Answer
$(-\infty,-1]\cup[1,\infty)$
Work Step by Step
We are given the function:
$f(x)=\sqrt{x^4-1}$.
First determine the domain on which the function is defined:
$x^4-1\geq 0$
$(x^2-1)(x^2+1)\geq 0$
$(x-1)(x+1)(x^2+1)\geq 0$
$D=(-\infty,-1]\cup[1,\infty)$
$f(x)$ is a composed function of the functions $x^4-1$ and $\sqrt x$. Both functions are continuous, therefore their composition is also continuous on the domain, so also left-continuous and right-continuous in $x=-1$ and $x=1$.
The interval of continuity is:
$(-\infty,-1]\cup[1,\infty)$