Answer
$f$ is continuous on $(-\infty,\infty)$.
Work Step by Step
\[f(x_=\left\{\begin{array}{ll}1-x^2\;\;\;\;,x\leq 1\\
\sqrt{x-1}\;\;\;\;,x>1\end{array}\right.\]
Clearly $f$ is continuous if $x\neq 1$, so we will check the continuity at $x=1$
\[\lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1}(1-x^2)=1-1^2=0\]
\[\lim_{x\rightarrow 1^{+}}f(x)=\lim_{x\rightarrow 1}\sqrt{x-1}=\sqrt{1-1}=0\]
\[\Rightarrow \lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1^{+}}f(x)=f(1)\]
$\Rightarrow f$ is continuous at $x=1$
Therefore $f$ is continuous on $(-\infty,\infty)$
Hence Proved.
