Question

Find all points of extrema on the interval [0,2pi] if y=x-cosx

Answer 1

So first differentiaite it y'=1+sin x to find all critical points put y'=0 you get sinx=-1 which is true at x=pi only in the given interval to check whether it is maximum you should ahve y''<0 but y''=cosx at x=pi cosx=0 hence y''=0 which does not satisfy the condtion y''<0 Hence there is no extreme point in the given interval....

Answer 2

You do know differentiation and its applications right?

Answer 3

what is the derivative of
x-cosx ?

Answer 4

yes, yes i do

Answer 5

what is
\[\frac{d}{dx}(x-\cos x)\] ?

Answer 6

sinx?

Answer 7

nvm its 1-sinx

Answer 8

not quite
remember derivatives of co-functions (cosine, cotangent, csc) are negative\[\frac{d}{dx}(x-\cos x)=\frac{d}{dx}x-\frac{d}{dx}\cos x=1+\sin x\]

Answer 9

oh ok so i equal that to zero then?

Answer 10

right
what do you get for x?

Answer 11

-pi over 2

Answer 12

but that's not in the interval [0,2pi] now is it?

Answer 13

what is the equivalent angle to -(pi/2) that is in the interval [0,2pi] ?

Answer 14

this is the unit circle correct?

Answer 15

yes
go around it the other way and get to the same point as -pi/2
what do you call that point going the other way (counterclockwise)?

Answer 16

0,1?

Answer 17

clockwise we get