fg(x)=f(g(x)): apply the right-hand function first. The input must belong to D_g, and its output must belong to D_f. Thus D_fg=x in D_g:g(x) in D_f. For a composite defined on all of D_g, require R_g D_f. Otherwise restrict the initial domain. With f(x)=sqrt(x-1), x 1, and g(x)=2x+3, x in R, fg(x)=sqrt(2x+2) requires x >= -1, while gf(x)=2sqrt(x-1)+3 requires x 1. Order changes both formula and domain.

\(\mathrm{fg}(x)=\mathrm{f}(\mathrm{g}(x))\): apply the right-hand function first. The input must belong to \(D_{\mathrm{g}}\), and its output must belong to \(D_{\mathrm{f}}\). Thus \(D_{\mathrm{fg}}=\{x\in D_{\mathrm{g}}:\mathrm{g}(x)\in D_{\mathrm{f}}\}\).

For a composite defined on all of \(D_{\mathrm{g}}\), require \(R_{\mathrm{g}}\subseteq D_{\mathrm{f}}\). Otherwise restrict the initial domain.

With \(\mathrm{f}(x)=\sqrt{x-1}\), \(x\ge1\), and \(\mathrm{g}(x)=2x+3\), \(x\in\mathbb R\), \(\mathrm{fg}(x)=\sqrt{2x+2}\) requires \(x\ge-1\), while \(\mathrm{gf}(x)=2\sqrt{x-1}+3\) requires \(x\ge1\). Order changes both formula and domain.

Order and composite domains

Order and composite domains