## Calculus: Early Transcendentals (2nd Edition)

$g(x)=\frac{1}{x^3-1}$ and $f(x)=\sqrt{x}$ Or: $g(x)=x^3-1$ and $f(x)=1/\sqrt{x}$ Domain: $(1,\infty)$
If we set $g(x)=\frac{1}{x^3-1}$ and $f(x)=\sqrt{x}$, then: $f \circ g=f(g(x))=\sqrt{\frac{1}{x^3-1}}=\frac{1}{\sqrt{x^3-1}}$ To find the domain of $h$, we need to make sure that $x^3-1>0$, so that we don't divide by 0 or take the square root of a negative number. $x^3-1>0$ for values of $x>1$. So the domain is: $(1,\infty)$. Note that other solutions are also possible. For example, $g(x)=x^3-1$ and $f(x)=1/\sqrt{x}$.