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[-/2 Points] DETAILS MY NOTES LARCALC12 3.3.017. ASK YOUR TEACHER PRACTICE Identify the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.) \[ f(x)=x+2 \sin (x), \quad 0
Ask by Chan Peters. in the United States
Mar 16,2025

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The function \( f(x) = x + 2\sin(x) \) is increasing on the intervals \( (0, \frac{2\pi}{3}) \) and \( (\frac{4\pi}{3}, 2\pi) \), and decreasing on the interval \( (\frac{2\pi}{3}, \frac{4\pi}{3}) \).

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To find where the function \( f(x) = x + 2 \sin(x) \) is increasing or decreasing, we start by calculating its derivative: \[ f'(x) = 1 + 2 \cos(x) \] Setting \( f'(x) = 0 \) gives us \( 1 + 2 \cos(x) = 0 \), leading to \( \cos(x) = -\frac{1}{2} \). The solutions in the interval \( (0, 2\pi) \) are \( x = \frac{2\pi}{3} \) and \( x = \frac{4\pi}{3} \). Consequently, we can determine the intervals by evaluating \( f'(x) \): - For \( (0, \frac{2\pi}{3}) \): \( f'(x) > 0 \) (increasing) - For \( (\frac{2\pi}{3}, \frac{4\pi}{3}) \): \( f'(x) < 0 \) (decreasing) - For \( (\frac{4\pi}{3}, 2\pi) \): \( f'(x) > 0 \) (increasing) Thus, \( f(x) \) is increasing on the intervals \( (0, \frac{2\pi}{3}) \) and \( (\frac{4\pi}{3}, 2\pi) \), and decreasing on \( (\frac{2\pi}{3}, \frac{4\pi}{3}) \).

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