Answer
(a) see below.
(b) $g(x)=\sqrt {x^4+x^2-6x+9}$
(c) $g(1)=\sqrt 5$
Work Step by Step
(a) $g(x)=\sqrt {f(x)}, f(x)\geq0$, let f($x_0)$ be a minimum of $f(x)$ in an open interval $(a,b)$ such that
for any point $x$ in $(a,b)$ except $x_0$, we have $f(x_0)\lt f(x)$. This will also be true for $g(x)$ as
$g(x_0)=\sqrt {f(x_0)}\lt \sqrt {f(x)}=g(x)$. We can also show it is true for a maximum.
(b) $g(x)=\sqrt {(x-3)^2+(x^2-0)^2}=\sqrt {x^4+x^2-6x+9}$
(c) Base on (a), the minimum of the function inside the radical can be found as $f(1)=5$, and the minimum
of $g(1)=\sqrt 5$ can thus be obtained.