Find the local maximum and local minimum valu | vedclass

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(D) Given $f(x) = \sin 2x - x$.Step $1$: Find the derivative $f'(x)$.$f'(x) = 2\cos 2x - 1$.Step $2$: Set $f'(x) = 0$ to find critical points.$2\cos 2

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None of these

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(D) Given $f(x) = \sin 2x - x$.Step $1$: Find the derivative $f'(x)$.$f'(x) = 2\cos 2x - 1$.Step $2$: Set $f'(x) = 0$ to find critical points.$2\cos 2x - 1 = 0 \implies \cos 2x = \frac{1}{2}$.In the interval $[0, \pi]$,$2x$ ranges from $[0, 2\pi]$.Thus,$2x = \frac{\pi}{3}$ or $2x = \frac{5\pi}{3}$.So,$x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$.Step $3$: Use the second derivative test.$f''(x) = -4\sin 2x$.At $x = \frac{\pi}{6}$,$f''(\frac{\pi}{6}) = -4\sin(\frac{\pi}{3}) = -4(\frac{\sqrt{3}}{2}) = -2\sqrt{3} At $x = \frac{5\pi}{6}$,$f''(\frac{5\pi}{6}) = -4\sin(\frac{5\pi}{3}) = -4(-\frac{\sqrt{3}}{2}) = 2\sqrt{3} > 0$. So,$x = \frac{5\pi}{6}$ is a point of local minimum.Step $4$: Calculate the values.Local maximum value $f(\frac{\pi}{6}) = \sin(\frac{\pi}{3}) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6} = \frac{3\sqrt{3} - \pi}{6}$.Local minimum value $f(\frac{5\pi}{6}) = \sin(\frac{5\pi}{3}) - \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} - \frac{5\pi}{6} = \frac{-3\sqrt{3} - 5\pi}{6}$.Since these values do not match the options provided,the correct answer is $D$.

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

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