Let $f$ be a function, $f(0, +\infty)\to\mathbb{R}$, $\displaystyle f(x)=\frac{\sqrt{x^4+16}}{x}$. Find the real values $m$ such that $\displaystyle f(x)+f\left(\frac{4}{x}\right)=m$ has exactly 2 solutions.
I observed that $\displaystyle f(x)+f\left(\frac{4}{x}\right)=2f(x)$. Afterwards, I observed that the function is deacrsing on $(0,2)$ and increasing on $(2,+\infty)$ and the limits to $0$ and to $+\infty$ are eequal with $+\infty$.
Any info will be really appreciated.
We want to find $m$ such that $2f(x) = m$ has two solutions. Notice this is equivalent to finding an $m$ such that $\sqrt{x^4 + 16} = \frac{mx}{2}$ has two solutions. Square both sides and rearrange to get $$x^4 - \frac{m^2}{4}x^2 + 16 = 0.$$ Then $$x^2 = \frac{\frac{m^2}{4} \pm \sqrt{\frac{m^4}{16}-64}}{2}.$$
Then $x$ will have exactly one solution (namely $\pm \frac{m}{2\sqrt{2}})$ if $\frac{m^4}{16} - 64 = 0$. This implies $m^4 = 2^{10}$ so the real solutions for $m$ are $\pm 4\sqrt{2}$.
When $m = 4\sqrt{2}$ we see that $x = 2$ is the solutions and when $m = -4\sqrt{2}$ we see that $x = -2$ is the solution. Now if $\frac{m^4}{16} - 64 < 0$ we get imaginary solutions.
Lastly if $\frac{m^4}{16} - 64 > 0$ so $ m > 4\sqrt{2}$ or $m < -4\sqrt{2}$ then we obtain two solutions. This coincides with your observation of where the function is decreasing and increasing. The local min and local max of the function $2f(x)$ occurs at $x = \pm 2$ which corresponds to $m = \pm 4\sqrt{2}$ so for values of $m$ above and below there are two solutions.