Question

@satellite73

Answer 1

yes?

Answer 2

\[\large \bf \frac{x+1}{(x-1)^2}\]
Find the value of x

Answer 3

?

Answer 4

sorry.....

Answer 5

is there an equation?

Answer 6

\[\large \bf \frac{x+1}{(x-1)^2}<1\]
Find the value of x

Answer 7

use this formula ( a+ b +^2 = a^2 +b^2 +2ab

Answer 8

oh crap

Answer 9

\[\large \bf \frac{x+1}{(x-1)^2}<1\\
\large \bf \frac{x+1}{(x-1)^2}-1<0\] of course
then subtract

Answer 10

ooh no hold the phone
easier than that!

Answer 11

\[\frac{x+1}{(x-1)^2}<1\] since \((x-1)^2>0\) then you can multiply both sides by \((x-1)^2\)

Answer 12

then

Answer 13

giving
\[x+1<(x-1)^2\]

Answer 14

multiply out on the right

Answer 15

\[x+1<x^2-2x+1\]

Answer 16

\[\large \bf x+1<x^2+1-2x\]
\[\large \bf -x^2+3x<0\]

Answer 17

make your life simpler and ue
\[x^2-3x>0\]

Answer 18

then,
\[\large \bf x^2-3x>0\]

Answer 19

right

Answer 20

\[\large \bf x(x-3)>0\]
\[\large \bf x>0,3\]

Answer 21

right?

Answer 22

hmm no

Answer 23

why?

Answer 24

well for one thing \(x>0,3\) makes no sense

Answer 25

so what is the correct way of writing?

Answer 26

your answer is two intervals, so you need either interval notation or inequality notation

Answer 27

oh!! k

Answer 28

like i said before, we are not done
we only know where it changes sign
you have to find where it is positive
i.e. where \(x(x-3)>0\)

Answer 29

so intervals are :-

\[\large \bf (-\infty,0)U(3,\infty)\]