Stuck here
- Thread starter achi_kun
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Please show us what you have tried and exactly where you are stuck.
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Please share your work/thoughts about this problem.
pka
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The question is \(\large \dfrac{1}{x}>2\) Please do not over complicate this.
Note that \(\large x\ne 0\) so there are two cases.
1) If \(x<0\) multiply by \(x\) to get \(1<2x\) or \(\dfrac{1}{2}<x<0\) that is a contradiction. Thus \(x\cancel{<\;}0\)
2) if \(x>0\) multiply by \(x\) to get \(1>2x\) or \(\dfrac{1}{2}>x>0\). Thus we have a solution. SEE HERE
Note that \(\large x\ne 0\) so there are two cases.
1) If \(x<0\) multiply by \(x\) to get \(1<2x\) or \(\dfrac{1}{2}<x<0\) that is a contradiction. Thus \(x\cancel{<\;}0\)
2) if \(x>0\) multiply by \(x\) to get \(1>2x\) or \(\dfrac{1}{2}>x>0\). Thus we have a solution. SEE HERE
Dr.Peterson
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I see two different problems there, both called #5. For one, they appear to be leading you through a method of solution, but it is rather confusing, particularly as there is at least one typo.
Which problem are you asking about? What have you tried?
Steven G
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If 0<a<b then 1/b< 1/a
1/x> 2 is the same as 2<1/x so 0<2 < 1/x. Then from the rule above x<1/2
Also you can think about it. You should know that 1/(1/2) = 2 and if you make 1/2 even smaller then the reciprocal will be even larger than 2. So x< 1/2
1/x> 2 is the same as 2<1/x so 0<2 < 1/x. Then from the rule above x<1/2
Also you can think about it. You should know that 1/(1/2) = 2 and if you make 1/2 even smaller then the reciprocal will be even larger than 2. So x< 1/2