21and for x<1-e;
function increases till max occurs at this bcoz logs bcum -ve in this interval....
(e+x)^(e+x) = (Î +x)^(Î +x)
then again dec till x>-e
18 Answers
21thik hai sunil....... but u cud do us rather the community a favour by deleting ur graf posts......
62yes it will..
then in that case.. can you just draw whatever conclusion is available from here?
62cool :)
good so far.
one more thing.. the second derivative is to be foudn to see the convexity.
2121OHKKKK.......
f'(x) = { (pi+x)-1ln(e+x) - (e+x)-1ln(pi+x) } / {[ln(e+x)]2}
denominatr > 0 always......
for x>1-e;
Both the logs are positive......
so numrtr < 0
as (e+x)^(e+x) < (pi+x)^(pi+x)
ther4 cont decrising in da intval 1-e<x<∞
62I think you are doing pretty well tapan..
You should now look at the derivative and the nature of the curve near the point (-e+1) on both sides..
Then you can move forward.
21Sir, ARe my points correct????
in this question ther is no symmetry......
wat else shud i look 4?? [7]
24yup we dont require any apology dear.........kindly remove these graphs from the threads so that others can give an honest attempt to the question....[1]
Hoping that this won't be repeated again[1][1]
11bhai log...
Sorry......
ab se no such graphs..
Ok.
Accept my apology....
21yah...... EUREKA IS SPOT ON!!!!
PL. MATE(SUNIL), U R SPOILIN THE FUN OF DA GRAF............. ITS A CHILD'S PLAY TO GENERATE GRAFS FRM DA COMP..... Y DONT U TRY IT URSELF...... U'LL NOT B ALLOWED UR LAPTOP INTO DA EXAM HALL IN IIT
24plzzzzzzzzzzz dont do this yaar.................dont post computer generated graphs.........plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz[2]
21max for the portion in -e<x<(1-Î )
occurs wen
(e+x)^(e+x) = (Î +x)^(Î +x)
ther r no other max/min in da graph
pl. correct me if i'm wrong..... koi javab to do... ?? [7]
21fol. are some of its char.....
1. functn defined only till x=-e (obviously)
2. y=not defined at x = -e + 1
3. y<0 for (1-Î )<x<(1-e)
more cumin soon..... lol
