Answer
There is no absolute minimum.
The absolute maximum is $f(0) = 1$
There is no local minimum.
There is no local maximum.
Work Step by Step
$f(x) = 1-\sqrt{x}$
Note that this function is only defined for values of $x$ such that $0 \leq x$
There is no absolute minimum.
The absolute maximum is $f(0) = 1$
There is no local minimum.
There is no local maximum.