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?

You must find the values of $x$ such that $\lfloor x\rfloor-1+x^2\geq0$.
The easy part:

*

*It's true for $x\geq1$, since $x^2\geq1$ and $\lfloor x\rfloor>0$.

*For $0\leq x<1$, it's false, because $\lfloor x\rfloor=0$ and $x^2<1$.

Now, the case $x<0$. First notice that for $x\in[n,n+1[$, for integer $n$, you have $\lfloor x\rfloor=n$, hence $\lfloor x\rfloor-1+x^2=x^2+n-1$. For $n<0$, its infimum, on $[n,n+1[$, is found when $x\to n+1$, and this infimum is $(n+1)^2+n-1=n^2+3n=n(n+3)$. Hence, for $n<-2$, it's nonnegative.
There are two intervals left, $[-2,-1[$ and $[-1,0[$.

*

*On $[-1,0[$, $\lfloor x\rfloor-1+x^2=x^2-2<0$ because $x^2\leq1$.

*On $[-2,-1[$, $\lfloor x\rfloor-1+x^2=x^2-3$. This is nonnegative for $x\leq-\sqrt{3}$, hence, for $-2\leq x\leq-\sqrt3$.

All in all, $\lfloor x\rfloor-1+x^2\geq0$ for $x\in]-\infty,-\sqrt3]\cup[1,+\infty[$.
