Find the range of each of the following functions: (where {.} and [.] represents fractional part and greatest integer part functions respectively)
`f(x)=1/(1+sqrt(x))`

Explanation

To find the range of the function \( f(x) = \frac{1}{1 + \sqrt{x}} \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
Let \( f(x) = y \). Thus, we have:
\[ y = \frac{1}{1 + \sqrt{x}} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ y(1 + \sqrt{x}) = 1 \]
This simplifies to:
\[ y + y\sqrt{x} = 1 \]

### Step 3: Isolate \( \sqrt{x} \)
Rearranging the equation to isolate \( \sqrt{x} \):
\[ y\sqrt{x} = 1 - y \]
\[ \sqrt{x} = \frac{1 - y}{y} \]

### Step 4: Square both sides to solve for \( x \)
Now, squaring both sides:
\[ x = \left(\frac{1 - y}{y}\right)^2 \]

### Step 5: Determine the conditions for \( y \)
Since \( \sqrt{x} \) must be non-negative (i.e., \( \sqrt{x} \geq 0 \)), the right-hand side must also be non-negative:
\[ \frac{1 - y}{y} \geq 0 \]
This inequality implies two conditions:
1. \( 1 - y \geq 0 \) (which gives \( y \leq 1 \))
2. \( y > 0 \) (since \( y \) is in the denominator)

### Step 6: Combine the conditions
From the above conditions, we conclude:
\[ 0 < y < 1 \]

### Conclusion
Thus, the range of the function \( f(x) = \frac{1}{1 + \sqrt{x}} \) is:
\[ (0, 1) \]