Question

find the smallest and largest values of the function f(r) = r - sin2r on the interval [0,π]

Answer 1

this one is easy

Answer 2

f'(r)= 1- 2 cos (2r)

Answer 3

and it equals to 0

Answer 4

1-2 cos(2r)=0
2 cos(2r)=1
cos(2r)=1/2

cos (pi/3) =1/2

thus r= pi/6

Answer 5

check left and right

Answer 6

hence cos2r = 1/2 so that \[r = \pi / 6 or r = \pi - (\pi/6) or 5\pi/6\]

Answer 7

mm but i guess it can be pi - (pi/6) and 5pi/6 too

Answer 8

f'(r)= 1- 2 cos (2r)
less than pi/6 , it is negative
more than pi/6 , it is positive

Answer 9

the fact that slope change from negative to positive mean pi/3 is min

Answer 10

hmm k

Answer 11

oops, I meant pi/6

Answer 12

:D

Answer 13

A plot is attached.

Answer 14

thanks :D