Answer
a) $\sqrt{-2x-1}$; $D=\left(-\infty, \dfrac{1}{2}\right]$
b) $1-2\sqrt {x-2}$; $D=\left[2,\infty\right)$
c) $\sqrt{\sqrt{x-2}-2}$; $D=\left[6,\infty\right)$
d) $4x-1$; $D=(-\infty,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\sqrt {x-2}$
$g(x)=1-2x$
Let's note:
$D_f$=the domain of $f$
$D_g$=the domain of $g$
We have:
$D_f=[2,\infty)$
$D_g=(-\infty,\infty)$
a) Determine $f\circ g$ and its domain $D_{f\circ g}$:
$(f\circ g)(x)=f(g(x))=f(1-2x)=\sqrt{1-2x-2}=\sqrt{-2x-1}$
As $D_f=[2,\infty)$, we must have:
$g(x)\geq 2$
$1-2x\geq 0$
$x\leq \dfrac{1}{2}$
$D_{f\circ g}=(-\infty,\infty)\cap\left(-\infty, \dfrac{1}{2}\right]=\left(-\infty, \dfrac{1}{2}\right]$
b) Determine $g\circ f$ and its domain $D_{g\circ f}$:
$(g\circ f)(x)=g(f(x))=g(\sqrt {x-2})=1-2\sqrt {x-2}$
As $D_f=[2,\infty)$, we must have:
$D_{g\circ f}=\left[2,\infty\right)$
c) Determine $f\circ f$ and its domain $D_{f\circ f}$:
$(f\circ f)(x)=f(f(x))=f(\sqrt {x-2})=\sqrt{\sqrt {x-2}-2}$
As $D_f=[2,\infty)$, we must also have:
$f(x)\geq 2$
$\sqrt{x-2}\geq 2$
$x-2\geq 4$
$x\geq 6$
$D_{f\circ f}=\left[2,\infty\right)\cap\left[6,\infty\right)=\left[6,\infty\right)$
d) Determine $g\circ g$ and its domain $D_{g\circ g}$:
$(g\circ g)(x)=g(g(x))=g(1-2x)=1-2(1-2x)=1-2+4x=4x-1$
$D_{g\circ g}=(-\infty,\infty)$
