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A couple problems are giving me trouble in finding the relative maxima/minima of the function. I think the problem stems from me possibly not finding all of the critical numbers of the function, but I don't see what I missed.

Given $f(x)= 5x + 10 \sin x$, I calculated the derivative as $5 + 10 \cos x$, and found the first critical number by this work:

$$5+ 10 \cos x=0$$ $$\frac{5}{5}+10 \cos x= 0-5 \Rightarrow 10 \cos x= -5$$ $$\frac{10 \cos x}{10}= \frac{-5}{10}\Rightarrow \cos x= -\frac{1}{2}$$ $$x= \arccos(-\frac{1}{2}) = \text{First critical number is }\frac{2\pi}{3}$$

That gave me the maxima of the formula, since $$f(\frac{2\pi}{3})= 5(\frac{2\pi}{3})+10 \sin(\frac{2\pi}{3})= \frac{10\pi}{3}+5\sqrt3$$

However, I need the other critical number to calculate the minima. Should I look for the value of $\arccos(\frac{1}{2})$?

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You made a typo at your second equation line, $\frac{5}{-5}$ instead of $5-5$. – Leonardo Fontoura Mar 27 '11 at 2:21
+1 for thinking before posting and showing some work – Ross Millikan Mar 27 '11 at 2:23
$\cos(\frac{4\pi}{3})=-\frac{1}{2}$ as well. – yunone Mar 27 '11 at 2:24
@yunone I was about to post that as an answer/hint. You beat me to it, so you should post it :P – Leonardo Fontoura Mar 27 '11 at 2:26
@Leonardo, it seems Didier has already beaten both of us to the punch! – yunone Mar 27 '11 at 2:29
up vote 1 down vote accepted

$\cos(x)=-1/2$ if and only if there exists an integer $n$ such that $x=2n\pi+2\pi/3$ or $x=2n\pi+4\pi/3$. Hence any of these real numbers $x$ may be (and in fact, is) a relative maximum or a relative minimum of $f$.

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so as long as cos(x)= $=-\frac{1}{2}$, all applicable values from the unit circle are valid? Also, where does the $2n\pi$ come from? – Jason Mar 27 '11 at 2:33
First question: yes, since the derivative will be zero at each of these points. But for each one, you will have to determine whether it is a local maximum or a local minimum. Second question: the cosine function has period $2\pi$ hence, if $f'(x)=0$ then $f'(x+2n\pi)=0$ for every integer $n$. – Did Mar 27 '11 at 2:42
Thanks for the explanation. – Jason Mar 27 '11 at 2:47

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