Question

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

\(f\left( x \right) = \sqrt x \) and \(g\left( x \right) = \sqrt(3){{1 - x}}\)

Short answer

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

(b) The composition \(g \circ f\) with \(f\left( x \right) = \sqrt x \) and \(g\left( x \right) = \sqrt(3){{1 - x}}\) is \(\sqrt(3){{1 - \sqrt x }}\) and has domain\(\left( {{\bf{0,}}\infty } \right)\).

(c) The composition \(f \circ f\) with \(f\left( x \right) = \sqrt x \) is \(\sqrt {\sqrt x } \) and has domain \(\left( {{\bf{0,}}\infty } \right)\).

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

Step-by-step solution

Given data

The given functions are,

\(f\left( x \right) = \sqrt x \) and \(g\left( x \right) = \sqrt(3){{1 - x}}\)

Composite function

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;

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

Determine the composite function \(f \circ g\) 

(a)

Evaluate \(f \circ g\)

\(\begin{array}{c}f \circ g = f\left( {g\left( x \right)} \right)\\ = f\left( {\sqrt(3){{1 - x}}} \right)\\ = \sqrt {\sqrt(3){{1 - x}}} \end{array}\)

The function is undefined when the argument of the square root is negative, that is \(x > 1\).

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

Determine the composite function \(g \circ f\) 

(b)

Evaluate\(g \circ f\)

\(\begin{array}{c}g \circ f = g\left( {\sqrt x } \right)\\ = \sqrt(3){{1 - \sqrt x }}\end{array}\)

The function is undefined when the argument of the square root is negative, that is \(x < 0\).

Hence, the composition \(g \circ f\) is \(\sqrt(3){{1 - \sqrt x }}\) with the domain \(\left( {0,\infty } \right)\).

Determine the composite \(f \circ f\) 

(c)

Evaluate\(f \circ f\)

\(\begin{array}{c}f \circ f = f\left( {f\left( x \right)} \right)\\ = f\left( {\sqrt x } \right)\\ = \sqrt {\sqrt x } \end{array}\)

The function is undefined when the argument of the square root is negative, that is \(x < 0\).

Hence, the composition \(f \circ f\) is \(\sqrt {\sqrt x } \) with the domain\(\left( {0,\infty } \right)\).

The composition \(g \circ g\) 

(d)

Evaluate\(g \circ g\)

\(\begin{array}{c}g \circ g = g\left( {g\left( x \right)} \right)\\ = g\left( {\sqrt(3){{1 - x}}} \right)\\ = \sqrt(3){{1 - \sqrt(3){{1 - x}}}}\end{array}\)

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

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