Question

Find the functions (a)\({\bf{f}} \circ {\bf{g}}\), (b)\({\bf{g}} \circ {\bf{f}}\), (c)\({\bf{f}} \circ {\bf{f}}\), (d) \({\bf{g}} \circ {\bf{g}}\) and their domains.

\({\bf{f}}\left( {\bf{x}} \right){\bf{ = 1 - 3x}}\) \({\bf{g}}\left( {\bf{x}} \right){\bf{ = cosx}}\)

Short answer

(a) The composition \({\bf{f}} \circ {\bf{g}}\) with \({\bf{f}}\left( {\bf{x}} \right){\bf{ = 1 - 3x}}\) and \({\bf{g}}\left( {\bf{x}} \right){\bf{ = cosx}}\) is \({\bf{1 - 3cosx}}\) and has domain\(\left( { - \infty ,\infty } \right)\).

(b) The composition \({\bf{g}} \circ {\bf{f}}\) with \({\bf{f}}\left( {\bf{x}} \right){\bf{ = 1 - 3x}}\) and \({\bf{g}}\left( {\bf{x}} \right){\bf{ = cosx}}\) is \({\bf{cos(1 - 3x)}}\) and has domain\(\left( { - \infty ,\infty } \right)\).

(c) The composition \({\bf{f}} \circ {\bf{f}}\) with \({\bf{f}}\left( {\bf{x}} \right){\bf{ = 1 - 3x}}\) is \({\bf{9}}{{\bf{x}}^{\bf{2}}}{\bf{ - 2}}\) and has domain\(\left( { - \infty ,\infty } \right)\).

(d) The composition \({\bf{g}} \circ {\bf{g}}\) with \({\bf{g}}\left( {\bf{x}} \right){\bf{ = cosx}}\) is \({\bf{cos(cosx)}}\) and has domain\(\left( { - \infty ,\infty } \right)\).

Step-by-step solution

Given data

The given functions are,

\(f\left( x \right) = 1 - 3x\)and\(g\left( x \right) = \cos x\)

Step 2:

A composite function is formed when one function is substituted into another function.For two functions\(f\left( x \right)\)and\(g\left( x \right)\), their composition is defined as;

Determine the composition function \({\bf{f}} \circ {\bf{g}}\) 

Evaluate\({\bf{f}} \circ {\bf{g}}\)

\(\begin{aligned}f \circ g &= f\left( {g\left( x \right)} \right)\\ &= f\left( {\cos x} \right)\\ &= 1 - 3\cos x\end{aligned}\)

This function is well defined everywhere. Thus, the domain is \(\left( { - \infty ,\infty } \right)\).

Hence, the composition \(f \circ g\) is \(1 - 3\cos x\) with the domain \(\left( { - \infty ,\infty } \right)\).

Determine the composition function \({\bf{g}} \circ {\bf{f}}\) 

Evaluate\({\bf{g}} \circ {\bf{f}}\)

\(\begin{aligned}g \circ f &= g\left( {f\left( x \right)} \right)\\ &= g\left( {1 - 3x} \right)\\ &= \cos (1 - 3x)\end{aligned}\)

This function is well defined everywhere. Thus, the domain is \(\left( { - \infty ,\infty } \right)\).

Hence, the composition \(g \circ f\) is \(\cos (1 - 3x)\) with the domain \(\left( { - \infty ,\infty } \right)\).

Determine the composition function \({\bf{f}} \circ {\bf{f}}\) 

Evaluate\({\rm{f}} \circ {\bf{f}}\)

\(\begin{aligned}f \circ f &= f\left( {f\left( x \right)} \right)\\ &= f\left( {1 - 3x} \right)\\ &= 1 - 3(1 - 3x)\\ &= 9{x^2} - 2\end{aligned}\)

This function is well defined everywhere. Thus, the domain is \(\left( { - \infty ,\infty } \right)\).

Hence, the composition \(f \circ f\) is \(9{x^2} - 2\) with the domain \(\left( { - \infty ,\infty } \right)\).

Determine the composition function \({\bf{g}} \circ {\bf{g}}\) 

Now evaluate\({\rm{g}} \circ {\rm{g}}\)

\(\begin{aligned}g \circ g &= g\left( {g\left( x \right)} \right)\\ &= g\left( {\cos x} \right)\\ &= \cos (\cos x)\end{aligned}\)

This function is well defined everywhere. Thus, the domain is\(\left( { - \infty ,\infty } \right)\).

Hence, the composition \(g \circ g\) is \(\cos (\cos x)\) with the domain\(\left( { - \infty ,\infty } \right)\).