Asked by Ashley

Find the equation of the tangent line to the curve below at point (0,3)
y=6/(1+e^-x)
Answered by Steve
y'(x) = 6e^-x / (1+e^-x)^2

y'(0) = 6/4 = 3/2

So, we have the line with slope 3/2 passing through (0,3)

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