The absolute minimum value, of the function f x = x 2 - x + 1 + x 2 - x + 1 , where t denotes the greatest integer function, in the interval - 1 , 2 , is

Step-by-step solution

f x = x 2 - x + 1 + x 2 - x + 1    x ∈ - 1 , 2 ∵ x 2 - x + 1 → x 2 - x + 1 > 0 ∴ f x = x 2 - x + 1 + x 2 - x + 1 Now, Consider g ( x ) = x 2 - x + 1 For the minimum value of g ( x ) , g ' ( x ) = 0 ⇒ 2 x - 1 = 0 ⇒ x = 1 2 x 2 - x + 1 attains its minimum value at x = 1 2 And min x 2 - x + 1 = 0   as   x 2 - x + 1 > 0 ⇒ f x attains its minimum at x = 1 2 So, f 1 2 = 3 4 + 0 = 3 4