- #1

- 103

- 0

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 ?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter garyljc
- Start date

- #1

- 103

- 0

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 ?

- #2

- 103

- 0

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 ?

- #3

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Hi garyljc!

Nope!

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

1/x < x

therefore x^2>1

…

am i correct ?

Nope!

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

- #4

- 25

- 0

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

- #5

- 103

- 0

thanks to both lizzie and tim

Share: