Answer
$fog$ is a one-to-one function.
Work Step by Step
- First, since $g(x)$ is one-to-one, according to definition:
$$g(x_1)\ne g(x_2)\hspace{1cm}\text{whenever}\hspace{1cm} x_1\ne x_2$$
- And since $g(x_1)\ne g(x_2)$, we can deduce
$$f(g(x_1))\ne f(g(x_2))$$
as $f$ is also one-to-one.
This, however, means that $fog(x_1)\ne fog(x_2)$ whenever $x_1\ne x_2$.
Therefore, $fog$ is a one-to-one function.
