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I am having a little trouble trying to figure out the following problem:

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

I have taken the derivative of the function, which I believe to be $1-2\sin x$. I know I have to get critical points, but this is where I am stuck because I do not know how to solve for $x$ in such a function as this. I feel as though $\pi /2$ is a critical point, but I am most likely wrong about that. Also, when I plug in the $0$ and $2\pi$, then $0$ comes out to be $-2$, however $2\pi$ comes out to be $2\pi -2$, which is also confusing me quite a bit.

I really just want to know how to get the critical points because the rest is simply plugging in the critical numbers and getting their values to determine the absolute maximum and minimum.

Any help is greatly appreciated!

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$1-2\sin(x)=0$ imples $\sin(x)=\frac{1}{2}$, so $x=\frac{\pi}{6}, \frac{5\pi}{6}$.

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  • $\begingroup$ Oh, that makes sense! Thank you @avid19! $\endgroup$ – camrymps Mar 29 '15 at 2:20

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