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?
 A: 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 $x\in]-\infty,-\sqrt3]\cup[1,+\infty[$.
A: $f(x)$ 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.
A: 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*}
A: The function under the radical is piecewise quadratic.
If we set $x:=i+f$ with $f$ in range $[0,1)$,
$$i-1+(i+f)^2$$ must be non negative.
It has a local minimum when $2(i+f)=0$, i.e. only when $i=f=0$. So all pieces are monotonous and it suffices to look at the endpoints to detect the negative values.
By solving the quadratic inequation
$$i-1+i^2\ge0$$
we have $i\notin(-\phi,1/\phi)$, or, as $i$ is an integer, $i\notin [-1,0]$. So we can have negatives with


*

*$i=-2\to[-2,-1)\to-3+(-2+f)^2=0\to f=2-\sqrt3$

*$i=0\to[0,1)\to-1+f^2=0\to$ no solution and the expression remains negative.
Putting these results together, the function is undefined for $$x\in(-\sqrt3,1).$$
