Q.Let $f(x) = \sqrt{x}$ (domain $x \geq 0$) and $g(x) = x + 4$ (domain $\mathbb{R}$). Find $(g \circ f)(x)$ and its domain, and $(f \circ g)(x)$ and its domain.

$(g\circ f)(x) = g(f(x)) = g(\sqrt{x}) = \sqrt{x}+4$. Since $f(x)=\sqrt{x}$ itself needs $x\geq0$, and $g$ accepts any real input, the domain of $g\circ f$ is exactly $x\geq0$ (inherited from $f$).

$(f\circ g)(x) = f(g(x)) = f(x+4) = \sqrt{x+4}$. Here $g(x)=x+4$ accepts any real $x$, but the OUTER function $f$ needs its input $x+4$ to be $\geq0$, i.e. $x\geq-4$. So the domain of $f\circ g$ is $x\geq-4$. …

For $g\circ f$, the domain restriction of the inner function $f$ carries straight through; for $f\circ g$, additionally require the OUTER function' …

• Applying the domain restriction $x\geq0$ to BOTH composites instead of recomputing it separately for each order. …