## Precalculus (10th Edition)

$f(x)=|x|$ and $g(x)=2x^2+3$
If $f(x)=|x|$ and $g(x)=2x^2+3$ then $(f\circ g)(x)=f(g(x))=f(2x^2+3)=|2x^2+3|=H(x).$