Let f (x) = x2-2x + 2, the minimum value of X ∈ [T, t + 1] (t ∈ R) be g (T), and find the expression of G (T)

F (x) = x2-2x + 2 = (x-1) 2 + 1, so the symmetry axis of the image is a straight line x = 1, and the opening of the image is upward. ① when t + 1 < 1, i.e. T < 0, f (x) is a decreasing function on [t, t + 1], so g (T) = f (T + 1) = T2 + 1; ② when t ≤ 1 ≤ T + 1, i.e. 0 ≤ t ≤ 1, the function f (x) gets the minimum value at the vertex, i.e. g (T) = f (1) = 1; ③ when t > 1, f (x) is on [T, t + 1] In conclusion, G (T) = T2 + 1 & nbsp;, & nbsp; & nbsp; T < 01 & nbsp;, & nbsp; & nbsp; 0 ≤ t ≤ 1t2 − 2T + 2 & nbsp;, t > 1

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