If frac{x^{2}-1}{x^{2}+1} geq 0, then x in | vedclass

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(A) Given the inequality $\frac{x^{2}-1}{x^{2}+1} \geq 0$. Since $x^{2} + 1 > 0$ for all real $x$,we can multiply both sides by $(x^{2} + 1)$ without

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$x \in(-\infty, -1] \cup [1, \infty)$

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(A) Given the inequality $\frac{x^{2}-1}{x^{2}+1} \geq 0$. Since $x^{2} + 1 > 0$ for all real $x$,we can multiply both sides by $(x^{2} + 1)$ without changing the inequality sign. This simplifies to $x^{2} - 1 \geq 0$. Factoring the expression,we get $(x - 1)(x + 1) \geq 0$. Using the wavy curve method (sign scheme),the expression is non-negative when $x \leq -1$ or $x \geq 1$. Thus,$x \in (-\infty, -1] \cup [1, \infty)$.

If $\frac{x^{2}-1}{x^{2}+1} \geq 0,$ then $x \in$

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