Let g(x)=1+x-[x]
`"and " f(x)={{:(-1","x lt 0),(0","x=0),(1","x gt 0):}`
Then, for all x, find f(g(x)).

Answer

To solve the problem, we need to find \( f(g(x)) \) where \( g(x) = 1 + x - [x] \) and \( f(x) \) is defined piecewise as follows:

\[ f(x) = \begin{cases} -1 & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 & \text{if } x > 0 \end{cases} \]

### Step 1: Analyze \( g(x) \)

The function \( g(x) = 1 + x - [x] \) can be rewritten as:

\[ g(x) = 1 + (x - [x]) = 1 + \{x\} \]

where \( \{x\} \) is the fractional part of \( x \). The fractional part \( \{x\} \) is always in the range \( [0, 1) \).

### Step 2: Determine the range of \( g(x) \)

Since \( \{x\} \) ranges from \( 0 \) to \( 1 \), we can find the range of \( g(x) \):

\[ g(x) = 1 + \{x\} \quad \Rightarrow \quad g(x) \in [1, 2) \]

### Step 3: Evaluate \( f(g(x)) \)

Now we need to evaluate \( f(g(x)) \) based on the range of \( g(x) \):

1. **For \( g(x) < 0 \)**: This is not possible since \( g(x) \) is always at least \( 1 \).
2. **For \( g(x) = 0 \)**: This is also not possible since \( g(x) \) is always at least \( 1 \).
3. **For \( g(x) > 0 \)**: Since \( g(x) \) is always in the range \( [1, 2) \), it is always greater than \( 0 \).

Thus, we can conclude that:

\[ f(g(x)) = 1 \quad \text{for all } x \]

### Final Answer

The final result is:

\[ f(g(x)) = 1 \quad \text{for all } x \]