Question

|1/x|>x

Looks simple enough, but really I...nvm >.<

Answer 1

isnt this the same as yesterday's

Answer 2

Not really, the x in this one is in inverse

Answer 3

\[|\frac{1}{x}|>x\] like that ?

Answer 4

exactly! :)

Answer 5

this one is probably not so bad, because it is true for all \(x<0\)

Answer 6

i.e. if x is negative, then it is certainly the case that \(|\frac{1}{x}|>x\) right? so now we only have to check what happens when x is positive, and when x is positive you have \(|\frac{1}{x}|=\frac{1}{x}\)

Answer 7

so your only real job is to solve
\[\frac{1}{x}>x\] for
\[\frac{1}{x}-x>0\]
\[\frac{1-x^2}{x}>0\]

Answer 8

we can ignore the denominator, because we know that \(x>0\)

Answer 9

http://openstudy.com/users/fatinatikah#/updates/4f96745ce4b000ae9ecc59c4

Answer 10

ahhhh I see what you did there :)

Answer 11

owh lol @saifoo.khan

Answer 12

;)

Answer 13

so only have to solve
\[1-x^2>0\] and this is a parabola that opens down, so it is positive between the zeros and negative outside them. therefore it is positive on in interval \((-1,1)\) and so your answer is
\((-\infty, 1)\) except not when \(x=0\) so i guess you could write
\[(-\infty, 0)\cup (0,1)\]

Answer 14

Thank you :D

Answer 15

yw