Question

solve the inequality
(1/x)-x>0
I know the anwer is 0 14 years ago

Answer 1

1 - x^2 > 0
x^2 < 1
(x+1)(x-1) < 0
so 0 < x < 1

Answer 2

did you get this?

Answer 3

but wouldn't you also get x<-1?

Answer 4

no I didn't

Answer 5

ok lets go slow

Answer 6

first of all you have to put the left hand side of the inequality as one fraction

Answer 7

oh yeah i was guessing im not sure of the answer

Answer 8

okay so we have a common denominator....

Answer 9

i screwed that up royally

Answer 10

\[\frac{1}{x}-x>0\]
\[\frac{1-x^2}{x}>0\]

Answer 11

that is the right way to add sorry

Answer 12

now factor the numerator, and you get
\[\frac{(1+x)(1-x)}{x}>0\]

Answer 13

with me so far?

Answer 14

yeah

Answer 15

ok so we have 3 factors to consider

\[1+x\]
\[x\]
\[1-x\] and they are 0 at the numbers
\[-1,0,1\] respectively

Answer 16

alright with you soo far

Answer 17

so we have to break up the real line into 4 different parts like this
___________-1_____________0______________1____________
and consider what happens on each part

Answer 18

perhaps the easiest thing to do in this case is make a sign chart for each of the factors.
\[1+x\] is zero at -1, positive for bigger numbers, negative for smaller. so i will write
\[1+x\]
------------------- -1 ++++++++++++++++++++++++++
so indicate is is negative to the left and positive to the right

Answer 19

\[x\] is evidently positive if x is positive (who is buried in grant's tomb) and negative if x is negative so the sign chart will look like

\[x\]
-------------------------------------- 0 ++++++++++++++++++++++
indicating it is positive to the left and negative to the right

Answer 20

\[1-x\] is the tricky one. it is 0 if x is 1, but it is positive to the RIGHT AND NEGATIVE TO THE LEFT of 1
so its sign chart will look like
\[1-x\]
++++++++++++++++++++++++++++++++ 1 ------------------------

Answer 21

now we pile them one on top of the other and get
------------------- -1 ++++++++++++++++++++++++++++++++++
--------------------------------------- 0 +++++++++++++++++++++++
+++++++++++++++++++++++++++++++++++++ 1 -------------------

Answer 22

if i have lost you i understand but i cannot think of a snap way to do this
now we just look. if
\[x<-1\] two factors are negative, one is positive so answer is positive

Answer 23

yeah it would be positive between 0 and 1 i got you!

Answer 24

so solution is...
\[x<-1\] or
\[0<x<1\]

Answer 25

tada!

Answer 26

one quick Q

Answer 27

whew ok
hope that one was clear

Answer 28

if you tried to make it a common denomintor wouldn't you have to multiply everything with x making it 1-x^/x not the way u wrote it?

Answer 29

i hope that is what i wrote
\[\frac{1}{x}-x=\frac{1}{x}-\frac{x^2}{x}=\frac{1-x^2}{x}\]

Answer 30

let me scroll up and look

Answer 31

yeah it is there. hope it is clear
the problem with these is that you are not free to multiply one side or the other by x because you don't know if x is positive or negative. so they require more work

Answer 32

no Im saying that when you multiplied the -x by 1/x dont you have to multiply the other parts by x too?

Answer 33

no
\[\frac{1}{x}\] is already over x

Answer 34

forget variables, let's do it with numbers
\[\frac{1}{3}-3\]

Answer 35

okay

Answer 36

you have to take
\[\frac{1-9}{3}\] right?

Answer 37

that is you need
\[\frac{1}{3}-\frac{9}{3}\] in order to subtract

Answer 38

oh okay that makes sense thanks :)

Answer 39

yw