Let f: R → R and g: R → R be defined as f(x) = \ cases _ e x, & x >...

Given: f\left(x\right) = \left\{\begin{cases}\log_{e}x, & x > 0 \\ {e}^{- x}, & x \le 0\end{cases}\right) and g\left(x\right) = \left\{$\begin{cases}x, & x \ge 0 \\ {e}^{x}, & x < 0\end{cases}$\right)

⇒ g\left(f\left(x\right)\right) = \left\{\begin{cases}f\left(x\right), & f\left(x\right) \ge 0 \\ {e}^{f\left(x\right)}, & f\left(x\right) < 0\end{cases}\right)</p><p>⇒ g\left(f\left(x\right)\right) = \left\{$\begin{cases}{e}^{- x}, & \left(- \infty, 0\right] \\ {e}^{\log x}, & \left(0, 1\right) \\ \log x, & \left[1, \infty\right)\end{cases}$\right)

Graph of $g\left(f\left(x\right)\right)$ is given below:

Range of $g$\left(f\left(x\right)\right) = [ 0, \infty ) \neCodomain

So, $g\left(f\left(x\right)\right)$ is many-one and into.