Will taking reciprocal necessarily change the inequality sign over here for all values of x?
#ReCiPrOcAl-inequality
stuck axle
exotic bloomBOT
stuck axle
magic cloud
if $a>b$ and a,b have same sign (and not 0 ofc) then $\frac 1a < \frac 1b$
idle prairieBOT
deep.
magic cloud
but if $a>b$ and a,b have different sign (and not 0 ofc) then $\frac 1a > \frac 1b$
idle prairieBOT
deep.
magic cloud
like for eg 2 > -3
so
1/2 > -1/3
but
if 420>69 then 1/420 < 1/69
wait 2/(x+1)>0 implies (x+1)/2 < 0 sus
both are different regions
it should’ve been x>-1
i think the guy just did mistake
@stuck axle
magic cloud
no i mean its not incorrect to take reciprocal
but
it should be
(x+1)/2>0
if we’re excluding 1
stuck axle
(x+1)/2>0
But for X<2 , x could be -3
So 2/x+1 = -1
Reciprocal of -1 is -1 which is less than 0
stuck axle
x<2 is independent condition
U gotta consider it for intersection ryt anyways
magic cloud
U gotta consider it for intersection ryt anyways
yes
-1<x<2
thats what ur region is
uhh did u get it?
stuck axle
Wait
I got smthg liek
X(x-1) /x+1 > 0
From the log inequation
Which implies x belongs to (-1, 0) U (1, ∞)
And then x <2
So (-1, 0) U (1, 2) so far
stuck axle
So (-1, 0) U (1, 2) so far
And now u can take reciprocal ryt
idle prairieBOT
stuck axle
And now u can take reciprocal ryt
Ok so 2/x+1 is positive
magic cloud
And now u can take reciprocal ryt
wait reciprocal of what exactly
magic cloud
Ok so 2/x+1 is positive
yes
exotic bloomBOT
Solved