Question

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

40. \(f\left( x \right) = \sqrt {{\bf{5}} - x} \) and \(g\left( x \right) = \sqrt {x - {\bf{1}}} \)

Short answer

(a) \(f \circ g\left( x \right) = \sqrt {5 - \sqrt {x - 1} } \) and domain of the function is \(\left( {1,26} \right)\).

(b) \(g \circ f\left( x \right) = \sqrt {\sqrt {5 - x} - 1} \) and domain of the function is \(\left( { - \infty ,4} \right)\).

(c) \(f \circ f\left( x \right) = \sqrt {5 - \sqrt {5 - x} } \) and domain of the function is \(\left( { - 20,5} \right)\).

(d) \(g \circ g\left( x \right) = \sqrt {\sqrt {x - 1} - 1} \) and domain of the function is \(\left( {2,\infty } \right)\).

Step-by-step solution

Find the function \(f \circ g\) and its domain

Thefunction\(f \circ g\left( x \right)\) can be evaluated as follows:

\(\begin{aligned}f \circ g\left( x \right) &= f\left( {\sqrt {x - 1} } \right)\\ &= \sqrt {5 - \sqrt {x - 1} } \end{aligned}\)

For the function \(f \circ g\left( x \right)\) to exist;

\(\begin{aligned}x - 1 &\ge 0\\x &\ge 1\end{aligned}\)

And,

\(\begin{aligned}5 - \sqrt {x - 1} &\ge 0\\\sqrt {x - 1} &\le 5\\0 &\le x - 1 \le 25\\1 &\le x \le 26\end{aligned}\)

So, the domain of the function is \(\left( {1,26} \right)\).

Find the function \(g \circ f\) and its domain

The function \(g \circ f\left( x \right)\) can be evaluated as follows:

\(\begin{aligned}g \circ f\left( x \right) &= g\left( {f\left( x \right)} \right)\\ &= g\left( {\sqrt {5 - x} } \right)\\ &= \sqrt {\sqrt {5 - x} - 1} \end{aligned}\)

For the function \(g \circ f\left( x \right)\) to exist;

\(\begin{aligned}5 - x &\ge 0\\x &\le 5\end{aligned}\)

And.

\(\begin{aligned}\sqrt {5 - x} - 1 &\ge 0\\5 - x &\ge 1\\x &\le 4\end{aligned}\)

So, thedomainof the function is \(\left( { - \infty ,4} \right)\).

Find the function \(f \circ f\) and its domain

The function \(f \circ f\left( x \right)\) can be evaluated as follows:

\(\begin{aligned}f \circ f\left( x \right) &= f\left( {f\left( x \right)} \right)\\ &= f\left( {\sqrt {5 - x} } \right)\\ &= \sqrt {5 - \sqrt {5 - x} } \end{aligned}\)

For the function \(f \circ f\left( x \right)\) to exist;

\(\begin{aligned}5 - x &\ge 0\\x &\le 5\end{aligned}\)

And,

\(\begin{aligned}5 - \sqrt {5 - x} &\ge 0\\\sqrt {5 - x} &\le 5\\0 &\le 5 - x \le 25\\ - 20 &\le x \le 5\end{aligned}\)

The domain of the function is \(\left( { - 20,5} \right)\).

Find the function \(g \circ g\) and its domain

The function \(g \circ g\left( x \right)\) can be evaluated as follows:

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

For the function \(g \circ g\left( x \right)\) to exist;

\(\begin{aligned}x - 1 &\ge 0\\x &\ge 1\end{aligned}\)

And,

\(\begin{aligned}\sqrt {x - 1} - 1 &\ge 0\\\sqrt {x - 1} &\ge 1\\x - 1 &\ge 1\\x &\ge 2\end{aligned}\)

The domain of the function is \(\left( {2,\infty } \right)\).