Range Math Example 4

Solution

1. 1
   Set $y = \frac{x^2 - 4}{x^2 + 1}$ and solve for $x^2$: $y(x^2 + 1) = x^2 - 4$, so $yx^2 + y = x^2 - 4$, giving $x^2(y - 1) = -4 - y$, thus $x^2 = \frac{-4 - y}{y - 1} = \frac{4 + y}{1 - y}$.
2. 2
   For real solutions, $x^2 \geq 0$: $\frac{4 + y}{1 - y} \geq 0$. This holds when both numerator and denominator share the same sign.
3. 3
   Case 1: $4 + y \geq 0$ and $1 - y > 0$ → $-4 \leq y < 1$. Case 2: $4 + y \leq 0$ and $1 - y < 0$ → $y \leq -4$ and $y > 1$, impossible. So range is $[-4, 1)$.