To determine the solution set for the inequality [tex]\(\frac{x}{4} \leq \frac{9}{x}\)[/tex], we will proceed through the analysis step by step.
1. Rewrite the inequality: [tex]\(\frac{x}{4} \leq \frac{9}{x}\)[/tex]
2. Clear the fraction: Multiply both sides by [tex]\(4x\)[/tex] (Note: [tex]\(x \neq 0\)[/tex] to avoid division by zero). The inequality becomes:
[tex]\[
x^2 \leq 36
\][/tex]
3. Solve the quadratic inequality:
[tex]\[
x^2 - 36 \leq 0
\][/tex]
This can be rewritten as:
[tex]\[
(x - 6)(x + 6) \leq 0
\][/tex]
4. Determine the critical points: The critical points, where the expression changes sign, are [tex]\(x = 6\)[/tex] and [tex]\(x = -6\)[/tex].
5. Analyze intervals: We will test the intervals determined by the critical points [tex]\([-6, 6]\)[/tex] to see where the inequality holds true:
- For [tex]\(x < -6\)[/tex] (e.g., [tex]\(x = -7\)[/tex]):
[tex]\[
(-7)^2 - 36 = 49 - 36 = 13 \quad (\text{positive})
\][/tex]
Therefore, the inequality [tex]\(x^2 - 36 \leq 0\)[/tex] does not hold.
- For [tex]\(-6 \leq x \leq 6\)[/tex]:
[tex]\[
x^2 - 36 \leq 0
\][/tex]
For any [tex]\(x\)[/tex] in this range, the value of [tex]\(x^2 - 36\)[/tex] is non-positive.
- For [tex]\(x > 6\)[/tex] (e.g., [tex]\(x = 7\)[/tex]):
[tex]\[
7^2 - 36 = 49 - 36 = 13 \quad (\text{positive})
\][/tex]
Therefore, the inequality [tex]\(x^2 - 36 \leq 0\)[/tex] does not hold.
6. Conclusion: The solution set of [tex]\((x - 6)(x + 6) \leq 0\)[/tex] is the interval that lies between the critical points where the inequality holds, inclusive of the critical points themselves.
Thus, the solution set of [tex]\(\frac{x}{4} \leq \frac{9}{x}\)[/tex] is:
[tex]\[
\boxed{[-6, 6]}
\][/tex]
