To determine the phase shift of the function \( f(x) = 2\sin(x + \frac{\pi}{6}) \), we can compare it to the standard form of a sine function, which is given by:
\[ f(x) = a \sin(b(x - c)) + d \]
In this standard form:
- \( a \) is the amplitude,
- \( b \) affects the period,
- \( c \) is the phase shift,
- \( d \) is the vertical shift.
For the function \( f(x) = 2\sin(x + \frac{\pi}{6}) \), we can rewrite it to match the standard form. Notice that:
\[ x + \frac{\pi}{6} = x - \left(-\frac{\pi}{6}\right) \]
This allows us to express the function as:
\[ f(x) = 2\sin\left(1\left(x - \left(-\frac{\pi}{6}\right)\right)\right) \]
Here, \( a = 2 \), \( b = 1 \), \( c = -\frac{\pi}{6} \), and there’s no vertical shift (\( d = 0 \)).
The phase shift \( c \) is determined from the transformation in the argument of the sine function, which results in a shift to the left by \( \frac{\pi}{6} \) because \( c \) is negative.
Thus, the phase shift of the function \( f(x) = 2\sin(x + \frac{\pi}{6}) \) is:
\[ \text{Phase shift} = -\frac{\pi}{6} \text{ (to the left)} \]
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