Originally Posted by
DINOCALC09 1st derivative = 2x - 5sinx
2nd derivative = 2 - 5cosx
set 2nd derivative = 0
so:
2 - 5cosx = 0
cosx = -2 / -5
x = arccos(-2/5) Mr F says: This should be arccos(2/5).
This is only satisfied once? Mr F says: There are two solutions over the interval 0 < x < 2pi. To see this, draw the graphs of y = cos x and y = 2/5 over 0 < x < 2pi. You should see two intersection points ......
And since the period of cos x is 2pi, you'll have two solutions over every interval of size 2 pi ..... So four solutions over 0 < x < 4pi, six solutions over 0 < x < 6pi ......, 2n solutions over 0 < x< 2npi
So the answer should be (B) ?