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 will follow these steps:

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

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ y(x^2 + 3) = 1 - x^2 \]
This simplifies to:
\[ yx^2 + 3y = 1 - x^2 \]

### Step 3: Rearrange the equation
Rearranging the equation, we get:
\[ yx^2 + x^2 = 1 - 3y \]
Factoring out \( x^2 \) from the left side:
\[ x^2(y + 1) = 1 - 3y \]

### Step 4: Solve for \( x^2 \)
Now, we can express \( x^2 \) as:
\[ x^2 = \frac{1 - 3y}{y + 1} \]

### Step 5: Determine the conditions for \( x^2 \)
Since \( x^2 \) must be non-negative (i.e., \( x^2 \geq 0 \)), we need:
\[ \frac{1 - 3y}{y + 1} \geq 0 \]

### Step 6: Analyze the inequality
To analyze this inequality, we will consider the numerator and denominator separately.

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

### Step 7: Find the critical points
The critical points from the above inequalities are:
- From \( 1 - 3y = 0 \), we get \( y = \frac{1}{3} \).
- From \( y + 1 = 0 \), we get \( y = -1 \).

### Step 8: Test intervals
We will test the intervals defined by these points:
- For \( y < -1 \): Both numerator and denominator are negative, so the fraction is positive.
- For \( -1 < y < \frac{1}{3} \): The numerator is positive, and the denominator is positive, so the fraction is positive.
- For \( y = \frac{1}{3} \): The fraction equals zero.
- For \( y > \frac{1}{3} \): The numerator is negative, and the denominator is positive, so the fraction is negative.

### Step 9: Conclusion
The valid range for \( y \) is:
\[ -1 < y \leq \frac{1}{3} \]

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