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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
Let the minimum value of function f (x) = x2-2x-1 in the interval [T, t + 1] be g (T), and find the range of G (T)
∵ f (x) = x2-2x-1 = (x-1) 2-2, ∵ axis of symmetry x = 1, vertex coordinates (1, - 2), as shown in the figure; f (x) monotonically decreases on (- ∞, 1), monotonically increases on (1, + ∞). When 0 ≤ t ≤ 1, G (T) = - 2; when t ≥ 1, G (T) = f (T) = T2 -
12. When 0
The original formula = (cosx / SiNx) (cosx / (cosx SiNx)) = 1 / TaNx (1-tanx) = 1 / (0.25 - (tanx-0.5) ^ 2)
When x = arctan (1 / 2), TaNx = 1 / 2, the original formula is the smallest, which is 4. X, about 0.463648 or 26.5651 degrees
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