(f g)(x)=f(g(x)): apply g first, then f. Usually f g!= g f. The identity function is I(x)=x. The natural domain of f g consists of inputs x in D_g for which g(x) in D_f. If the composition is required on all of D_g, check R_g D_f. Example For f(x)=sqrt(x) and g(x)=x-2, (f g)(x)=sqrt(x-2), x 2; (g f)(x)=sqrt(x)-2, x 0. Their ranges are respectively [0,infinity) and [-2,infinity).

\((\mathrm{f}\circ \mathrm{g})(x)=\mathrm{f}(\mathrm{g}(x))\): apply \(\mathrm{g}\) first, then \(\mathrm{f}\). Usually \(\mathrm{f}\circ \mathrm{g}\ne \mathrm{g}\circ \mathrm{f}\). The identity function is \(\mathrm{I}(x)=x\).

The natural domain of \(\mathrm{f}\circ \mathrm{g}\) consists of inputs \(x\in D_\mathrm{g}\) for which \(\mathrm{g}(x)\in D_\mathrm{f}\). If the composition is required on all of \(D_\mathrm{g}\), check \(R_\mathrm{g}\subseteq D_\mathrm{f}\).

For \(\mathrm{f}(x)=\sqrt{x}\) and \(\mathrm{g}(x)=x-2\), \((\mathrm{f}\circ \mathrm{g})(x)=\sqrt{x-2}\), \(x\ge2\); \((\mathrm{g}\circ \mathrm{f})(x)=\sqrt{x}-2\), \(x\ge0\). Their ranges are respectively \([0,\infty)\) and \([-2,\infty)\).

Functions

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Composite functions

Composite functions