Answer
$(f \circ g)(x)=f(g(x))=f(x+1)=\sqrt {x+1}$
$(g \circ f)(x)=g(f(x))=g(\sqrt{x})=\sqrt{x} + 1$
$D_{f \circ g} = [-1, \infty)$
$D_{g \circ f} = [0, \infty)$
Work Step by Step
The domain $D_f$ of $f(x)$ is $[0,\infty)$ and the domain $D_g$ of $g(x)$ is $\mathbb{R}$.
The composite functions $f \circ g$ and $g \circ f$ are defined by
$(f \circ g)(x)=f(g(x))=f(x+1)=\sqrt {x+1}$ and
$(g \circ f)(x)=g(f(x))=g(\sqrt{x})=\sqrt{x} + 1$
The domain of $f \circ g$ is defined by:
$D_{f \circ g} = \{x \in D_g : g(x) \in D_f\} = \{x \in \mathbb{R} : x+1 \in [0, \infty)\}$
that is, $D_{f \circ g} = \{x \in \mathbb{R} : x \ge -1\}$
Hence, $D_{f \circ g} = [-1, \infty)$.
The domain of $g \circ f$ is defined by:
$D_{g \circ f} = \{x \in D_f : f(x) \in D_g\} = \{x \in [0, \infty) : \sqrt{x} \in \mathbb{R}\}$
Hence, $D_{g \circ f} = [0, \infty)$.
