## University Calculus: Early Transcendentals (3rd Edition)

$g(x)$ is one-to-one.
$g(x)$ is one-to-one. We will prove this by contradiction. Suppose $g(x)$ is not one-to-one, thus there exist $x_1\ne x_2$ such that $g(x_1)=g(x_2)$, but then $f(g(x_1))=f(g(x_2))$, which contradicts the fact that $f \circ g$ is one-to-one.