Order

The compositions $\blue{g(\green{h(x)})}$ and $\green{h( \blue{g(x)})}$ are generally not equal to each other. We can, for example, choose $\blue{x^2}$ and $\green{x+1}$. This gives us:

$$\begin{array}{crl}\blue{g( \green{h(x)})} &=& \blue{(\green{x+1})^2}=x^2+2x+1\\ \green{h(\blue{g(x)})} &=& x^2+1\end{array}$$

This means $\blue{g( \green{h(x)})}$ and $\green{h( \blue{g(x)})}$ are different functions.

Notation

When we write $\blue{g(x)}=\blue{\sqrt{x}}$ and $\green{h(x)}=\green{x+3}$, the variable $x$ in $\blue{g(x)}$ has nothing to do with the variable $x$ in $\green{ h(x) }$. The $x$ in these functions only has meaning when combined with $\blue{g(x)}= \ldots$ or $\green{ h(x) }= \ldots$. We might as well write $\blue{g(h)} = \blue{\sqrt{h}}$. Now $h$ is the variable. Regardless of what variable we use for $\green{g(x)}$, the formula $\blue{g(}\green{h(x)}\blue{)}=\blue{\sqrt{\green{x+3}}}$ still holds.

Graph
Say $\blue{g(h)}=\blue{h^4}$ and $\green{h(x)}=\green{x^2-5x}$. Then $f(x)=\blue{g(}\green{h(x)}\blue{)}=\blue{\left(\green{x^2 - 5x} \right)^{4}}$.

The composition looks as follows.

Note that we used the alternative notation as mentioned in the subbox "Notation".

Alternative notation

We can prevent confusion between different $x$'s by writing the function $g(x)$ as $g(h)$. By doing this, the examples above can be rewritten as:

$$\begin{array}{llll}f(x)=&\sqrt{x+2}&=\blue{g(}\green{h(x)}\blue{)}\quad\text{gives}\quad\blue{g(h)}=\blue{\sqrt{h}}\quad\text{and}\quad\green{h(x)}=\green{x+2}\\\\f(x)=&\sin(x^2)&=\blue{g(}\green{h(x)}\blue{)}\quad\text{gives}\quad\blue{g(h)}=\blue{\sin(h)}\quad\text{and}\quad\green{h(x)}=\green{x^2}\\\\f(x)=&(x^3-4x)^6&=\blue{g(}\green{h(x)}\blue{)}\quad\text{gives}\quad\blue{g(h)}=\blue{h^6}\quad\text{and}\quad\green{h(x)}=\green{x^3-4x}\end{array}$$