<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] Important fact: the minimum of the function is NOT the same as where $f'(x) = 0$. The latter is a local minimum - a point that it is minimal compared to it's local points, but taking as a simple example the function $f(x) = x^3$ on $[-1,1]$ the function is minimised at $-1$ despite this not being a turning point.
Despite this, in your example the local minimum happens to also be the global minimum. To solve the equation $x^2(4 + \cos(x ))= 9\pi^2$, the simplest way is to spot that $x = \frac{3\pi}{2}$ is a solution.  As far as I'm aware, this kind of equation is not normally soluble by standard non-numerical methods, but the logic behind the guess is that if $\cos(x) = 0$, we're left with the equation $4x^2 = 9\pi^2$, and the solution to this gives $\cos(x) = 0$.
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