What is the range of the function f that is defined by
`f(x) = {{:(3^((1)/(x^(2)+1))", if " x ge 0),(5x+3", if " x lt 0):}` ?

Options

A. `y le 0`
B. `0 lt y lt 3`
C. `y ge 3`
D. `3 le y le 5`

Explanation

To find the range of the function defined as:

\[ f(x) = \begin{cases} 3^{\frac{1}{x^2 + 1}} & \text{if } x \geq 0 \\ 5x + 3 & \text{if } x < 0 \end{cases} \]

we will analyze each piece of the function separately.

### Step 1: Analyze the first part of the function \( f(x) = 3^{\frac{1}{x^2 + 1}} \) for \( x \geq 0 \)

1. **Determine the behavior of \( x^2 + 1 \)**:
- For \( x \geq 0 \), \( x^2 \) is always non-negative, hence \( x^2 + 1 \geq 1 \).
- Therefore, \( \frac{1}{x^2 + 1} \) will be at most \( 1 \) (when \( x = 0 \)) and approaches \( 0 \) as \( x \) increases.

2. **Evaluate the range of \( 3^{\frac{1}{x^2 + 1}} \)**:
- When \( x = 0 \), \( f(0) = 3^{\frac{1}{1}} = 3 \).
- As \( x \) increases, \( \frac{1}{x^2 + 1} \) decreases towards \( 0 \), thus \( f(x) \) approaches \( 3^0 = 1 \).
- Therefore, the range of this part is \( 1 < f(x) \leq 3 \).

### Step 2: Analyze the second part of the function \( f(x) = 5x + 3 \) for \( x < 0 \)

1. **Determine the behavior of \( 5x + 3 \)**:
- As \( x \) approaches \( 0 \) from the left (i.e., \( x \to 0^- \)), \( f(x) \) approaches \( 5(0) + 3 = 3 \).
- As \( x \) decreases (becomes more negative), \( 5x + 3 \) will decrease without bound.

2. **Evaluate the range of \( 5x + 3 \)**:
- The function \( 5x + 3 \) can take any value less than \( 3 \) since it decreases indefinitely as \( x \) becomes more negative.
- Therefore, the range of this part is \( f(x) < 3 \).

### Step 3: Combine the ranges from both parts

- From the first part, we have \( 1 < f(x) \leq 3 \).
- From the second part, we have \( f(x) < 3 \).

Combining these results, we find that the overall range of the function \( f(x) \) is:

\[ 1 < f(x) < 3 \]

### Final Answer

The range of the function \( f \) is:

\[ (1, 3) \]