Answer
Local maximum: $(0,1)$
local minimum: $(-1,0).$
$(b)$
Absolute maximum: none.
Absolute minimum: $(-1,0).$
$(c)$
See graph.
Work Step by Step
$f(x)=(x+1)^{2},\quad x\in(-\infty,0]$
$f'(x)=2(x+1)$
$f$ and $f'$ are defined on $(-\infty,0]$
$f'(x)=0$ for $ x=-1$: critical point.
$f'(-2)=-2\lt 0$
$f'(-0.5)=0.25\gt 0$
$f(-1)=0,\quad f(0)=1$
$f':\quad \begin{array}{cccc}
-\infty & & -1 & & 0 & \\
( & -- & | & ++ & ] & \\\hline
& \searrow & & \nearrow & 1 & \\
& & 0 & & & \\
& & & & &
\end{array}$
$(a)$
Local maximum: $(0,1)$
local minimum: $(-1,0).$
$(b)$
Absolute maximum: none.
Absolute minimum: $(-1,0).$
$(c)$
See graph.
