what is the solution set of $\frac{x}{4} leq \frac{9}{x}$?

Answer:

First, rewrite the inequality $\frac{x}{4}\leq\frac{9}{x}$ as $\frac{x}{4}-\frac{9}{x}\leq0$. Then find a common - denominator: $\frac{x^{2}-36}{4x}\leq0$. Factor the numerator: $\frac{(x - 6)(x + 6)}{4x}\leq0$. The critical points are $x=-6,x = 0,x = 6$. We consider the intervals $(-\infty,-6),(-6,0),(0,6),(6,\infty)$.

• For $x\in(-\infty,-6)$, let $x=-7$. Then $\frac{(-7 - 6)(-7 + 6)}{4\times(-7)}=\frac{(-13)\times(-1)}{-28}=-\frac{13}{28}\leq0$, so $(-\infty,-6]$ is part of the solution set.
• For $x\in(-6,0)$, let $x=-1$. Then $\frac{(-1 - 6)(-1+6)}{4\times(-1)}=\frac{(-7)\times5}{-4}=\frac{35}{4}>0$, so $(-6,0)$ is not part of the solution set.
• For $x\in(0,6)$, let $x = 1$. Then $\frac{(1 - 6)(1 + 6)}{4\times1}=\frac{(-5)\times7}{4}=-\frac{35}{4}\leq0$, so $(0,6]$ is part of the solution set.
• For $x\in(6,\infty)$, let $x = 7$. Then $\frac{(7 - 6)(7 + 6)}{4\times7}=\frac{1\times13}{28}>0$, so $(6,\infty)$ is not part of the solution set.

The solution set is $x\leq - 6$ or $0<x\leq6$. In interval notation, $(-\infty,-6]\cup(0,6]$. Looking at the number - line graphs, the correct one is the graph where the line is solid at $x=-6$ and $x = 6$ and open at $x = 0$, and the shaded regions are $x\leq - 6$ and $0<x\leq6$.

Explanation:

Step1: Rewrite the inequality

$\frac{x}{4}-\frac{9}{x}\leq0$

Step2: Find a common - denominator

$\frac{x^{2}-36}{4x}\leq0$

Step3: Factor the numerator

$\frac{(x - 6)(x + 6)}{4x}\leq0$

Step4: Identify critical points

$x=-6,x = 0,x = 6$

Step5: Test intervals

Test $(-\infty,-6),(-6,0),(0,6),(6,\infty)$ using sample points.