11some one reply please
I gave a graph of the day
y=\frac{x^{2}+x+1}{x^{2}-x+1}
When i was seeing the solution by users, I noticed something very interesting!
you have to explain something very interesting that is going on in this graph... the graph has been solved already on the GOD of 10-03-09
Not something extraordinary.. but a very small thing that helps you to analyse the whole thing beautifully.
-
UP 0 DOWN 0 0 20
20 Answers
62no palani..
it is not..
that you have jumped to conclude because.. maxima is 3 and minima s 1/3
so you should have said 1/9th.. not 1/3 times..
but even that is not a correct conclusion
62The thing i wanted to hear was..
f(x) = 1/f(-x)
so what is happening here is that at -ve values of x, function is inverse at positivie values
so the behaviour of the graph on the -ve x axis can be found easily by just plotting the graph on the +ve x axis!
11is the GOD same as this one
i think that is y=(x+1)2/(x-1)2
about this graph when x is real y belongs to (1/3,3)
62@philip
that is correct..
something more than that?
A very very simple thing... !!
62@philip
that is correct..
something more than that?
A very very simple thing... !!
1for f(-x) > 1 ; f(x) < 1
for f(-x) < 1 ; f(x) >1
im almost sure u're not looking for this [4] bcoz it is too obvious
1n this graph wen
x<0
the eqn is is inverted
i.e
numerator becomes denomi
& vice versa
62@prjith.. no i was not looking at that :D
There is something unique here.. which palani already pointed out..
see carefully...
f(x)f(-x) = 1
62@akand.. yes you are right.. but that is true for all "Good" functions..
I mean most functions that we see other than [x] kind!
1the graph is not defined at x=0 and x=1....[1][1][1]..ami rite bhaiyya.......
62@philip..
see carefully
for symmetry at (a,b) if (x,y) is a solution then (a-x,b-y) has also to be a solution..
1what i saw was that
it was symmetric about the point (0,1)
dont see this