use both the first and second derivative test to show that f(x)=sin^2 x has a relative minimum of x=0
i keep getting that x=0 has no min or max because it equals zero in the second derivative test this is what im doing see if you can see what im doing wrong
f '(x)=2cosx and i set that equal to zero and i get x=0 then
f ''(x)=-sinx which is equal to zero when i plug x in and that gives me no min or max