# Thread: inequality in intervals

1. ## inequality in intervals

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2. $\frac{1}{x} < 4$.

First, it should be abundantly obvious that $x \neq 0$.

To solve this inequality for $x$, you will need to consider two cases, the first being when $x$ is positive, and the second when $x$ is negative, since multiplying/dividing by a negative reverses the inequality sign.

Case 1: $x > 0$.

$\frac{1}{x} < 4$

$1 < 4x$

$\frac{1}{4} < x$.

Since $x > 0$ and $x > \frac{1}{4}$, putting it together gives $x > \frac{1}{4}$, or $x \in \left(\frac{1}{4}, \infty\right)$.

Case 2: $x < 0$

$\frac{1}{x} < 4$

$1 > 4x$

$\frac{1}{4} > x$.

So $x < 0$ and $x < \frac{1}{4}$. Putting it together gives $x < 0$, or $x \in (-\infty, 0)$.

Therefore, our final solution is

$x \in (-\infty, 0) \cup \left(\frac{1}{4}, \infty\right)$. You can check this by graphing the functions $y = \frac{1}{x}$ and $y = 4$ and making sure that everything below the line $y = 4$ are the $x$ values listed above.