Is My Graphing Method for (1/x)<x<1 Correct?

[garyljc]

Hi all ,
could someone help me out with this
(1/x)<x<1

i tried by drawing the graphs
and separated it in 2 equations
but the answer that i found was -1<x<1
is it correct ?

 

[garyljc]

i got a new breakthrough haha
this equation does not have a solution set

correct me if I'm wrong
if first do the first 2 steps of the equation ,it follows that 1/x < x
therefore x^2>1
second equation will be that x<1
therefore there is no solution
am i correct ?

 

[tiny-tim]

Hi garyljc!

1/x < x
therefore x^2>1
…
am i correct ?

Nope!

Hint: If a < b and c < 0, is ac < bc?

 

[lizzie]

1/x<x
1/x - x<0
(1-x^2)/x<0
(x^2 - 1)/x>0

case 1: numerator and denominator are positive
numerator
x^2-1>0
(x+1)(x-1)>0
By wavy curve method
x<-1 or x>1
denominator
x>0
so solution for this case is x>1
but x<1 (given in the question)
hence no solution in this case.

case 2: numerator and denominator are negative
x^2-1<0
(x+1)(x-1)<0
By wavy curve method
-1<x<1
x<0
so solution for this case is -1<x<0
but x<1
hence solution for this case is -1<x<0

SO THE ANSWER IS -1<X<0

 

[garyljc]

thanks to both lizzie and tim