# Solve the inequality $\frac{2}{x} > 3 x$ [closed]

I've been slowly losing my mind on how to do this question, any help would be greatly appreciated.

Solve the inequality $$\frac{2}{x} > 3 x.$$

## closed as off-topic by Xander Henderson, zz20s, Eevee Trainer, mrtaurho, Claude LeiboviciJan 14 at 7:08

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• Draw a picture!!! – aidangallagher4 Jan 14 at 0:04

To solve the equation $$\frac{2}{x} > 3x$$ we could multiply be $$x$$ to obtain $$2 > 3x^2.$$ But you have to be careful! If $$x < 0$$, the inequality is reversed. In other words: $$\frac{2}{x} > 3x \iff \begin{cases} 2 < 3x^2 & x < 0 \\ 2 > 3x^2 & x > 0. \end{cases}$$ Notice the equation isn't defined for $$x = 0$$, because you can't divide by $$0$$.

Case 1: $$x > 0$$. Then we have $$2> 3x^2 \iff x^2 < \frac{2}{3} \iff x \in \left(- \sqrt{ \frac{2}{3}}, \sqrt{ \frac{2}{3}}\right).$$ But, since $$x > 0$$, our first solution interval is $$I_1 := \left(0, \sqrt{ \frac{2}{3}}\right)$$.

Case 2: $$x < 0$$. Then we have $$2 < 3x^2 \iff x^2 > \frac{3}{2} \iff x > \sqrt{\frac{3}{2}} \quad \text{and} \qquad x < -\sqrt{\frac{3}{2}}$$ But again, since $$x < 0$$, our second (and final) solution interval is $$I_2 := \left(- \infty, -\sqrt{\frac{3}{2}}\right)$$.

So our solution is $$I_1 \cup I_2 =\left(- \infty, -\sqrt{\frac{3}{2}}\right) \cup \left(0, \sqrt{ \frac{2}{3}}\right)$$.

We observe that necessarily $$x\neq 0$$ whence we can rewrite the inequality as $$\frac{2}{x}-3x>0\iff\frac{2x-3x^3}{x^2}>0\iff 2x-3x^3>0\iff x(2-3x^2)>0$$ which you can solve.

Usually you're supposed to give some context and explain what you've tried so far. However, I will take your word that you have been "slowly losing" your mind and give an answer.

Since this is a nonlinear inequality, there is no "cookie cutter" way to solve it. But you can get an idea of what to do by graphing $$y_1=2/x$$ and $$y_2=3x$$, and observing where $$y_1>y_2$$.

See here for a graph with $$y_1=2/x$$ in red and $$y_2=3x$$ in blue. Where does the red curve lie above the blue curve?

Well, you can see that there are two intersection points. But the red lies above the blue from $$-\infty$$ up to the first intersection point, and then again from zero to the second intersection point.

To find the intersection points, set $$2/x=3x$$ and solve.

Can you write down the final answer?