Asked by TayB
Find an equation of the tangent line to the curve
y = 6/(1+e^−x)at the point (0,3).
y = 6/(1+e^−x)at the point (0,3).
Answers
Answered by
Steve
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.