`f(x)=e^(x)/([x+1]),x ge 0`

Answer

To find the range of the function \( f(x) = \frac{e^x}{\lfloor x \rfloor + 1} \) where \( x \geq 0 \), we will follow these steps:

### Step 1: Understand the components of the function
The function consists of two parts:
- The numerator \( e^x \), which is an exponential function that increases rapidly as \( x \) increases.
- The denominator \( \lfloor x \rfloor + 1 \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). This means that for any integer \( n \), when \( n \leq x < n+1 \), \( \lfloor x \rfloor = n \).

### Step 2: Analyze the behavior of \( f(x) \)
Since \( x \geq 0 \), we can analyze \( f(x) \) in intervals defined by integers:
- For \( 0 \leq x < 1 \): \( \lfloor x \rfloor = 0 \) so \( f(x) = \frac{e^x}{1} = e^x \).
- For \( 1 \leq x < 2 \): \( \lfloor x \rfloor = 1 \) so \( f(x) = \frac{e^x}{2} \).
- For \( 2 \leq x < 3 \): \( \lfloor x \rfloor = 2 \) so \( f(x) = \frac{e^x}{3} \).
- And so on for \( n \leq x < n+1 \): \( f(x) = \frac{e^x}{n+1} \).

### Step 3: Determine the minimum and maximum values in each interval
1. **For \( 0 \leq x < 1 \)**:
- \( f(x) = e^x \) is increasing, so the minimum value at \( x = 0 \) is \( f(0) = e^0 = 1 \) and approaches \( e^1 = e \) as \( x \) approaches 1.

2. **For \( 1 \leq x < 2 \)**:
- \( f(x) = \frac{e^x}{2} \) is also increasing. The minimum value at \( x = 1 \) is \( f(1) = \frac{e^1}{2} = \frac{e}{2} \) and approaches \( \frac{e^2}{2} \) as \( x \) approaches 2.

3. **For \( 2 \leq x < 3 \)**:
- \( f(x) = \frac{e^x}{3} \) is increasing. The minimum value at \( x = 2 \) is \( f(2) = \frac{e^2}{3} \) and approaches \( \frac{e^3}{3} \) as \( x \) approaches 3.

4. **Continuing this pattern**:
- For \( n \leq x < n+1 \), the minimum value is \( \frac{e^n}{n+1} \) and approaches \( \frac{e^{n+1}}{n+1} \) as \( x \) approaches \( n+1 \).

### Step 4: Identify the overall range
- As \( n \) increases, \( \frac{e^n}{n+1} \) approaches 0 for large \( n \) and the maximum value approaches infinity.
- The minimum value across all intervals starts from \( 1 \) (at \( x = 0 \)) and increases without bound as \( x \) increases.

### Conclusion
The range of \( f(x) \) is \( \left[ 1, \infty \right) \).