Answer
$f\circ g$ is also one-to-one.
(see proof below)
Work Step by Step
If we take any two distinct values for x such that $x_{1}\neq x_{2},$ since $g$ is one-to-one, it follows that $g(x_{1})\neq g(x_{2}).$
Now, $g(x_{1})$ and$ g(x_{2})$ are two distinct values in the domain of f. Being different, it follows that $f[g(x_{1})]\neq f[g(x_{2})],$ because f is one-to-one.
So, since $f\circ g $ is defined as $f(g(x))$, from $x_{1}\neq x_{2}$, it follows that $(f\circ g)(x_{1})\neq(f\circ g)(x_{2}),$
This means that $f\circ g$ is also one-to-one.
