EVT Find the maximum and minimum values of \( f(x)=x=2 \sin x \) on the interval \( [0,2 \pi] \)

To find the maximum and minimum values of the function \( f(x) = x - 2 \sin x \) on the interval \( [0, 2\pi] \), we must first identify the critical points by taking the derivative and setting it to zero. 1. **Find the derivative:** \[ f'(x) = 1 - 2 \cos x \] 2. **Set the derivative to zero and solve for critical points:** \[ 1 - 2 \cos x = 0 \implies 2 \cos x = 1 \implies \cos x = \frac{1}{2} \] The solutions for \( \cos x = \frac{1}{2} \) in the interval \( [0, 2\pi] \) are: \[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \] 3. **Evaluate \( f(x) \) at the critical points and endpoints:** - At \( x = 0 \): \[ f(0) = 0 - 2 \sin(0) = 0 \] - At \( x = \frac{\pi}{3} \): \[ f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - 2 \sin\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - 2 \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \sqrt{3} \] - At \( x = \frac{5\pi}{3} \): \[ f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} - 2 \sin\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} - 2 \cdot \left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{3} + \sqrt{3} \] - At \( x = 2\pi \): \[ f(2\pi) = 2\pi - 2 \sin(2\pi) = 2\pi \] 4. **Compare the evaluated values:** - \( f(0) = 0 \) - \( f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - \sqrt{3} \) - \( f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sqrt{3} \) - \( f(2\pi) = 2\pi \) 5. **Determine maxima and minima:** - The minimum value occurs at \( f(0) = 0 \). - The maximum value occurs at \( f(2\pi) = 2\pi \). Thus, the maximum value of \( f(x) \) on the interval \( [0, 2\pi] \) is \( 2\pi \), and the minimum value is \( 0 \).