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Find the equation of the tangent line to the graph of the function

f(x)=x-3/3x-5 at x=1

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f'(x) = 4/(3x-5)^2
f'(1) = 4/4 = 1

f(1) = -2/-2 = 1
now you have a point (1,1) and a slope (1)

(y-1) = 1(x-1)

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