Find the absolute maximum and minimum of the function f(x)=sin^2x+cosx

Hello : 
f(x) = sin²(x) +cos(x)
the dirivate is : f'(x) = 2sin(x) cos(x) -sin(x) because : 
( sin²(x) )' =  2sin(x) cos(x)   and (cos(x))' = - sin(x)

f'(x) = 0  :  sin(x) (2cos(x) - 1) = 0
sin(x) = 0 : x = k π    k in Z
2cos(x) - 1 =0 : 2cos(x) =1
 cos(x) = 1/2 = cos( π /3)
 x =  π /3 +2k π  or   x = - π /3 +2k π 
in intervall :  [0 , 2π] : 
k=0 : x= 0   f(0) = 1   ( minimum)
k=1 :  x = π    f(π) = -1 (minimum)
x=  π /3  f (  π /3 ) = 1.25 ( maximum)

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.

a - base

b - exponent

so,

the base is 8the exponent is 3the exponent tells you to multiply 8 together three timesthe expanded form is 8*8*8