Question

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

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

Short answer

(a) The composition \(f \circ g\) is \(\frac{{x + 1}}{{x + 2}} + \frac{{x + 2}}{{x + 1}}\) with the domain\(x \in \mathbb{R} - \left\{ { - 1, - 2} \right\}\).

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

(c) The composition \(f \circ f\) is \(\frac{{{{\left( {{x^2} + 1} \right)}^2} + {x^2}}}{{x\left( {{x^2} + 1} \right)}}\) with the domain\(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\).

(d) The composition \(g \circ g\) is \(\frac{{2x + 3}}{{3x + 5}}\) with the domain\(\left( { - \infty , - \frac{5}{3}} \right) \cup \left( { - \frac{5}{3},\infty } \right)\).

Step-by-step solution

Given data

The provided functions are;

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

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;

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

The composition \({\bf{f}} \circ {\bf{g}}\) 

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

\(\begin{aligned}f \circ g &= f\left( {g\left( x \right)} \right)\\ &= f\left( {\frac{{x + 1}}{{x + 2}}} \right)\\ &= \frac{{x + 1}}{{x + 2}} + \frac{{x + 2}}{{x + 1}}\end{aligned}\)

This function is defined everywhere except at \(x = - 1, - 2.\)

Therefore the domain is\(x \in \mathbb{R} - \left\{ { - 1, - 2} \right\}\).

Hence, the composition \(f \circ g\) is \(\frac{{x + 1}}{{x + 2}} + \frac{{x + 2}}{{x + 1}}\) with the domain\(x \in \mathbb{R} - \left\{ { - 1, - 2} \right\}\)

The composition \({\bf{g}} \circ {\bf{f}}\) 

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

\(\begin{aligned}g \circ f &= g\left( {f\left( x \right)} \right)\\ &= g\left( {x + \frac{1}{x}} \right)\\ &= \frac{{x + \frac{1}{x} + 1}}{{x + \frac{1}{x} + 2}}\\ &= \frac{{{x^2} + x + 1}}{{{x^2} + 2x + 1}}\\ &= \frac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}}}\end{aligned}\)

This function is defined everywhere except at\(x = - 1\).

Therefore the domain is\(\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)\).

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

The composition \({\bf{f}} \circ {\bf{f}}\) 

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

\(\begin{aligned}f \circ f &= f\left( {f\left( x \right)} \right)\\ &= f\left( {x + \frac{1}{x}} \right)\\ &= x + \frac{1}{x} + \frac{1}{{x + \frac{1}{x}}}\\ &= \frac{{{x^2} + 1}}{x} + \frac{x}{{{x^2} + 1}}\\ &= \frac{{{{\left( {{x^2} + 1} \right)}^2} + {x^2}}}{{x\left( {{x^2} + 1} \right)}}\end{aligned}\)

The above function is not defined at\(x = 0\).

Therefore the domain is\(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\)

Hence, the composition \(f \circ f\) is \(\frac{{{{\left( {{x^2} + 1} \right)}^2} + {x^2}}}{{x\left( {{x^2} + 1} \right)}}\) with the domain\(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\).

The composition \({\bf{g}} \circ {\bf{g}}\) 

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

\(\begin{aligned}g \circ g &= g\left( {g\left( x \right)} \right)\\ &= g\left( {\frac{{x + 1}}{{x + 2}}} \right)\\ &= \frac{{\frac{{x + 1}}{{x + 2}} + 1}}{{\frac{{x + 1}}{{x + 2}} + 2}}\\ &= \frac{{2x + 3}}{{3x + 5}}\end{aligned}\)

This function is defined everywhere except\(x = - \frac{5}{3}\).

Thus, the domain is\(\left( { - \infty , - \frac{5}{3}} \right) \cup \left( { - \frac{5}{3},\infty } \right)\).

Hence, the composition \(g \circ g\) is \(\frac{{2x + 3}}{{3x + 5}}\) with the domain\(\left( { - \infty , - \frac{5}{3}} \right) \cup \left( { - \frac{5}{3},\infty } \right)\).