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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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    and I upvoted you both, thanks!2011-03-27

1 Answers 1

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$\cos(x)=-1/2$ 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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    Thanks for the explanation.2011-03-27