Let $f ( x ) = x - \sin x$ on $[ 0,2 \pi ]$ The absolute minimum and maximum of ƒ are located respectively at x =

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To find the absolute minimum and maximum of $f(x) = x - \sin(x)$ on the interval $[0, 2\pi]$, we first observe that $f'(x) = 1 - \cos(x)$. The critical points occur where $f'(x) = 0$, which happens when $\cos(x) = 1$. This occurs at $x = 0$ and $x = 2\pi$ within the given interval, but since these are the endpoints of the interval, we don't find any critical points in the interior that would contribute to local minima or maxima. Therefore, we only need to evaluate $f(x)$ at the endpoints of the interval to find the absolute extrema. At $x = 0$, $f(0) = 0 - \sin(0) = 0$. At $x = 2\pi$, $f(2\pi) = 2\pi - \sin(2\pi) = 2\pi$. Since $f(x)$ is increasing over the interval $[0, 2\pi]$ (because $f'(x) \geq 0$ for all $x$ in $[0, 2\pi]$), the absolute minimum is at $x = 0$ and the absolute maximum is at $x = 2\pi$.

Let f(x)=x−sin⁡xf ( x ) = x - \sin xf(x)=x−sinx On [0,2π][ 0,2 \pi ][0,2π]

Let $f ( x ) = x - \sin x$ On $[ 0,2 \pi ]$