Given that the function f (x) satisfies f (x) = 2F (2-x) - x2 + 8x-8 on R, then the tangent equation of the curve y = f (x) at point (1, f (1)) is () A. y=2x-1B. y=xC. y=3x-2D. y=-2x+3 | Math enthusiast | Mathematical art

Given that the function f (x) satisfies f (x) = 2F (2-x) - x2 + 8x-8 on R, then the tangent equation of the curve y = f (x) at point (1, f (1)) is () A. y=2x-1B. y=xC. y=3x-2D. y=-2x+3

∵ f (x) = 2F (2-x) - x2 + 8x-8, ∵ f (1) = 2F (1) - 1 ∵ f (1) = 1 ∵ f ′ (x) = - 2F ′ (2-x) - 2x + 8 ∵ f ′ (1) = - 2F ′ (1) + 6 ∵ f ′ (1) = 2 according to the geometric meaning of derivative, the tangent equation of curve y = f (x) at point (1, f (1)) is Y-1 = 2 (x-1), that is y = 2x-1, so a

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