Calculus 1

Find an equation of the tangent line to the curve
y = 6/(1+e^−x)at the point (0,3).

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  1. y = 6/(1+e^-x) = 6e^x/(1+e^x)
    y' = 6e^x/(1+e^x)^2
    so, at (0,3) the slope is 3/2

    Now you have a point and a slope, so the line is

    y-3 = 3/2 (x-0)

    see the graphs at

    http://www.wolframalpha.com/input/?i=plot+y%3D+6%2F%281%2Be^-x%29%2C+y%3D3%2F2+x%2B3

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