MA005: Calculus I

Basic functions are often combined with each other to describe more complicated situations. Here we will consider the composition of functions, functions of functions.

Definition: The composite of two functions $f$ and $g$ , written $f \circ g$, is $f \circ g(x) ≡ f( g(x) )$.

The domain of the composite function $f \circ g(x) = f( g(x) )$ consists of those x–values for which $g(x)$ and $f( g(x) )$ are both defined - we can evaluate the composition of two functions at a point x only if each step in the composition is defined.

If we think of our functions as machines, then composition is simply
a new machine consisting of an arrangement of the original machines. The composition $f \circ g$ of the function machines $f$ and $g$ shown in Fig. 5(a) is an arrangement of the machines so that the original input $x$ goes into machine $g$ , the output from machine $g$ becomes the input into machine $f$, and the output from machine $f$ is our final output. The composition of the function machines $f \circ g(x) = f( g(x) )$ is only valid if $x$ is an allowable input into $g (x$ is in the domain of $g)$ and if $g(x)$ is then an allowable input into $f$ . The composition $g \circ f$ involves arranging the machines so the original input goes into $f$, and the output from $f$ then becomes the input for $g$ (Fig. 5(b) ).

Example 2: For $f(x) = x – 2 , g(x) = x^2$, and $h(x) = \left\{ \begin{array}{ll} 3x & \mbox { if x < 2 } \\ x – 1 & \mbox { if 2 ≤ x }\end{array} \right.$, evaluate $f \circ g(3)$, $g \circ f(6)$, $f \circ h(2)$ and $h \circ g(–3)$. Find the equations and domains of $f \circ g(x)$ and $g \circ f(x)$.

Solution: $f \circ g(3) = f( g(3) ) = f( 3^2 ) = f( 9 ) = \sqrt{9 – 2} = \sqrt7 ≈ 2.646$
$g \circ f(6) = g( f(6) ) = g(\sqrt{6 – 2} ) = g( \sqrt 4 ) = g( 2 ) = 2^2 = 4$
$f \circ h(2) = f( h(2) ) = f( 2 – 1 ) = f( 1 ) = \sqrt{1 – 2} = \sqrt{–1}$ which is undefined
$h \circ g(–3) = h( g(–3) ) = h( 9 ) = 9 – 1 = 8$.

$f \circ g(x) = f( g(x) ) = f( x^2 ) = x^2 – 2$, and the domain of $f \circ g$ is those x–values for which $x^2 – 2 ≥ 0$ so the domain of $f \circ g$ is all $x$ such that $x ≥ 2$ or $x ≤ – 2$.

$g \circ f(x) = g( f(x) ) = g(\sqrt{ x – 2} ) = { x – 2 }^2 = x – 2$, but we can evaluate the first piece, $f$, of the composition only if $f(x) = x – 2$ is defined, so the domain of $g \circ f$ is all $x ≥ 2$.

Practice 4: For $f(x) = \frac{x}{x–3}$ , $g(x) = \sqrt{1+x}$ , and $h(x) = \left\{ \begin{array}{ll} 2x & \mbox { if  x ≤ 1}\\ 5 – x & \mbox { if 1 < x}\end{array} \right.$.
Evaluate $f \circ g(3)$, $f \circ g(8)$, $g \circ f(4)$, $f \circ h(1)$, $f \circ h(3)$, $f \circ h(2)$ and $h \circ g(–1)$. Find the equations for $f \circ g(x)$ and $g \circ f(x)$.