How to solve $|1+1/x| > 2$

Question

I am having a problem with this question:

Find all real $x$ that satisfies $|1 + 1/x| > 2$.

This is clearly not defined in $x = 0$. By my logic, it should be solved with:

$1+1/x > 2$ or $1+1/x < -2$

But the result I am getting from this is wrong. ($x<1$)

Correct result is

$-1/3 < x < 1$

How to solve a problem like this? Why is this logic not working here:

$|x| > 2$

$x > 2$ or $x < -2$

Answer 1

I suspect you failed "switch" signs for negative values.

Either $1 + \frac 1x < -2$ or $1 + \frac 1x > 2$.

If $1 +\frac 1x <-2$ then $\frac 1x < -3$ and $x$ is negative.

So $1 > -3x$ (inequality flipped because $x < 0$)

$-\frac 13 < x$ (ditto for $-3$). And as $x$ is negative.

$-\frac 13 < x < 0$

OR

$1 +\frac 1x > 2$ so $\frac 1x > 1$ and $x$ is positive and $x < 1$. So $0 < x < 1$.

So $-\frac 13 < x < 0$ or $0 < x <1$ or $x \in (-\frac 13,0) \cup (0,1)$.

Answer 2

Taking from $$|1+\frac1x|>2\implies 1+\frac1x>2\ \text{or}\ 1+\frac1x<-2,$$ this gives us two cases:

Case 1: \begin{align} 1+\frac1x&>2\\ \frac1x-1&>0\\ \frac{1-x}x&>0\implies 0<x<1 \end{align}

Case 2: \begin{align} 1+\frac1x&<-2\\ 3+\frac1x&<0\\ \frac{3x+1}{x}&<0\implies-\frac13<x<0 \end{align}

We can then take the union of the two sets, and the final answer is $\boxed{-\dfrac13<x<1, x\neq0}$. Your mistake probably comes from solving the rational inequality incorrectly.

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Edit: here is how I would solve rational inequalities, in general.

Let $f(x)=\dfrac{3x+1}{x}$. The zeroes of the numerator and denominator are $-\dfrac13$ and $0$. Use these to form bounds of intervals.

\begin{array}{|r|c|c|c|} \hline & 3x+1 & x & \text{sign of } f(x) \\ \hline x<-\frac13 & - & - & +\\ -\frac13<x<0 & + & - & -\\ x>0 & + & + & + \\ \hline \end{array}

Since we want $f(x)=\dfrac{3x+1}{x}<0$, it follows that the solution is $-\dfrac13<x<0$ from the table above.

Answer 3

We have $$1+\frac1x > 2 \iff \frac1x > 1 \iff 0 < x < 1$$

and similarly

$$1+\frac1x < -2 \iff \frac1x < -3 \iff -\frac13 < x < 0$$

Therefore we have

$$\left|1+\frac1x\right| > 2 \iff 1+\frac1x > 2 \text{ or } 1+\frac1x < -2 \iff -\frac13 < x < 0 \text{ or } 0 < x < 1$$ which is equivalent to $$x \in \left\langle -\frac13, 1\right\rangle\setminus \{0\}.$$

Answer 4

\begin{align} |1+1/x|>2&\iff 1+1/x>2 \text{ or } (1+1/x)<-2\\ &\iff 0<x<1 \text{ or } -\frac13<x<0. \end{align}

Answer 5

Let's follow your logic and see where it leads us. Start with the first case

$$1 + \frac{1}{x} > 2.$$

This means that $\frac{1}{x} > 2 - 1 = 1 \implies 0 < x < 1.$ Note that you got $x < 1,$ which is not true, because if $x < 0,$ we clearly have that $\frac{1}{x} < 0 < 1.$

On the other hand, your second case has

$$1 + \frac{1}{x} < -2.$$

In this case, $\frac{1}{x} < -2 - 1 = -3.$ Note that in this case, $x$ must be negative. Solving the inequality as such, we have that $0 > x > -\frac{1}{3}.$

We take the union of these two cases for the complete solution set: $-\frac{1}{3} < x < 1, x \ne 0.$

Answer 6

Alternatively, $x\ne 0$ and: $$|x+1|>2|x| \iff (x+1)^2>(2x)^2 \iff (x-1)(3x+1)<0 \iff -1/3<x<1$$ Hence: $$x\in (-1/3,0)\cup (0,1)$$

Answer 7

If you square both sides, you obtain $$\left(1+1/x\right)^2>2^2,$$ or $$(1+1/x)^2-2^2>0,$$ which factors as $$(1+1/x-2)(1+1/x+2)>0,$$ or $$(1/x-1)(1/x+3)>0.$$ This implies $1/x<-3$ or $1/x>1.$ This implies $0>x>-1/3$ or $0<x<1.$ Hence the solution is the set of all $x\ne 0$ in $(-1/3,1).$

Answer 8

$x\not =0$;

1) $1+1/x >2$;

2) $1+1/x <-2$;

1) $1/x >1$;

Then $x >0$:

$1>x$, i.e $0<x<1.$

2) $1/x < -3$;

Then $x<0$:

$1>-3x$ ; $-1/3 <x$ ; i.e $-1/3<x<0$;

3) Altogether: $x \in (-1/3,0)\cup (0,1)$.

Answer 9

If you divide it by .2 or .5 or basically values < 1 you get higher values:

1 + (1 / .2) = 6

1 + (1 / .5) = 3

I'm not a mathematician so this may be missing the point of your equation, but this does make it true I believe since the order of operation is division before addition