Answer
a) $\sqrt{2x+3}$; $D=\left[-\dfrac{3}{2},\infty\right)$
b) $2\sqrt x+3$; $D=\left[0,\infty\right)$
c) $\sqrt[4] x$; $D=\left[0,\infty\right)$
d) $4x+9$; $D=(-\infty,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\sqrt x$
$g(x)=2x+3$
Let's note:
$D_f$=the domain of $f$
$D_g$=the domain of $g$
We have:
$D_f=[0,\infty)$
$D_g=(-\infty,\infty)$
a) Determine $f\circ g$ and its domain $D_{f\circ g}$:
$(f\circ g)(x)=f(g(x))=f(2x+3)=\sqrt{2x+3}$
As $D_f=[0,\infty)$, we must have:
$g(x)\geq 0$
$2x+3\geq 0$
$x\geq -\dfrac{3}{2}$
$D_{f\circ g}=(-\infty,\infty)\cap\left[-\dfrac{3}{2},\infty\right)=\left[-\dfrac{3}{2},\infty\right)$
b) Determine $g\circ f$ and its domain $D_{g\circ f}$:
$(g\circ f)(x)=g(f(x))=g(\sqrt x)=2\sqrt x+3$
As $D_f=[0,\infty)$, we must have:
$D_{g\circ f}=\left[0,\infty\right)$
c) Determine $f\circ f$ and its domain $D_{f\circ f}$:
$(f\circ f)(x)=f(f(x))=f(\sqrt x)=\sqrt{\sqrt x}=\sqrt[4] x$
As $D_f=[0,\infty)$ and $f(x)=\sqrt x\geq 0$, we must have:
$D_{f\circ f}=\left[0,\infty\right)$
d) Determine $g\circ g$ and its domain $D_{g\circ g}$:
$(g\circ g)(x)=g(g(x))=g(2x+3)=2(2x+3)+3=4x+9$
$D_{g\circ g}=(-\infty,\infty)$
