To solve the inequality [tex]\(\frac{x}{x^2 - 16} \leq 0\)[/tex], let's go through a detailed, step-by-step procedure.
1. Factor the Denominator:
[tex]\[
x^2 - 16 \text{ can be factored as } (x - 4)(x + 4).
\][/tex]
Therefore, the inequality becomes:
[tex]\[
\frac{x}{(x - 4)(x + 4)} \leq 0
\][/tex]
2. Identify Critical Points:
The critical points are values of [tex]\(x\)[/tex] that make the numerator or the denominator zero:
[tex]\[
x = 0, \quad x = 4, \quad x = -4
\][/tex]
3. Determine the Sign of the Expression in Each Interval:
We need to test the sign of [tex]\(\frac{x}{(x - 4)(x + 4)}\)[/tex] in each of the following intervals defined by the critical points:
[tex]\[
(-\infty, -4), \quad (-4, 0), \quad (0, 4), \quad (4, \infty)
\][/tex]
4. Test Each Interval:
- For [tex]\(x \in (-\infty, -4)\)[/tex], choose [tex]\(x = -5\)[/tex]:
[tex]\[
\frac{-5}{(-5 - 4)(-5 + 4)} = \frac{-5}{(-9)(-1)} = \frac{-5}{9} < 0
\][/tex]
- For [tex]\(x \in (-4, 0)\)[/tex], choose [tex]\(x = -1\)[/tex]:
[tex]\[
\frac{-1}{(-1 - 4)(-1 + 4)} = \frac{-1}{(-5)(3)} = \frac{-1}{-15} = \frac{1}{15} > 0
\][/tex]
- For [tex]\(x \in (0, 4)\)[/tex], choose [tex]\(x = 1\)[/tex]:
[tex]\[
\frac{1}{(1 - 4)(1 + 4)} = \frac{1}{(-3)(5)} = \frac{1}{-15} = -\frac{1}{15} < 0
\][/tex]
- For [tex]\(x \in (4, \infty)\)[/tex], choose [tex]\(x = 5\)[/tex]:
[tex]\[
\frac{5}{(5 - 4)(5 + 4)} = \frac{5}{(1)(9)} = \frac{5}{9} > 0
\][/tex]
5. Include Critical Points in the Solution:
Check the values at critical points:
- At [tex]\(x = -4\)[/tex], the denominator is zero, so it is undefined.
- At [tex]\(x = 0\)[/tex]:
[tex]\[
\frac{0}{0 - 16} = 0, \quad\text{ which is } \leq 0
\][/tex]
- At [tex]\(x = 4\)[/tex], the denominator is zero, so it is undefined.
6. Combine the Results:
The expression is less than or equal to zero in the intervals where the test point gave a negative value and the point [tex]\(x = 0\)[/tex]:
[tex]\[
(-\infty, -4) \cup \{0\} \cup (0, 4)
\][/tex]
Therefore, the solution to the inequality [tex]\(\frac{x}{x^2 - 16} \leq 0\)[/tex] is:
[tex]\[
\boxed{(-\infty, -4) \cup [0] \cup (0, 4)}
\][/tex]
