Ask a New Question
Menu

Find the equation of the tangent line to the curve below at point (0,3)

y=6/(1+e^-x)

1 answer

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
Similar Questions
  1. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 1 answer
  2. Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each
    1. answers icon 3 answers
  3. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 0 answers
  4. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 3 answers
more similar questions