<s>[INST] Range of an inverse trigonometric function Find the range of $f(x)=\arccos\sqrt {x^2+3x+1}+\arccos\sqrt {x^2+3x}$
My attempt is:I first found domain,
$x^2+3x\geq0$
$x\leq-3$ or $x\geq0$...........(1)
$x^2+3x+1\geq0$
$x\leq\frac{-3-\sqrt5}{2}$ or $x\geq \frac{-3+\sqrt5}{2}$...........(2)
From (1) and (2),
domain is $x\leq-3$ or $x\geq0$
but could not solve further..Any help will be greatly appreciated.
 [/INST] You do not have the correct domain. We must also have $-1\leq\sqrt{x^2+3x+1}\leq1$ and $-1\leq\sqrt{x^2+3x}\leq1$. In other words, $\sqrt{x^2+3x+1}\leq1$ and $\sqrt{x^2+3x}\leq1$, since they are positive. 
Thus $x^2+3x+1\leq1$ (squaring is allowed since both are positive), or $x^2+3x\leq0$, this gives $-3 \leq x \leq 0$. Together with $x \leq -3$ or $x \geq 0$, we get that the only numbers that give a well defined value are $x=0$ or $x=-3$. 
This gives $\arccos(\sqrt{1})+\arccos(\sqrt{0})$ in both cases, so the range is $\frac{1}{2}\pi$.
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