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

Answer

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

### Step 1: Evaluate the function at the lower bound
First, we need to find the value of the function at the minimum point of the domain, which is \( x = 0 \).

\[ f(0) = \frac{e^0}{0 + 1} = \frac{1}{1} = 1 \]

### Step 2: Analyze the behavior of the function as \( x \) increases
Next, we need to analyze how the function behaves as \( x \) increases. As \( x \) approaches infinity, we need to look at the limits of the function.

\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{e^x}{x + 1} \]

### Step 3: Determine the limit as \( x \) approaches infinity
As \( x \) increases, the numerator \( e^x \) grows exponentially while the denominator \( x + 1 \) grows linearly. Thus, we can conclude that:

\[ \lim_{x \to \infty} f(x) = \infty \]

### Step 4: Conclusion about the range
Since \( f(0) = 1 \) and \( f(x) \) approaches infinity as \( x \) increases, the function is continuous and increasing for \( x \geq 0 \). Therefore, the range of \( f(x) \) is from the minimum value at \( x = 0 \) to infinity.

Thus, the range of \( f(x) \) is:

\[ [1, \infty) \]

### Final Answer
The range of the function \( f(x) = \frac{e^x}{x + 1} \) for \( x \geq 0 \) is \( [1, \infty) \).

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