Answer
(a) f(x) -> {x : x $\ne$ -1} = (-$\infty$, -1) U (-1, +$\infty$)
g(x) -> (-$\infty$, +$\infty$)
(b) f(x) -> {x : x $\geq$ 0} = [0, +$\infty$)
g(x) -> {x : x $\geq$ 0} = [0, +$\infty$}
Work Step by Step
(a) The denominator, in f(x), must be $\ne$ 0, so x $\ne$ -1. There are no restrictions for x in g(x).
(b) If x $\geq$ 0 the denominator never vanishes. It's only necessary to know that $\sqrt x$ is valid for every x $\geq$ 0.
