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?

Let $g(x) = x^2 + \lfloor{x}\rfloor - 1$.

Then the domain of $f$ is $\{x \in \mathbb{R} \mid g(x) \ge 0\}$.

Note that for all $x \in \mathbb{R}$, we have $x-1 < \lfloor{x}\rfloor \le x$, hence for all $x \in \mathbb{R}$,
\begin{align*}
g(x) &= x^2 + \lfloor{x}\rfloor - 1\\[4pt]
&\ge x^2 + (x-1) - 1\\[4pt]
&=x^2 + x - 2\\[4pt]
&=(x+2)(x-1)\\[4pt]
\end{align*}
Since $(x+2)(x-1) \ge 0$ if $x \le -2\;\,$or$\;x \ge 1$, it follows that
$$(-\infty,-2] \cup [1,\infty)$$
is a subset of the domain of $f$.

It remains to find which parts of the interval $(-2,1)$ are in the domain of $f$.

Analyze it piecewise . . .

If $-2 < x < -1$, then
\begin{align*}
g(x) &= x^2 + \lfloor{x}\rfloor - 1
\qquad\qquad\qquad\qquad\qquad\;\\[4pt]
&= x^2 + (-2) - 1\\[4pt]
&=x^2 - 3\\[4pt]
\end{align*}
hence the interval $(-2,-\sqrt{3}]$ is in the domain of $f$, but not the rest of the interval $(-2,-1)$.

If $-1 \le x < 0$, then
\begin{align*}
g(x) &= x^2 + \lfloor{x}\rfloor - 1\\[4pt]
&= x^2 + (-1) - 1\\[4pt]
&=x^2 - 2\\[4pt]
&<0\qquad\text{[since $x^2-2<0\;\;$if$\;\,-1\le x < 0$]}\\[4pt]
\end{align*}
If $0 \le x < 1$, then
\begin{align*}
g(x) &= x^2 + \lfloor{x}\rfloor - 1\\[4pt]
&= x^2 + (0) - 1\\[4pt]
&=x^2 - 1\\[4pt]
&<0\qquad\text{[since $x^2-1 < 0\;\;$if$\;\,0 \le x < 1$]}
\;\;\;\\[4pt]
\end{align*}
Thus, the domain of $f$ is
\begin{align*}
&(-\infty,-2] \cup (-2,-\sqrt3] \cup [1,\infty)\\[8pt]
&=(-\infty,-\sqrt3] \cup [1,\infty)\\[4pt]
\end{align*}
