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I'm reviewing some high school basic algebra and I would like to know when this function:

$$2x-1-\frac{1}{x}$$

is positive.

Solution

Suppose $x\gt0$, then $2x-1-\frac{1}{x}\ge0\Leftrightarrow 2x^2-x-1\ge 0\Leftrightarrow x\ge 1$.

Suppose $x\lt 0$, then $2x-1-\frac{1}{x}\ge0\Leftrightarrow 2x^2-x-1\le 0\Leftrightarrow -1/2\le x\lt 0$

Then $2x-1-\frac{1}{x}\ge 0 \Leftrightarrow x\ge 1$ and $-1/2\le x\lt 0$.

Am I right?

Thanks

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Since $x\neq0$, you can also try multiplying with $x^2$ instead of x –  Thanos Darkadakis Jul 14 '14 at 10:56
    
@ThanosDarkadakis good remark, this simplify everything, can you post this as an answer so that I can upvote it? –  user85493 Jul 14 '14 at 11:01

3 Answers 3

up vote 6 down vote accepted

Since $x\neq0$, instead if taking 2 cases for $x\gt0$ and $x\lt0$, you know that your function $f(x)$ has the same sign with $x^2f(x)$.

$x^2f(x)=x^2(2x-1-\frac{1}{x})=2x^3-x^2-x=2x(x-1)(x+\frac12)$

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Your answer is nearly right.

It should be $-1/2<x<0$ or $x>1$. Think about it. How can $x$ satisfy both of these inequalities at the same time?

Also note the strict inequalities: it should be $<$ rather than $\leq,$ and $>$ rather than $\geq$.

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yes, you're right thanks for point me these out. –  user85493 Jul 14 '14 at 10:52
    
How do you solve this inequality without the assumptions $x\gt 0$ and $x\lt 0$? –  user85493 Jul 14 '14 at 10:55

$2x-1-\frac{1}{x}=\frac{2x^2-x-1}{x}=\frac{(x-1 )(x+\frac{1}{2})}{x}$

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Why so large beginning? –  Ruslan Jul 14 '14 at 13:19

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