Asked by Ashley
Find the equation of the tangent line to the curve below at point (0,3)
y=6/(1+e^-x)
y=6/(1+e^-x)
Answers
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
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
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