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Find the equation of the line tangent to the graph of f(x) = 11 − x + 4 ln (2x−1) at x = 1.

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f'(x) = -1 + 2/(2x-1)

f(1) = 11-1+4+ln(1) = 14
f'(1) = -1 + 2/1 = 1

so, now you have a point and a slope, so the tangent line is
y-14 = 1(x-1)

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