Is this true, for all $\displaystyle x$ , $\displaystyle x^2 \ge 0$?
If that is true add 4 to both sides, $\displaystyle 4+x^2 \ge 4$.
Therefore, $\displaystyle \sqrt{4+x^2}$ is defined for all $\displaystyle x$.
Therefore, the domain of the function is all $\displaystyle x$.
To find the domain,
$\displaystyle 1-x\ge 0$
$\displaystyle -x \ge -1$
$\displaystyle \boxed{x \le 1}$
To find the range,
$\displaystyle f(1)=\sqrt{1-1}+1^2=1$
$\displaystyle f(0)=\sqrt{1-0}+0^2=1$
$\displaystyle f(-1)=\sqrt{1-^-1}+(-1)^2=\sqrt{2}+1$
Therefore, $\displaystyle \boxed{f(x) \ge 1}$