(i) Determine whether composite functions \(fg\) and \(gf\) exists
\(f : x \rightarrow 2x + 1, \: x \in R, \: x > 4\)
Domain of \(f\)
\(D_f : x > 4\)
Range of \(f\)
For \(x = 4, \: y = 9\)
For \(x \rightarrow +\sim, \: y \rightarrow +\sim\)
\(R_f : y > 9\)
\(g : x \rightarrow \sqrt{x}, \: \: x \in R, \: x \geq 0\)
Domain of \(g\)
\(D_g : x \geq 0\)
Range of \(g\)
\(R_g : y \geq 0\)
Composite function \(fg\)
\(R_g \nsubseteq D_f\), hence \(fg\) does not exist.
Composite function \(gf\)
\(R_f \subseteq D_g\), hence \(gf\) exists.
(ii) Determine the composite function at (i), its domain and range
Composite function \(gf\)
substitute \(f\) into \(g\)
\begin{equation*}
\begin{split}
gf & = g (2x + 1) \\\\
gf & = \sqrt{2x + 1}
\end{split}
\end{equation*}
\(gf : x \rightarrow \sqrt{2x + 1}, \: x > 4\)
Range of \(gf\)
For \(x \rightarrow 4, \: y \rightarrow 3\)
For \(x \rightarrow +\sim, \: y \rightarrow +\sim\)
\(R_{gf} : y > 3\)
Note:
Domain of \(gf\) is same with domain of \(f\)