Given the function f (x) = x2-2ax, find the minimum value g (a) of F (x) in the interval [- 1,1]

Given the function f (x) = x2-2ax, find the minimum value g (a) of F (x) in the interval [- 1,1]

The symmetry axis of F (x) = x2-2ax = (x-a) 2-A2 & nbsp; is x = A. when - 1 ≤ a ≤ 1, the minimum value of F (x) in the interval [- 1, 1] is g (a) = f (a) = - A2. When a < - 1, G (a) = f (- 1) = 1 + 2A. When a > 1, G (a) = f (1) = 1-2a. In conclusion, the minimum value of F (x) in the interval [- 1, 1] is g (a) = - A2 & nbsp;, & nbsp; - 1 ≤ a ≤ 11 + 2A & nbsp; & nbsp; ,a<−11−2a , a>1.