It is known that the function f (x) = 1 / 3 to the power of X, X belongs to [- 1,1], and the minimum value of function g (x) = f (x) 2-2af (x) + 3 is h (a) For H (a), I calculate three expressions in three cases, but many of my classmates calculate - 6. Why?

It is known that the function f (x) = 1 / 3 to the power of X, X belongs to [- 1,1], and the minimum value of function g (x) = f (x) 2-2af (x) + 3 is h (a) For H (a), I calculate three expressions in three cases, but many of my classmates calculate - 6. Why?

Given the function f (x) = (1 / 3) ^ x, X ∈ [- 1,1], the minimum value of function g (x) = [f (x)] &# 178; - 2AF (x) + 3 is h (a), and then H (a) is obtained
G (x) = [f (x)] ² - 2AF (x) + 3 = [f (x) - A] ² - A & #178; + 3 ≥ 3-A & #178; = H (a), when f (x) = a, the equal sign holds
Because - 1 ≤ x ≤ 1, ﹥ 3 ≥ f (x) ≥ 1 / 3, we get 3 ≥ a ≥ 1 / 3, ﹥ 3-3 ﹥ 178; = - 6 ≥ H (a) ≥ 3 - (1 / 3) ﹥ 178; = 26 / 9
The minimum value of H (a) is - 6