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 of finding \( f(g(x)) \) where \( g(x) = 1 + x - [x] \) and \( f(x) \) is defined as:

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

we will follow these steps:

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

The function \( g(x) = 1 + x - [x] \) can be interpreted as follows:
- \( [x] \) is the greatest integer less than or equal to \( x \).
- Therefore, \( x - [x] \) gives the fractional part of \( x \), which is always in the range \( [0, 1) \).
- Thus, \( g(x) = 1 + (x - [x]) \) will always yield a value in the range \( [1, 2) \) because:
- The minimum value occurs when \( x \) is an integer (making \( g(x) = 1 + 0 = 1 \)).
- The maximum value occurs just before \( x \) reaches the next integer (making \( g(x) \) approach \( 2 \) but never reach it).

### Step 2: Analyze \( f(g(x)) \)

Now, we need to evaluate \( f(g(x)) \). Since we determined that \( g(x) \) always yields values in the interval \( [1, 2) \), we can analyze \( f(x) \) based on this range:

- Since \( g(x) \) is always greater than \( 0 \) (specifically, \( g(x) \geq 1 \)), we can conclude:
- \( g(x) > 0 \) implies \( f(g(x)) = 1 \) (because \( f(x) = 1 \) when \( x > 0 \)).

### Conclusion

Thus, for all \( x \):

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