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

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

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

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    $\begingroup$ Draw a picture!!! $\endgroup$ – aidangallagher4 Jan 14 at 0:04
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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 $I_1 \cup I_2 =\left(- \infty, -\sqrt{\frac{3}{2}}\right) \cup \left(0, \sqrt{ \frac{2}{3}}\right)$.

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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.

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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?

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