Consider a situation where a hybrid car consumes fuel at a rate of \(5.2\) L per \(100\) km, and gas sells at \( $ 1.35\) per litre.

If \(x\) is the number of kilometers driven, \(C\) is the cost of the gas in dollars, and \(n\) is the number of litres, then

• the cost in dollars for \(n\) litres of fuel is given by \(C(n) = 1.35n\);
• the number of litres required to drive \(x\) kilometres is \(n(x) = \dfrac{5.2x}{100} = 0.052x\); and
• the cost of the fuel to travel \(x\) kilometres depends on the number of litres required.

Thus, the cost of the fuel as a function of the distance travelled is given by \(C(n(x))\).

This situation demonstrates another way in which functions can be combined. This process is referred to as the composition of functions.

When the output of one function is used as the input for another function, the resulting function is called a composite of the two functions.

The function, \(f\) composed with \(g\), is denoted by

and defined by \(\left(f \circ g\right)(x) = f\left(g(x)\right)\).

The function \(f \circ g\) is obtained by substituting \(g(x)\) for all occurrences of \(x\) in \(f(x)\).

The domain of \(\left(f \circ g\right)(x)\) is the set of all values of \(x\) in the domain of \(g\) that produce an output, \(g(x)\), which is a value in the domain of \(f\).

In This Module

• We will expand our understanding of composite functions, combining a variety of functions studied in this course.
• We will determine the composition of two functions numerically, algebraically, and graphically.
• We will identify when the value of a composite function exists by identifying its domain.

• We will look at ways of expressing a given function as a composition of two or more functions.

Given the table of values for \(f(x)\) and \(g(x)\), find:

a. \((g \circ f)(3)~\)\(= g\left(f(3)\right)\)

b. \(f(g(2))\)
c. \(\left(f \circ f\right)(4)\)

Solution

a. From the table, we know that \(f(3) = 1\), so

Given the table of values for \(f(x)\) and \(g(x)\), find:

a. \((g \circ f)(3)~\)\(= g\left(f(3)\right)\)

b. \(f(g(2))\)
c. \(\left(f \circ f\right)(4)\)

Solution

b. Since \(g(2) = 5\),

c. Since \(f(4) = 3\),

Given the table of values for \(f(x)\) and \(g(x)\), find:

a. \((g \circ f)(3)~\)\(= g\left(f(3)\right)\)

b. \(f(g(2))\)
c. \(\left(f \circ f\right)(4)\)

Solution

Consider \(f(g(1))\).

Since \(g(1) = 6\), then \(f(g(1)) = f(6)\).

However, \(f\) is not defined when \(x = 6\).

That is, \(x = 6\) is not in the domain of \(f\).

Thus, \(f(g(1))\) is not defined. We can therefore conclude that \(x = 1\) is not in the domain of \(f \circ g\).

Recall: The domain of \(\left(f \circ g\right)(x)\) is the set of all values of \(x\) in the domain of \(g\) that produce an output, \(g(x)\), which is in the domain of \(f\).

The domain of \(\left(f \circ g\right)(x)\) is \(\left\{ 2, 3, 4, 5 \right\}\).