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.

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

Short answer

(a) The composition \({\bf{f}} \circ {\bf{g}}\) with \({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}{{\bf{x}}^{\bf{2}}}{\bf{ - 1}}\) and \({\bf{g}}\left( {\bf{x}} \right){\bf{ = 2x + 1}}\) is \({\bf{4}}{{\bf{x}}^{\bf{2}}}{\bf{ + 2x}}\) and has domain\(\left( { - \infty ,\infty } \right)\).

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

(c) The composition \({\bf{f}} \circ {\bf{f}}\) with \({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}{{\bf{x}}^{\bf{2}}}{\bf{ - 1}}\) and \({\bf{g}}\left( {\bf{x}} \right){\bf{ = 2x + 1}}\) is \({{\bf{x}}^{\bf{4}}}{\bf{ - 2}}{{\bf{x}}^{\bf{2}}}\) and has domain\(\left( { - \infty ,\infty } \right)\).

(d) The composition \({\bf{g}} \circ {\bf{g}}\) with \({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}{{\bf{x}}^{\bf{2}}}{\bf{ - 1}}\) and \({\bf{g}}\left( {\bf{x}} \right){\bf{ = 2x + 1}}\) is \({\bf{4x + 3}}\) and has domain\(\left( { - \infty ,\infty } \right)\).

Step-by-step solution

Given data

The given functions are,

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

Step 2:

For two functions\(f\left( x \right)\)and\(g\left( x \right)\), their composition is defined as;

\({\bf{fog = f}}\left( {{\bf{g}}\left( {\bf{x}} \right)} \right)\;\;\;\;\;.....\left( {\bf{1}} \right)\)

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

From equation (1),

\(\begin{aligned}f \circ g &= f\left( {2x + 1} \right)\\ &= {\left( {2x + 1} \right)^2} - 1\\ &= 4{x^2} + 2x + 1 - 1\\ &= 4{x^2} + 2x\end{aligned}\)

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

Hence, the composition function \(f \circ g\) is \(4{x^2} + 2x\) with the domain\(\left( { - \infty ,\infty } \right)\).

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

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

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

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

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

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

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

\(\begin{aligned}f \circ f &= f\left( {f\left( x \right)} \right)\\ &= f\left( {{x^2} - 1} \right)\\ &= {\left( {{x^2} - 1} \right)^2} - 1\\ &= {x^4} + 1 - 2{x^2} - 1\\ &= {x^4} - 2{x^2}\end{aligned}\)

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

Hence, the composition function \(f \circ f\) is \({x^4} - 2{x^2}\) with the domain\(\left( { - \infty ,\infty } \right)\).

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

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

\(\begin{aligned}g \circ g &= g\left( {g\left( x \right)} \right)\\ &= g\left( {2x + 1} \right)\\ &= 2\left( {2x + 1} \right) + 1\\ &= 4x + 3\end{aligned}\)

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

Hence, the composition \(g \circ g\) is \(4x + 3\) with the domain\(\left( { - \infty ,\infty } \right)\).