Answer
$(a)$
Local maximum: $(1,1)$
local minimum: $(2,0).$
$(b)$
Absolute maximum: none.
Absolute minimum: $(2,0).$
$(c)$
Work Step by Step
$g(x)=x^{2}-4x+1,\quad x\in[1,\infty)$
$g'(x)=2x-4=2(x-2)$
$g$ and $g'$ are defined on $[1,\infty).$
$g'(x)=0$ for $ x=2$: critical point.
$g'(1.5)=-1\lt 0$
$g'(3)=1\gt 0$
$g(1)=1,\quad g(2)=0$
$g':\quad \begin{array}{llllll}
1 & & 2 & & \infty & \\
[ & -- & | & ++ & ) & \\
\hline 1 & \searrow & & \nearrow & & \\
& & 0 & & & \\
& & & & &
\end{array}$
$(a)$
Local maximum: $(1,1)$
local minimum: $(2,0).$
$(b)$
Absolute maximum: none.
Absolute minimum: $(2,0).$
$(c)$
See graph.
