Answer
a) $x+2$; $D=\left[2,\infty\right)$
b) $\sqrt{x^2+2}$; $D=(-\infty,\infty)$
c) $x^4+8x^2+20$; $D=\left(-\infty,\infty\right)$
d) $\sqrt{\sqrt{x-2}-2}$; $D=[6,\infty)$
Work Step by Step
We are given the functions:
$f(x)=x^2+4$
$g(x)=\sqrt{x-2}$
Let's note:
$D_f$=the domain of $f$
$D_g$=the domain of $g$
We have:
$D_f=(-\infty,\infty)$
$D_g=[2,\infty)$
a) Determine $f\circ g$ and its domain $D_{f\circ g}$:
$(f\circ g)(x)=f(g(x))=f(\sqrt{x-2})=(\sqrt{x-2})^2+4=x-2+4=x+2$
As $D_g=[2,\infty)$, we have:
$D_{f\circ g}=[2,\infty)\cap(-\infty,\infty)=[2,\infty)$
b) Determine $g\circ f$ and its domain $D_{g\circ f}$:
$(g\circ f)(x)=g(f(x))=g(x^2+4)=\sqrt {x^2+4-2}=\sqrt{x^2+2}$
As $D_g=[2,\infty)$, we must have:
$f(x)\geq 2$
$x^2+4\geq 2$
$x^2\geq -2$
$x\in(-\infty,\infty)$
$D_{g\circ f}=(-\infty,\infty)$
c) Determine $f\circ f$ and its domain $D_{f\circ f}$:
$(f\circ f)(x)=f(f(x))=f(x^2+4)=(x^2+4)^2+4=x^4+8x^2+20$
As $D_f=(-\infty,\infty)$, we have:
$D_{f\circ f}=(-\infty,\infty)$
d) Determine $g\circ g$ and its domain $D_{g\circ g}$:
$(g\circ g)(x)=g(g(x))=g(\sqrt{x-2})=\sqrt{\sqrt{x-2}-2}$
As $D_g=[2,\infty)$ we must have:
$g(x)\geq 2$
$\sqrt{x-2}\geq 2$
$x-2\geq 4$
$x\geq 6$
$D_{g\circ g}=[2,\infty)\cap[6,\infty)=[6,\infty)$
