Question

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

37. \(f\left( x \right) = \frac{{\bf{1}}}{{\sqrt x }}\) and \(g\left( x \right) = x + {\bf{1}}\)

Short answer

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

(b) \(g \circ f\left( x \right) = \frac{1}{{\sqrt x }} + 1\) and domain of the function is \(\left( {0,\infty } \right)\).

(c) \(f \circ f\left( x \right) = \sqrt(4){x}\) and domain of the function is \(\left( {0,\infty } \right)\).

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

Step-by-step solution

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

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

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

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

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

And,

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

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

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

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

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

For the function \(g \circ f\left( x \right)\) to exist, \(x > 0\).

So, thedomainof the function is \(\left( {0,\infty } \right)\).

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

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

\(\begin{aligned}f \circ f\left( x \right) &= f\left( {f\left( x \right)} \right)\\ &= f\left( {\frac{1}{{\sqrt x }}} \right)\\ &= \frac{1}{{\left( {\sqrt {\frac{1}{{\sqrt x }}} } \right)}}\\ &= \sqrt(4){x}\end{aligned}\)

As \(x > 0\), so the domain of the function is \(\left( {0,\infty } \right)\).

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

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

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

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