Let the function f (x) = g (2x-1) + X, the tangent equation of the square curve y = g (x) at point (1, G (1)) be y = 2x + 1, then the tangent equation of the curve y = f (x) at point (1, f (1)) is y = 2x + 1 The tangent equation needs to be written in detail

f'(x)=2g'(x)+1=2x+1
So G '(x) = x, that is g (x) = x & # 178;, so f (x) = (2x-1) &# 178; + x = 4x & # 178; - 3x + 1
f'(x)=8x-3
f'(1)=5
f(1)=2
So tangent slope k = 5, over (1,2)
So the tangent equation is y = 5x-3

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