Answer :
Let's solve the inequality [tex]\(\frac{x^2 - 1}{x^2 + 1} \leq 0\)[/tex] step-by-step.
1. Identify the critical points:
The inequality involves a rational expression, so we first identify the points where the numerator and the denominator are zero.
- The numerator [tex]\(x^2 - 1\)[/tex] is zero when [tex]\(x^2 = 1\)[/tex], which gives [tex]\(x = \pm 1\)[/tex].
- The denominator [tex]\(x^2 + 1\)[/tex] is never zero, since [tex]\(x^2 + 1 > 0\)[/tex] for all real [tex]\(x\)[/tex].
2. Understand the behavior of the expression:
Since the denominator [tex]\(x^2 + 1\)[/tex] is always positive, the sign of the expression [tex]\(\frac{x^2 - 1}{x^2 + 1}\)[/tex] is determined by the numerator [tex]\(x^2 - 1\)[/tex].
Let's analyze the sign of [tex]\(x^2 - 1\)[/tex]:
- For [tex]\(x^2 - 1 < 0\)[/tex], the expression is negative. This happens when [tex]\(|x| < 1\)[/tex].
- For [tex]\(x^2 - 1 = 0\)[/tex], the expression is zero. This happens at [tex]\(x = \pm 1\)[/tex].
- For [tex]\(x^2 - 1 > 0\)[/tex], the expression is positive. This happens when [tex]\(|x| > 1\)[/tex].
3. Determine the intervals:
Based on the step 2 analysis,
- Since we want [tex]\(\frac{x^2 - 1}{x^2 + 1} \leq 0\)[/tex],
- The expression equals 0 at [tex]\(x = -1\)[/tex] and [tex]\(x = 1\)[/tex].
- The expression is negative when [tex]\(|x| < 1\)[/tex], meaning in the interval [tex]\(-1 < x < 1\)[/tex].
4. Combine the intervals:
Combining the intervals where the expression is negative and where it is zero, we find that the expression is less than or equal to zero within the interval [tex]\(-1 \leq x \leq 1\)[/tex].
Thus, the solution to the inequality [tex]\(\frac{x^2 - 1}{x^2 + 1} \leq 0\)[/tex] is [tex]\(-1 \leq x \leq 1\)[/tex].
So, the final answer is:
[tex]\[ \boxed{-1 \leq x \leq 1} \][/tex]
1. Identify the critical points:
The inequality involves a rational expression, so we first identify the points where the numerator and the denominator are zero.
- The numerator [tex]\(x^2 - 1\)[/tex] is zero when [tex]\(x^2 = 1\)[/tex], which gives [tex]\(x = \pm 1\)[/tex].
- The denominator [tex]\(x^2 + 1\)[/tex] is never zero, since [tex]\(x^2 + 1 > 0\)[/tex] for all real [tex]\(x\)[/tex].
2. Understand the behavior of the expression:
Since the denominator [tex]\(x^2 + 1\)[/tex] is always positive, the sign of the expression [tex]\(\frac{x^2 - 1}{x^2 + 1}\)[/tex] is determined by the numerator [tex]\(x^2 - 1\)[/tex].
Let's analyze the sign of [tex]\(x^2 - 1\)[/tex]:
- For [tex]\(x^2 - 1 < 0\)[/tex], the expression is negative. This happens when [tex]\(|x| < 1\)[/tex].
- For [tex]\(x^2 - 1 = 0\)[/tex], the expression is zero. This happens at [tex]\(x = \pm 1\)[/tex].
- For [tex]\(x^2 - 1 > 0\)[/tex], the expression is positive. This happens when [tex]\(|x| > 1\)[/tex].
3. Determine the intervals:
Based on the step 2 analysis,
- Since we want [tex]\(\frac{x^2 - 1}{x^2 + 1} \leq 0\)[/tex],
- The expression equals 0 at [tex]\(x = -1\)[/tex] and [tex]\(x = 1\)[/tex].
- The expression is negative when [tex]\(|x| < 1\)[/tex], meaning in the interval [tex]\(-1 < x < 1\)[/tex].
4. Combine the intervals:
Combining the intervals where the expression is negative and where it is zero, we find that the expression is less than or equal to zero within the interval [tex]\(-1 \leq x \leq 1\)[/tex].
Thus, the solution to the inequality [tex]\(\frac{x^2 - 1}{x^2 + 1} \leq 0\)[/tex] is [tex]\(-1 \leq x \leq 1\)[/tex].
So, the final answer is:
[tex]\[ \boxed{-1 \leq x \leq 1} \][/tex]
