# Composition of Functions

Suppose that ${y}={f{{\left({u}\right)}}}={\ln{{\left({u}\right)}}}$ and ${u}={g{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}$. Since ${y}$ is a function of ${u}$ and ${u}$ is afunction of ${x}$ the we obtain that ${y}$ is a function of ${x}$: ${y}={f{{\left({u}\right)}}}={f{{\left({g{{\left({x}\right)}}}\right)}}}={f{{\left({\sin{{\left({x}\right)}}}\right)}}}={\ln{{\left({\sin{{\left({x}\right)}}}\right)}}}$.

The procedure is called composition because the new function is composed of the two given functions $f$ and $g$.

If we are given two functions $f$ and $g$, we start with a number $x$ in the domain of $g$ and find value $g\left(x\right)$. If this number is in the domain of $f$, then we can calculate the value of ${f{{\left({g{{\left({x}\right)}}}\right)}}}$. The result is a new function ${h}{\left({x}\right)}={f{{\left({g{{\left({x}\right)}}}\right)}}}$ obtained by substituting $g$ into $f$.

Definition. Given two functions ${f{}}$ and ${g{}}$, the composite function ${f{\circ}}{g{}}$ is defined by ${\left({f{\circ}}{g}\right)}{\left({x}\right)}={f{{\left({g{{\left({x}\right)}}}\right)}}}$.

Domain of composite functions is set of all ${x}$ such that ${x}$ is in domain of ${g{}}$ and ${g{{\left({x}\right)}}}$ is in domain of ${f{}}$.

Example 1. Find ${f{\circ}}{g{}}$, ${g{\circ}}{f{}}$, ${f{\circ}}{f{}}$, ${g{\circ}}{g{}}$ if ${f{{\left({x}\right)}}}={{x}}^{{3}}$ and ${g{{\left({x}\right)}}}=\frac{\sqrt{{{x}}}}{{{x}-{1}}}$.

${f{\circ}}{g{=}}{f{{\left({g{{\left({x}\right)}}}\right)}}}={f{{\left(\frac{\sqrt{{{x}}}}{{{x}-{1}}}\right)}}}={{\left(\frac{\sqrt{{{x}}}}{{{x}-{1}}}\right)}}^{{3}}$

${g{\circ}}{f{=}}{g{{\left({f{{\left({x}\right)}}}\right)}}}={g{{\left({{x}}^{{3}}\right)}}}=\frac{\sqrt{{{{x}}^{{3}}}}}{{{{x}}^{{3}}-{1}}}$.

Notice, that in general ${f{\circ}}{g{\ne}}{g{\circ}}{f{}}$.

${f{\circ}}{f{=}}{f{{\left({f{{\left({x}\right)}}}\right)}}}={f{{\left({{x}}^{{3}}\right)}}}={{\left({{x}}^{{3}}\right)}}^{{3}}={{x}}^{{9}}$

${g{\circ}}{g{=}}{g{{\left({g{{\left({x}\right)}}}\right)}}}={g{{\left(\frac{\sqrt{{{x}}}}{{{x}-{1}}}\right)}}}=\frac{\sqrt{{\frac{\sqrt{{{x}}}}{{{x}-{1}}}}}}{{\frac{\sqrt{{{x}}}}{{{x}-{1}}}-{1}}}$.

Note, that we can take composition of more than two functions: ${\left({f{\circ}}{g{\circ}}{h}\right)}{\left({x}\right)}={f{{\left({g{{\left({h}{\left({x}\right)}\right)}}}\right)}}}$.

Example 2. Find ${f{\circ}}{g{\circ}}{h}$ if ${f{{\left({x}\right)}}}={{x}}^{{2}}$, ${g{{\left({x}\right)}}}=\frac{{x}}{{{x}-{1}}}$ and ${h}{\left({x}\right)}={\sin{{\left({x}\right)}}}$.

${f{\circ}}{g{\circ}}{h}={f{{\left({g{{\left({h}{\left({x}\right)}\right)}}}\right)}}}={f{{\left({g{{\left({\sin{{\left({x}\right)}}}\right)}}}\right)}}}={f{{\left(\frac{{\sin{{\left({x}\right)}}}}{{{\sin{{\left({x}\right)}}}-{1}}}\right)}}}={{\left(\frac{{\sin{{\left({x}\right)}}}}{{{\sin{{\left({x}\right)}}}-{1}}}\right)}}^{{2}}$.

Sometimes we need to perform inverse task, in other words given composite function, we need to find functions it is formed from.

Example 3. Find f, g, and h if ${f{\circ}}{g{\circ}}{h}={{\left({\cos{{\left({x}+{9}\right)}}}\right)}}^{{2}}$.

The formula for ${f{\circ}}{g{\circ}}{h}$ says: first add 9, then take the cosine of the result, and finally square.

So, we take ${h}{\left({x}\right)}={x}+{9}$, ${g{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}$ and ${f{{\left({x}\right)}}}={{x}}^{{2}}$.

Then

${f{\circ}}{g{\circ}}{h}={f{{\left({g{{\left({h}{\left({x}\right)}\right)}}}\right)}}}={f{{\left({g{{\left({x}+{9}\right)}}}\right)}}}={f{{\left({\cos{{\left({x}+{9}\right)}}}\right)}}}={{\left({\cos{{\left({x}+{9}\right)}}}\right)}}^{{2}}$.

It is worth noting, that characteristic of function as composite is not connected with natural functional dependence of ${f{}}$ and ${x}$, it is just a way to represent this dependence.

For example, let ${f{=}}\sqrt{{{1}-{{u}}^{{2}}}}$ for ${u}\in{\left[-{1},{1}\right]}$ and ${y}={\sin{{\left({x}\right)}}}$ for ${x}\in{\left[-\frac{\pi}{{2}},\frac{\pi}{{2}}\right]}$ then ${\left({f{\circ}}{g}\right)}{\left({x}\right)}={f{{\left({g{{\left({x}\right)}}}\right)}}}={f{{\left({\sin{{\left({x}\right)}}}\right)}}}=\sqrt{{{1}-{{\left({\sin{{\left({x}\right)}}}\right)}}^{{2}}}}={\cos{{\left({x}\right)}}}$. Here ${\cos{{\left({x}\right)}}}$ is represented as composite function.