Solving a super basic inequality

Question

Can you please help me find where the problem arises in the solution?

$1 + \frac{1}{x} \geq 0$

My attempt:

$1 + \frac{1}{x} \geq 0$

$1 \geq - \frac{1}{x}$

$x \geq -1$

The answer is $x \in (- \infty, -1] \bigcup (0, \infty)$.

I understand why the answer is so, but I made this mistake while solving a problem and didn't notice, so I want to understand how to approach problems like this.

Thanks!

Does this answer your question? How to solve inequality $1-\frac{1}{x}\geq 0$ correctly?. The reasoning is exactly the same. Found using Approach0. — Toby Mak, Dec 27 '20 at 07:53

score 6 · Answer 1 · answered Dec 26 '20 at 13:34

6

You cannot just multiply by $x$ since it can be negative too. So, below is the right approach: $$1 + \frac 1x \ge 0 \tag{$x\ne 0$} \\ \frac{x+1}{x} \ge 0 \iff x(x+1) \ge 0 \\\implies x \in (- \infty, -1] \cup (0, \infty)$$

answered Dec 26 '20 at 13:34

VIVID

11,604

score 2 · Accepted Answer · answered Dec 26 '20 at 13:36

2

Multiplying by a negative number reverses the inequality, so you need to consider the cases separately:

$$x\gt 0 \implies x\ge -1 \ \text{(this is always true)}\\ x\lt 0 \implies x\le -1 $$ by your second line and taking the intersection gives $x\gt 0$ or $x\le -1$.

answered Dec 26 '20 at 13:36

Vishu

14,469

1

I actually tried this, but didn't realize that because I get $x \geq -1$ when $x > 0$, I need to take $x >0$, thank you so much! Now everything is clear. – Ani Khachatryan Dec 26 '20 at 13:44
@AniKhachatryan :) – Vishu Dec 26 '20 at 13:49

score 1 · Answer 3 · answered Dec 26 '20 at 13:35

1

If $x$ is negative, when you multiply on both sides of the equality by $x$ you would have to change the direction of the inequality. Avoid doing this when you don't know if $x$ is positive.

answered Dec 26 '20 at 13:35

Clerni

438

Solving a super basic inequality

3 Answers3