Domain of $f(x)=\sqrt{\lfloor x\rfloor-1+x^2}$ I drew the number line and tested with different values, getting the correct domain $(-\infty,-\sqrt3)\cup[1,\infty)$. However, how do I solve this faster by manipulating the function?

$f(x)$ is not real-valued if the expression $g(x)$ inside the square root is negative; i.e., $$g(x) = \lfloor x \rfloor - 1 + x^2 < 0.$$  Since $$x-1 < \lfloor x \rfloor \le x,$$ we find $$(x+2)(x-1) < g(x) \le x^2+x-1.$$  Hence we are assured to be outside the domain if $x^2 + x - 1 < 0$, and assured to be inside the domain if $(x+2)(x-1) > 0$.  The first condition excludes the interval $$x \not\in \left(\frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right),$$ and the second includes $$x \in (-\infty,-2] \cup [1,\infty).$$  Since this does not resolve the fate of values not contained in either of these intervals, we now must look at the behavior of $\lfloor x \rfloor$ and $g(x)$ here.  For $x \in [-2 , (-1-\sqrt{5})/2)$, we have $$\lfloor x \rfloor = -2, \quad g(x) = x^2 - 3,$$ thus $g(x) \ge 0$ for $x < -\sqrt{3}$.
For $x \in [(-1+\sqrt{5})/2, 1)$, we have $$\lfloor x \rfloor = 0, \quad g(x) = x^2 - 1,$$ and there is no modification; $g(x) \ge 0$ when $x \ge 1$.  Therefore the complete domain is $$(-\infty, -\sqrt{3}] \cup [1,\infty).$$  Note that your answer contains a sign error.
