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Find an equation of the tangent line to the curve

y = 6/(1+e^−x)at the point (0,3).

1 answer

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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