Suppose `f:RRtoRR^(+)` is defined by `f(x)=(x^(2))/(1+x^(2))`. What is the range of the function?

Options

A. `[0,1)`
B. `[0,1]`
C. `(0,1]`
D. `(0,1)`

Explanation

To find the range of the function \( f(x) = \frac{x^2}{1 + x^2} \), we will analyze the behavior of this function step by step.

### Step 1: Understand the function
The function is defined as:
\[ f(x) = \frac{x^2}{1 + x^2} \]
where \( x \) is a real number. We note that \( f(x) \) maps to positive real numbers, as the denominator \( 1 + x^2 \) is always positive.

### Step 2: Determine the minimum value
Since \( x^2 \geq 0 \) for all real \( x \), the minimum value of \( f(x) \) occurs when \( x = 0 \):
\[ f(0) = \frac{0^2}{1 + 0^2} = \frac{0}{1} = 0 \]
Thus, the minimum value of \( f(x) \) is 0.

### Step 3: Determine the maximum value
Next, we need to find the maximum value of \( f(x) \). As \( x \) increases (positively or negatively), \( x^2 \) increases, and we analyze the limit of \( f(x) \) as \( x \) approaches infinity:
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2}{1 + x^2} = \lim_{x \to \infty} \frac{1}{\frac{1}{x^2} + 1} = \frac{1}{0 + 1} = 1 \]
Thus, as \( x \) approaches infinity, \( f(x) \) approaches 1.

### Step 4: Determine if 1 is included in the range
To determine if \( f(x) \) can actually equal 1, we set up the equation:
\[ \frac{x^2}{1 + x^2} = 1 \]
This implies:
\[ x^2 = 1 + x^2 \]
which simplifies to:
\[ 0 = 1 \]
This is a contradiction, indicating that \( f(x) \) can never actually reach 1.

### Step 5: Conclusion on the range
From the analysis, we find that:
- The minimum value of \( f(x) \) is 0.
- The maximum value approaches 1 but does not include it.

Thus, the range of the function \( f(x) \) is:
\[ [0, 1) \]

### Final Answer
The range of the function \( f(x) = \frac{x^2}{1 + x^2} \) is \( [0, 1) \).

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