Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm looking for hints on how to efficiently solve this inequality: $$\left( \frac {|x|-|1-x|}{|x|} \right)^{2x-1} \gt \left(\frac {|x|-|1-x|}{|x|} \right)^{8-x} $$

share|improve this question
    
I think you may want to separate for two cases: $x>0$ and $x<0$ –  Elimination Jul 30 at 10:44
1  
@elimination Actually probably also in $1-x>0$ and $1-x<0$ –  dreamer Jul 30 at 10:56
1  
Right, thank you @dreamer –  Elimination Jul 30 at 10:57
1  
We're in the process of removing the homework tag, so I've removed it from your question. –  anorton Jul 30 at 11:16

2 Answers 2

Set $y = \dfrac{|x|-|1-x|}{|x|}= 1 - \lvert \frac1x-1 \rvert$.

Note $y > 1 $ is not possible, and $y \in [0, 1]$ when $x \ge \frac12$. Otherwise $y$ is negative, (so what?).

So the only case to consider is $y \in [0, 1]$, which means $2x-1< 8-x \implies ...$

share|improve this answer
    
Hahah, cough don't look at the edit history cough ;) –  dreamer Jul 30 at 11:15
    
@dreamer started writing a hint and ended up with short answer, so had to edit :( –  Macavity Jul 30 at 11:18
    
Haha, I know, I was just kidding :) –  dreamer Jul 30 at 11:22
    
exponential function wlays has a base whihc is greater then zero and not equal to one, and hence your said first lmitation (right?) –  Bak1139 Jul 30 at 11:47
1  
@Bak1139 $f(x)=1^x = 1$ is trivially a constant, though technically a valid function. In the above problem $y=1$ is not a solution purely because the inequality is strict. –  Macavity Jul 30 at 12:00

One can distinguish multiple cases

  • $x>0, \ 1-x>0$
  • $x>0, \ 1-x<0$
  • $x<0, \ 1-x>0$
  • $x<0, \ 1-x<0$

As an example,when $x<0$ and $1-x<0$ we have that the inequality reads $\left(\frac{1-x-x}{-x}\right)^{2x-1}>\left(\frac{1-x-x}{-x}\right)^{8-x}$ or equivalently $\left(\frac{2x-1}{x}\right)^{2x-1}>\left(\frac{2x-1}{x}\right)^{8-x}$

It then follows that this is true for $2x-1>8-x$ or equivalently $3x>9 \rightarrow x>3$, but this is not possible since we found this solution under the assumption that $x<0$.

You could do the same for each case. This is the most intuitive solution, but Macavity's way is nicer.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.