`f(x)={(log_(e)x",",0lt x lt 1),(x^(2)-1 ",",x ge 1):}` and `g(x)={(x+1",",x lt 2),(x^(2)-1 ",",x ge 2):}`.
Then find `g(f(x)).`

Explanation

To find \( g(f(x)) \) where

\[ f(x) = \begin{cases} \log_e x & \text{for } 0 < x < 1 \\ x^2 - 1 & \text{for } x \geq 1 \end{cases} \]

and

\[ g(x) = \begin{cases} x + 1 & \text{for } x < 2 \\ x^2 - 1 & \text{for } x \geq 2 \end{cases} \]

we will evaluate \( g(f(x)) \) by considering the two cases for \( f(x) \).

### Step 1: Evaluate \( f(x) \) for \( 0 < x < 1 \)

In this case, \( f(x) = \log_e x \).

We need to check if \( f(x) < 2 \) or \( f(x) \geq 2 \):

- Since \( \log_e x \) is negative for \( 0 < x < 1 \), it follows that \( f(x) < 2 \).

Thus, we will use the first case of \( g(x) \):

\[ g(f(x)) = g(\log_e x) = \log_e x + 1 \]

### Step 2: Evaluate \( f(x) \) for \( x \geq 1 \)

In this case, \( f(x) = x^2 - 1 \).

Next, we need to check if \( f(x) < 2 \) or \( f(x) \geq 2 \):

- For \( x = 1 \), \( f(1) = 1^2 - 1 = 0 < 2 \).
- For \( x = 2 \), \( f(2) = 2^2 - 1 = 3 \geq 2 \).
- For \( x > 2 \), \( f(x) = x^2 - 1 \) will also be greater than 2.

Thus, we can conclude:
- For \( 1 \leq x < 2 \), \( f(x) < 2 \) and we use the first case of \( g(x) \):

\[ g(f(x)) = g(x^2 - 1) = (x^2 - 1) + 1 = x^2 \]

- For \( x \geq 2 \), \( f(x) \geq 2 \) and we use the second case of \( g(x) \):

\[ g(f(x)) = g(x^2 - 1) = (x^2 - 1)^2 - 1 \]

### Final Result

Combining both cases, we have:

\[ g(f(x)) = \begin{cases} \log_e x + 1 & \text{for } 0 < x < 1 \\ x^2 & \text{for } 1 \leq x < 2 \\ (x^2 - 1)^2 - 1 & \text{for } x \geq 2 \end{cases} \]