Question

Solve for 'x': $-1 \leq \frac{1+x^2}{2x} \leq 1$

Answer 1

x ∈ {-1, 1}

Answer 2

The compound inequality $-1 \leq \frac{1+x^2}{2x} \leq 1$ is equivalent to $|\frac{1+x^2}{2x}| \leq 1$. Since $1+x^2 \geq 0$, this simplifies to $\frac{1+x^2}{2|x|} \leq 1$. Rearranging gives $x^2 - 2|x| + 1 \leq 0$. Let $y=|x|$, so $y^2 - 2y + 1 \leq 0$, which is $(y-1)^2 \leq 0$. Since a square must be non-negative, the only possibility is $(y-1)^2 = 0$, which means $y=1$. Substituting back $|x|=y$, we get $|x|=1$, which implies $x=1$ or $x=-1$. These values satisfy the condition $x \neq 0$.