I asked a question earlier about open and closed intervals, so why did my teacher say this was wrong?
Determine any values of c in the interval [0,2pi] for which f'(x) = 0.
f(x) = cos x
My answer was:
f'(x) = -sin x
set equal to 0
-sin x = 0
x = 0, pi, 2pi
0/7 points: 0 and 2pi are not acceptable answers