Range of the function `f(x)=(x^(2))/(x^(2)+1)` is

Options

A. (-1,0)
B. (-1,1)
C. [0,1)
D. (1,1)

Answer

To find the range of the function \( f(x) = \frac{x^2}{x^2 + 1} \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
We start by letting \( y = f(x) \):
\[ y = \frac{x^2}{x^2 + 1} \]

### Step 2: Rearrange the equation
Next, we rearrange the equation to express \( x^2 \) in terms of \( y \):
\[ y(x^2 + 1) = x^2 \]
This simplifies to:
\[ yx^2 + y = x^2 \]
Rearranging gives:
\[ yx^2 - x^2 + y = 0 \]
Factoring out \( x^2 \):
\[ x^2(y - 1) + y = 0 \]

### Step 3: Solve for \( x^2 \)
Now, we can isolate \( x^2 \):
\[ x^2(y - 1) = -y \]
Thus, we have:
\[ x^2 = \frac{-y}{y - 1} \]

### Step 4: Analyze the expression for \( x^2 \)
For \( x^2 \) to be non-negative (since \( x^2 \geq 0 \)), we need:
\[ \frac{-y}{y - 1} \geq 0 \]
This inequality holds true when both the numerator and denominator are either both positive or both negative.

### Step 5: Determine the conditions on \( y \)
1. **Numerator**: \( -y \geq 0 \) implies \( y \leq 0 \).
2. **Denominator**: \( y - 1 < 0 \) implies \( y < 1 \).

Thus, we need \( y \leq 0 \) and \( y < 1 \). However, since \( y \) represents the output of the function \( f(x) \), we can see that \( f(x) \) is always non-negative because \( x^2 \geq 0 \). Therefore, we need to consider \( y \geq 0 \).

### Step 6: Combine the conditions
From the analysis:
- \( y \geq 0 \) (since \( f(x) \) is non-negative)
- \( y < 1 \)

### Step 7: Conclusion
Thus, the range of the function \( f(x) = \frac{x^2}{x^2 + 1} \) is:
\[ [0, 1) \]