6i cant get the ans ...someone please show it!!!
15 Answers
6
1
the part of curve in IIIrd quadrant is the inverted image of part of curve in Ist quadrant
6how did u work this out...
am getting the part of 3rd quad as a mirror image in 2d quad...
1i don't think the graph exists in 2nd quadrant
there is no -ve value of x for which y is +ve so the graph can't exist in 2nd quadrant
you also confirm this from othersas i am not so good in graphs
6at x=0 it f(x) will be ∞ because curve is not defined
take
lim x+1/x
x→0
which equals infinte
if i solved correctly...:P
6i agree to the fact that it exists in 3rd and 1st quad...
my prob is
am not geting theCURVES alright...
can someone help
6oh i got it man!!thanzz
i also drew the graph..so posting it!!
1rather tanuj's post must be pinked ...very nice and precise graph
6well i never found that x=y can also be a asymptote!!
23y = x + 1x
or (y-x) = 1x
since y = 1x has asymptote y = 0 ,
u can say that the former one has y-x=0 as asymptote
