Find all the possible values of `f(x) =(1-x^2)/(x^2+3)`

Explanation

To find all the possible values of the function \( f(x) = \frac{1 - x^2}{x^2 + 3} \), we can follow these steps:

### Step 1: Set the function equal to a variable
Let \( f(x) = y \). Therefore, we have:
\[ y = \frac{1 - x^2}{x^2 + 3} \]

### Step 2: Rearrange the equation
Multiply both sides by \( x^2 + 3 \) to eliminate the denominator:
\[ y(x^2 + 3) = 1 - x^2 \]
This simplifies to:
\[ yx^2 + 3y = 1 - x^2 \]

### Step 3: Combine like terms
Rearranging gives:
\[ yx^2 + x^2 = 1 - 3y \]
Factoring out \( x^2 \):
\[ x^2(y + 1) = 1 - 3y \]
Thus, we can express \( x^2 \) as:
\[ x^2 = \frac{1 - 3y}{y + 1} \]

### Step 4: Determine conditions for \( x^2 \)
Since \( x^2 \) must be non-negative (as it is a square), we require:
\[ \frac{1 - 3y}{y + 1} \geq 0 \]

### Step 5: Analyze the inequality
The fraction \( \frac{1 - 3y}{y + 1} \) is non-negative when both the numerator and denominator are either both positive or both negative.

1. **Numerator**: \( 1 - 3y \geq 0 \) implies \( y \leq \frac{1}{3} \)
2. **Denominator**: \( y + 1 > 0 \) implies \( y > -1 \)

### Step 6: Find critical points
The critical points from the inequalities are:
- From \( 1 - 3y = 0 \), we get \( y = \frac{1}{3} \)
- From \( y + 1 = 0 \), we get \( y = -1 \)

### Step 7: Test intervals
We will test the intervals determined by the critical points \( y = -1 \) and \( y = \frac{1}{3} \):
- For \( y < -1 \): Choose \( y = -2 \)
\[ \frac{1 - 3(-2)}{-2 + 1} = \frac{1 + 6}{-1} = -7 \quad (\text{negative}) \]
- For \( -1 < y < \frac{1}{3} \): Choose \( y = 0 \)
\[ \frac{1 - 3(0)}{0 + 1} = \frac{1}{1} = 1 \quad (\text{positive}) \]
- For \( y > \frac{1}{3} \): Choose \( y = 1 \)
\[ \frac{1 - 3(1)}{1 + 1} = \frac{1 - 3}{2} = \frac{-2}{2} = -1 \quad (\text{negative}) \]

### Step 8: Conclusion
The function \( f(x) \) is non-negative in the interval \( (-1, \frac{1}{3}] \). However, since \( y = -1 \) makes the denominator zero, it is excluded from the solution set.

Thus, the possible values of \( f(x) \) are:
\[ y \in (-1, \frac{1}{3}] \]