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Find the critical points of: $y = x + \cos(x)$

$y' = 1 -\sin(x)$

$\sin(x) = 1$ for stationary points.

$\therefore x = \frac{\pi}{2} + 2\pi k , k \in \mathbb{Z}$

I had trouble finding the value/s for $y$ since I wasn't sure of its period or how I would sub $x$'s period back into the original equation.

How do I find the values for $y$?

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$y$ is not periodic, so it has no period. The values of the $\cos$ summand are, of course, periodic with period $2\pi$. But notice that if $\sin(a) = 1$, then $\cos(a) = 0$ (since $1 = \sin^2(a)+\cos^2(a)$), so when you plug in these values into $y$, the summand $\cos(x)$ will, necessarily, have value $0$. –  Arturo Magidin Mar 26 '12 at 3:24
    
@ArturoMagidin: So, $y = x$ since the $cos(x)$ term is always = 0 at stationary points? I think that clears things up a lot, if I'm not missing anything else. –  stariz77 Mar 26 '12 at 3:30

2 Answers 2

up vote 3 down vote accepted

You're essentially there: $y = x + \cos(x) = \frac\pi2 + 2\pi k + \cos(\frac\pi2 + 2\pi k) = \frac\pi2 + 2\pi k$. There are infinitely many $y$-values, one for each $k \in \mathbb{Z}$.

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$y(\frac{\pi}{2} + 2\pi k) = \frac{\pi}{2} + 2\pi k + \operatorname{cos}(\frac{\pi}{2} + 2\pi k) = \frac{\pi}{2} + 2\pi k $

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