IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

chilbi.chillz

find inverse of f(X) = x + sinx & area b/w f(X) & f inverse x ??

sunip_the_mini_mastermind

area is 8 sq.units

chilbi.chillz

plz gimme da complete ans ... i mean tell me hw u did it

sunip_the_mini_mastermind

f(x)=x+sinx will give the graph of sinx about y=x...its inverse will be its reflection about the line y=x...i hope now u will be able to do it...

neeraj_agarwal_1990

inverse is given by x=y+siny (i don't think we can express y in terms of x)

and yes...u'll have to draw the graph of x+sinx which would vary about x+sinx and its inverse is the reflection about y=x line...

chilbi.chillz

yaar i knw all dis stuff dat inverse is da reflection abt line y = x bt & point of intersection of f(x) & f inverse x will lie on dis line only

bt tell me hw to proceed ... if ne 1 knws to do it by graph then plz do it n tell me da complete soln.

neeraj_agarwal_1990

then do u wanna know how to make the graph???

y=x+sinx
find the intersection of y with x-axis
y=0 => sinx =-x they intersect at only x=0..

so graph passes through origin

its an odd function....symmetric abt. origin

y'=1+cosx
it changes sign at every cosx=-1,i.e x=(2n+1)pi

y''=-sinx
so shape changes from convex to concave and concave to convex after every -sinx=0 ,i.e. x=2npi

also, for x between 0 and pi, y''<0 , so shape is convex upwards...and b/w pi and 2pi, y''>0,so concave upwards and so on...

i hope u can now make the graph....and u'll then easily draw the inverse by taking reflection abt. y=x...

now area would come out to be 4.integral [(x)-(x+sinx)] from 0 to pi

i have taken 4 times the area b/w y=x and y=x+sinx due to symmetry

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