<s>[INST] Local minimum of $f(x) = 4x + \frac{9\pi^2}{x} + \sin x$ What's the minimum value of the function
$$f(x) = 4x + \frac{9\pi^2}{x} + \sin x$$
for $0 < x < +\infty$? The answer should be $12\pi - 1$, but I get stuck with the expression involving both $\cos x$ and $x^2$ in the derivative. Taking the derivative, we have:
$$f'(x) = 4 - \frac{9\pi^2}{x^2} + \cos x.$$
In order to find the local extrema of the function, we set $f'(x) = 0$. Therefore,
\begin{align}
4 - \frac{9\pi^2}{x^2} + \cos x &= 0 \\
4x^2 - 9\pi^2 + x^2 \cos x &= 0 \\
x^2 (4 + \cos x) &= 9\pi^2.
\end{align}
However, I'm not sure what to do from here or if, indeed, I'm doing it right at all. Any help would be appreciated.
 [/INST] If I may interfer in the discussion, the first derivative being $$f'(x) = 4 - \frac{9\pi^2}{x^2} + \cos x$$ you find the minimum for $x=\frac{3\pi}{2}$ just because, for this specific value, the cosine is zero. I can bet that the problem was set with this in mind.
But, let us change slightly your function to $$f(x) = 4x + \frac{8\pi^2}{x} + \sin x$$ Now, $$f'(x) = 4 - \frac{8\pi^2}{x^2} + \cos x$$ and the problem becomes quite different and you cannot solve anymore $f'(x)=0$ using elementary methods. Instead numerical methods should be used and Newton is probably the simplest root-finder for the solution of non linear equation. Starting from a "reasonable" guess $x_0$, Newton updates are given by $$x_{n+1}=x_n-\frac{f'(x_n)}{f''(x_n)}$$ So, in the case of the modified problem, using $x_0=\frac{3\pi}{2}\simeq 4.71239$, the successive iterates will be $4.53525$, $4.54083$, $4.54084$ which is the solution of $f'(x)=0$ for six significant figures.
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