What is the period of the function f(x) = ln(2 + sin 2x)?
π
The period of a function \(f(x)\) is the smallest positive value \(T\) such that \(f(x + T) = f(x)\) for all \(x\) in the domain of \(f\). For trigonometric functions, this concept is fundamental.
We are asked to find the period of the function \(f(x) = \ln(2 + \sin 2x)\). This function is composed of an inner trigonometric function, \(\sin 2x\), and outer functions, adding 2 and taking the natural logarithm (\(\ln\)).
The periodicity of \(f(x)\) is primarily determined by the periodicity of its trigonometric component, \(\sin 2x\).
The standard sine function, \(\sin \theta\), has a period of \(2\pi\). This means \(\sin(\theta + 2\pi) = \sin \theta\) for all \(\theta\).
For a function of the form \(\sin(ax + b)\) or \(\cos(ax + b)\), the period is given by \(\frac{2\pi}{|a|}\). In our case, the trigonometric component is \(\sin 2x\). Here, \(a = 2\).
Using the formula, the period of \(\sin 2x\) is:
\(T_{\sin 2x} = \frac{2\pi}{|a|} = \frac{2\pi}{|2|} = \frac{2\pi}{2} = \pi\)
So, the function \(\sin 2x\) repeats its values every \(\pi\) units. This means \(\sin(2(x + \pi)) = \sin(2x + 2\pi)\). Since the sine function has a period of \(2\pi\), \(\sin(2x + 2\pi) = \sin(2x)\).
Now consider the full function \(f(x) = \ln(2 + \sin 2x)\).
Therefore, the period of the function \(f(x) = \ln(2 + \sin 2x)\) is the same as the period of \(\sin 2x\).
We found that the period of \(\sin 2x\) is \(\pi\).
Let's verify: \(f(x + \pi) = \ln(2 + \sin(2(x + \pi))) = \ln(2 + \sin(2x + 2\pi))\).
Since \(\sin(\theta + 2\pi) = \sin \theta\), we have \(\sin(2x + 2\pi) = \sin 2x\).
So, \(f(x + \pi) = \ln(2 + \sin 2x) = f(x)\).
This confirms that \(\pi\) is a period. The period of \(\sin 2x\) was \(\pi\), and the outer functions did not reduce this period. Thus, the smallest positive period is \(\pi\).
The period of the function \(f(x) = \ln(2 + \sin 2x)\) is \(\pi\).
| Function Component | Rule/Period |
|---|---|
| \(\sin \theta\) | Period \(2\pi\) |
| \(\sin(ax)\) | Period \(\frac{2\pi}{|a|}\) |
| \(\sin(2x)\) | Period \(\frac{2\pi}{|2|} = \pi\) |
| \(g(x) + c\) (if \(g(x)\) has period T) | Period T |
| \(h(g(x))\) (if \(g(x)\) has period T and \(h\) is monotonic) | Period T |
| \(\ln(2 + \sin 2x)\) | Period of \(\sin 2x\), which is \(\pi\) |
| Function | Standard Period |
|---|---|
| \(\sin x\) | \(2\pi\) |
| \(\cos x\) | \(2\pi\) |
| \(\tan x\) | \(\pi\) |
| \(\csc x\) | \(2\pi\) |
| \(\sec x\) | \(2\pi\) |
| \(\cot x\) | \(\pi\) |
| \(A \cdot f(Bx + C) + D\) where \(f(x)\) has period T | \(\frac{T}{|B|}\) (for sin, cos, sec, csc with T=\(2\pi\)) \(\frac{T}{|B|}\) (for tan, cot with T=\(\pi\)) |
When dealing with functions that are combinations of different types, the period is often determined by the component with the largest period, or the least common multiple (LCM) of the individual periods if they are summed or multiplied. However, in a composition like \(f(g(x))\), if \(g(x)\) has a period \(T_g\), the period of \(f(g(x))\) is often \(T_g\) itself, especially if \(f\) is a single-valued function that doesn't introduce its own, potentially smaller, period (like absolute value can sometimes do) or alter the fundamental frequency of repetition. For monotonic functions like \(\ln(y)\) applied to a periodic input \(y\), the period remains unchanged.
In our case, \(g(x) = 2 + \sin 2x\), and \(f(y) = \ln(y)\). The period of \(\sin 2x\) is \(\pi\), so the period of \(g(x) = 2 + \sin 2x\) is also \(\pi\). Since \(\ln(y)\) is monotonic for \(y > 0\), the function \(f(g(x)) = \ln(2 + \sin 2x)\) retains the period of \(g(x)\), which is \(\pi\).
It is important to find the smallest positive period. Our calculation for \(\sin 2x\) yields the fundamental period \(\pi\). Since the outer functions don't compress the cycle, \(\pi\) is indeed the smallest positive value for which \(f(x + T) = f(x)\).
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