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I want to minimize a real valued function $$f(x) = \frac{2}{3} \cos(x)+1+ \frac{1}{9}\cos^2(x),$$ but it turns out that the minimum occurs at $x = \pi - \arccos 3$.

I am not sure what to conclude from this fact. Does the minimum now occur at $x = \pi$, or how should I interpret the fact that $\arccos$ is only defined in the interval $[-1,1]$?

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One would need to know the original problem to see what's going on. Perhaps it is an endpoint maximum. Perhaps it is a mistake. Could you please edit your question to provide more information? – André Nicolas Jun 3 '12 at 15:26
@André Nicolas I made the problem more concrete – George Jun 3 '12 at 15:31
Is your function $f(x) = \frac{2}{3}\cos(x)+1+ \frac{1}{9}\cos^2(x)$? – Babak S. Jun 3 '12 at 15:35
@Babak Sorouh yes – George Jun 3 '12 at 15:37
@did: You can answer a question by posting an answer. – Gigili Jun 3 '12 at 15:40
up vote 3 down vote accepted

If we take $$f'(x)=-\frac 2 3 \sin x - \frac 2 9 \cos x \sin x$$ and set it equal to zero, we find that all critical points occur when $\sin x =0$ or $\cos x = -3$. The latter is impossible (over the reals), so all critical points occur when $\sin x =0$. This implies that $x=n \pi$, where $n\in \mathbb Z$. Examining the original equation, it is clear that the minimums occur when $n$ is odd or, equivalently, for $x=2k\pi+\pi$ for $k\in\mathbb Z$.

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Assuming the cosines are on the numerator:$$f'(x)=-\frac{\sin x+6\sin x\cos x}{27}=0\Longleftrightarrow x=k\pi\,,\,k\in\mathbb{Z}\,,\,\,or\,\,\cos x=-\frac{1}{6}$$I can't see how you got $\,\arccos 3\,$ there...

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Actually $x$ need not be a multiple of $2\pi$, just of $\pi$ in order for $\sin x$ to be zero. – process91 Jun 3 '12 at 15:44
Indeed so. Thanks – DonAntonio Jun 3 '12 at 16:08

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