Question

Find absolute max and min on f(x)= sin^2x-cosx

Answer 1

Answer:

F'(X) = 2 sin (X) cos (x) + sin(x) = 0

=> sin(X) [ 2 cos (X) + 1 ] = 0

=> sin (X) = 0 , cos (x) = - 1/2

=> when sin(X) = 0 then cos (X) = 1 and F(X) = -1 (min)

=> when cos(X) = - 1/2 then sin(X) = √3/2 and F(X) = (3/4) + (1/2) = 5/4 (max)