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How would I find the intervals in which f decreases and decreases in the following function.

$f(x)=2x-\cos(x)$ defined on $[0,2 \pi$]

taking the derivative I have $2+sin(x)=0$

the $sin(x)=-2$ but I am stuck here.

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up vote 3 down vote accepted

Your derivation is correct!

$\sin(x) \neq 2$ for all $x\in \mathbb R$. There are no zeros.

But we can still proceed! Recall: $$\sin(x) \leq |1|\quad\text{for all}\;\;x \in \mathbb{R}. \quad\text{I.e., }\;-1 \leq \sin x \leq 1$$

So your derivative will evaluate to between $2 - (-1) = 3$ and $2 - 1 = 1$ for all $x \in \mathbb R$. That is, for all real $x$, $$f'(x) \in [1, 3] > 0$$

Hence, the derivative is always positive, hence the function $f(x)$ is always increasing.

Here's visual confirmation, compliments of WolframAlpha:

enter image description here

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Yes it is always increasing so I would write f increases on (-infinity,infinity) – Fernando Martinez Mar 2 '13 at 21:19
Exactly, that's correct! – amWhy Mar 2 '13 at 21:20
I see it makes sense thanks. – Fernando Martinez Mar 2 '13 at 21:21

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