at the graph, if we count the number of cycles in an interval. Suppose we take the interval from \(x = 0\) to \(x = 2\pi\). If we assume the function repeats its pattern. Wait, no, another approach: the period is the horizontal distance over which the function completes one full oscillation.
If we look at the graph, from one peak to the next peak (one full cycle). Let's assume the function is \(y=\sin(Bx)\). The period \(T\) satisfies \(y(x + T)=y(x)\). If we consider the standard unit - circle based definition (for sine and cosine functions).
Wait, actually, if we use the formula \(T=\frac{2\pi}{|B|}\). But since we are given a graph, we can count: if we assume the function is \(y = A\sin(Bx)\), and we know that for \(y=\sin(x)\) the period is \(2\pi\). If we look at the given graph, in the interval \(x\in[0, 2\pi]\), the number of cycles \(n = 3\). So \(nT=2\pi\), \(T=\frac{2\pi}{3}\)
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\(\frac{2}{3}\pi\)