Daily Graph 20-02-09

one thing for sure.....it will oscillate at n Ï€/2 where n ε Z⁺

for y=sin(cos(tan x)))

proceeding in very orthodox manner i guess

the dom is r-(nπ/2) an
the range is (-sin1,sin1)

then calculating the nature of the graph

1)pts of discontinuty n x Î /2 as tan goes undef. there
2)pt. of maxima x=0 as cos 0=1
3)similarly pt. of minima

4)behavoiur 0 <x<Ï€/2

tanx increases
cos(tanx) decreases
sin(cos(tan x))) increases

5) behavoiur π/2<x<π
tanx increases
cos(tanx) decreases
sin(cos(tan x))) decreases

this is for sin(tanx)

doms : R - (n+.5)Î

functn is increasing on : tan^-1(-Î /2)<x<tan^-1(Î /2)
functn is decreasing on : tan^-1(Î /2)<x<tan^-1(3Î /2)

at x=nÎ y = 0;

pases thru origin

range : [-1,1](obvoiusly)

range is but obviously -1 to 1 but i think that domain is R-n∩/2

the inc.-dec. pattern folows da same order for multiples....

basically da idea is 4 jus 1 full circle!!!

is it rite??

Sir, are my specifications here rite???

tapan's dom and range are rite....

as x comes close to π/2...it will oscillate very close....discontinuity at π/2....the pattern will continue....

not that goodlooking.....

sir, pl. chk!!

y=sin(cos(tan(x))

sir aapke kya vichaar hain