Question

Find the functions (a)\(f \circ g\), (b) \(g \circ f\),(c)\(f \circ f\), and (d) \(g \circ g\)

and their domains.

\(f\left( x \right) = \frac{x}{{1 + x}}\) \(g\left( x \right) = sin2x\)

Short answer

(a) The composition \(f \circ g\) with \(f\left( x \right) = \frac{x}{{1 + x}}\) and \(g\left( x \right) = sin2x\) is \(\frac{{sin2x}}{{1 + sin2x}}\) and has domain\(\left( { - \infty ,\infty } \right)\;{\rm{where}}\;x \ne \frac{{n\pi }}{4}\;\;,\;\;n = ... - 11, - 9, - 3, - 1,5,7,13...\).

(b) The composition \(g \circ f\) with \(f\left( x \right) = \frac{x}{{1 + x}}\) and \(g\left( x \right) = sin2x\) is \(sin\left( {\frac{{2x}}{{1 + x}}} \right)\) and has domain\(\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)\).

(c) The composition \(f \circ f\) with \(f\left( x \right) = \frac{x}{{1 + x}}\) is \(\frac{x}{{1 + 2x}}\) and has domain\(\left( { - \infty , - \frac{1}{2}} \right) \cup \left( { - \frac{1}{2},\infty } \right)\)

(d) The composition \(g \circ g\) with \(g\left( x \right) = sin2x\) is \(sin2\left( {sin2x} \right)\) and has domain\(\left( {{\bf{ - }}\infty {\bf{,}}\infty } \right)\).

Step-by-step solution

Given data

The given functions are;

\(f\left( x \right) = \frac{x}{{1 + x}}\)and \(g\left( x \right) = \sin 2x\)

Step 2:

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

\({\bf{f}} \circ {\bf{g = f}}\left( {{\bf{g}}\left( {\bf{x}} \right)} \right)\)

The composition \(f \circ g\) 

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

\(\begin{array}{c}f \circ g = f\left( {g\left( x \right)} \right)\\ = f\left( {\sin 2x} \right)\\ = \frac{{\sin 2x}}{{1 + \sin 2x}}\end{array}\)

The function is undefined when \(\sin 2x = - 1\) i.e.

\(x = ..., - \frac{{11\pi }}{4}, - \frac{{9\pi }}{4}, - \frac{{3\pi }}{4}, - \frac{\pi }{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4},\frac{{13\pi }}{4},\frac{{15\pi }}{4},...\)

Hence, the composition \(f \circ g\) is \(\frac{{\sin 2x}}{{1 + \sin 2x}}\) with the domain\(\left( { - \infty ,\infty } \right)\;\;\;\;\;x \ne \frac{{n\pi }}{4}\;\;,\;\;n = ... - 11, - 9, - 3, - 1,5,7,13...\).

The composition \(g \circ f\) 

From equation (1),

\(\begin{array}{c}g \circ f = g\left( {\frac{x}{{1 + x}}} \right)\\ = \sin \left( {\frac{{2x}}{{1 + x}}} \right)\end{array}\)

This function is undefined when\(x = - 1\).

Hence, the composition \(g \circ f\) is \(\sin \left( {\frac{{2x}}{{1 + x}}} \right)\) with the domain\(\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)\).

The composition \(f \circ f\) 

From equation (1),

\(\begin{array}{c}f \circ f = f\left( {\frac{x}{{1 + x}}} \right)\\ = \frac{{\frac{x}{{1 + x}}}}{{1 + \frac{x}{{1 + x}}}}\\ = \frac{x}{{1 + 2x}}\end{array}\)

This function is undefined when\(2x = - 1\)that is \(x = - \frac{1}{2}\).

Hence, the composition \(f \circ f\) is \(\frac{x}{{1 + 2x}}\) with the domain\(\left( { - \infty , - \frac{1}{2}} \right) \cup \left( { - \frac{1}{2},\infty } \right)\).

The composition \(g \circ g\) 

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

\(\begin{array}{c}g \circ g = g\left( {g\left( x \right)} \right)\\ = g\left( {\sin 2x} \right)\\ = \sin 2\left( {\sin 2x} \right)\end{array}\)

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

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