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