The difference between the greatest and least | vedclass

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(A) Given function is $f(x) = -x + \sin 2x$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.First,find the critical points by setting $f'(x) = 0$.$f

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$\pi$

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(A) Given function is $f(x) = -x + \sin 2x$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.First,find the critical points by setting $f'(x) = 0$.$f'(x) = -1 + 2 \cos 2x = 0$$\cos 2x = \frac{1}{2}$Since $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$2x \in [-\pi, \pi]$.Thus,$2x = \pm \frac{\pi}{3}$,which gives $x = \pm \frac{\pi}{6}$.Now,evaluate $f(x)$ at the critical points and the endpoints:$f(-\frac{\pi}{2}) = -(-\frac{\pi}{2}) + \sin(-\pi) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \approx 1.57$$f(-\frac{\pi}{6}) = -(-\frac{\pi}{6}) + \sin(-\frac{\pi}{3}) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} \approx 0.52 - 0.866 = -0.346$$f(\frac{\pi}{6}) = -(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) = -\frac{\pi}{6} + \frac{\sqrt{3}}{2} \approx -0.52 + 0.866 = 0.346$$f(\frac{\pi}{2}) = -(\frac{\pi}{2}) + \sin(\pi) = -\frac{\pi}{2} + 0 = -\frac{\pi}{2} \approx -1.57$Comparing these values,the greatest value is $M = \frac{\pi}{2}$ and the least value is $m = -\frac{\pi}{2}$.The difference is $M - m = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

The difference between the greatest and least values of the function $f(x) = -x + \sin 2x$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is

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