Composite functions

\((\mathrm{f}\circ \mathrm{g})(x)=\mathrm{f}(\mathrm{g}(x))\): 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 \(R_\mathrm{g}\subseteq D_\mathrm{f}\).

Example

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 \([-2,\infty)\).

Following the input

For a table or graph, read \(\mathrm g(x)\) first and use that output as the next input to \(\mathrm f\). If the next input is absent from the domain of \(\mathrm f\), the composition is undefined at the original input.

To solve \((\mathrm f\circ\mathrm g)(x)=k\), first find every allowed value \(u\) for which \(\mathrm f(u)=k\), then solve \(\mathrm g(x)=u\). A many-to-one outer function can give several intermediate values; keep all that the inner function actually attains.

For an invertible composition \(\mathrm h=\mathrm g\circ\mathrm f\), the statement \(\mathrm h^{-1}(a)=b\) is equivalent to \(a=\mathrm h(b)\). This can avoid deriving a full inverse rule when only one value is needed.

To find the range of a composition, apply the outer function only to the outputs attained by the inner function. Do not automatically copy the outer function’s full range.