## Calculus: Early Transcendentals (2nd Edition)

$g(x)=\frac{\sqrt{2}}{x^6+x^2+1}$ and $f(x)=x^2$ The domain of $h$ is $\mathbb{R}$.
If we set $g(x)=\frac{\sqrt{2}}{x^6+x^2+1}$ and $f(x)=x^2$, then: $f \circ g=f(g(x))=(\frac{\sqrt{2}}{x^6+x^2+1})^2=\frac{2}{(x^6+x^2+1)^2}$ The domain of $h$ is $\mathbb{R}$ (because the denominator is never 0).