Find all possible values of expressions `(2+x^2)/(4-x^2)`

Explanation

To find all possible values of the expression \(\frac{2+x^2}{4-x^2}\), we will follow these steps:

### Step 1: Set up the equation
Let \( y = \frac{2 + x^2}{4 - x^2} \).

### Step 2: Cross-multiply
To eliminate the fraction, we cross-multiply:
\[ y(4 - x^2) = 2 + x^2 \]
This simplifies to:
\[ 4y - yx^2 = 2 + x^2 \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ 4y - 2 = x^2 + yx^2 \]
Factoring out \( x^2 \) from the right side:
\[ 4y - 2 = x^2(1 + y) \]

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

### Step 5: Determine conditions for \( x^2 \)
Since \( x^2 \) cannot be negative, we require:
\[ \frac{4y - 2}{1 + y} \geq 0 \]

### Step 6: Analyze the inequality
This inequality holds true when both the numerator and denominator are either both positive or both negative.

1. **Numerator \( 4y - 2 \geq 0 \)**:
\[ 4y \geq 2 \implies y \geq \frac{1}{2} \]

2. **Denominator \( 1 + y > 0 \)**:
\[ y > -1 \]

### Step 7: Combine conditions
Since \( y \) must be greater than \(-1\) and also greater than or equal to \(\frac{1}{2}\), we take the intersection:
\[ y \geq \frac{1}{2} \]

### Step 8: Consider the case when \( y < -1 \)
If \( y < -1 \), the sign of the denominator changes, thus:
\[ 4y - 2 \leq 0 \implies y \leq \frac{1}{2} \]
But since we assumed \( y < -1 \), we have:
\[ y < -1 \]

### Step 9: Combine results
From the two cases, we conclude:
1. For \( y \geq \frac{1}{2} \): This gives us the interval \([\frac{1}{2}, \infty)\).
2. For \( y < -1 \): This gives us the interval \((-\infty, -1)\).

### Final Result
Thus, the possible values of the expression \(\frac{2+x^2}{4-x^2}\) are:
\[ (-\infty, -1) \cup \left[\frac{1}{2}, \infty\right) \]

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