Find the equation of the tangent line to the curve below at point (0,3)
y=6/(1+e^-x)
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
Well, I would tell you the equation of the tangent line, but I'm afraid it would just go in one ear and out the other.
To find the equation of the tangent line to the curve at point (0,3), we need to find the slope of the tangent line.
First, let's find the derivative of the curve. The derivative of y = 6/(1+e^-x) can be determined using the chain rule.
Let u = 1 + e^(-x). Taking the derivative with respect to x, we get:
du/dx = d/dx(1 + e^(-x))
= d/dx(1) + d/dx(e^(-x))
= 0 - e^(-x)
= -e^(-x)
Next, we can use the quotient rule to find the derivative of y = 6/u:
dy/dx = (u * d/dx(6) - 6 * du/dx) / u^2
= (0 - 6 * (-e^(-x))) / (1 + e^(-x))^2
= (6e^(-x) / (1 + e^(-x))^2
To find the slope of the tangent line at (0,3), we substitute x = 0 into the derivative:
dy/dx = (6e^(0)) / (1 + e^(0))^2
= 6 / (1 + 1)^2
= 6 / 4
= 3/2
Therefore, the slope of the tangent line at (0,3) is 3/2.
Now we can find the equation of the tangent line using the slope-intercept form, y = mx + b, where m is the slope and (0,3) is a point on the line.
Using the point-slope form, we have:
y - 3 = (3/2)(x - 0)
y - 3 = (3/2)x
y = (3/2)x + 3
Therefore, the equation of the tangent line to the curve y = 6/(1 + e^-x) at point (0,3) is y = (3/2)x + 3.
To find the equation of the tangent line to the curve at point (0,3), we can start by finding the slope of the tangent line.
The slope of the tangent line can be found using the derivative of the function.
Let's find the derivative of the function y = 6/(1 + e^(-x)):
1. Start by applying the quotient rule to the function. The quotient rule states that if we have a function of the form f(x) = g(x)/h(x), then its derivative is given by [g'(x)h(x) - g(x)h'(x)] / [h(x)]^2.
2. Apply the quotient rule to y = 6/(1 + e^(-x)).
Let g(x) = 6 and h(x) = (1 + e^(-x)).
Then, g'(x) = 0 (since 6 is a constant) and h'(x) = e^(-x) (using the chain rule).
3. Substitute these values into the quotient rule formula to find the derivative of y:
y' = [0 * (1 + e^(-x)) - 6 * e^(-x)] / [(1 + e^(-x))]^2
Simplify the numerator:
y' = - 6e^(-x) / (1 + e^(-x))^2
Now that we have obtained the derivative, we can find the slope of the tangent line at point (0,3) by plugging in x = 0 into the derivative:
y' = -6e^(0) / (1 + e^(0))^2
= -6 / (1 + 1)^2
= -6 / 4
= -3/2
So, the slope of the tangent line at point (0,3) is -3/2.
Now that we have the slope, we can use the point-slope form of a line to find the equation of the tangent line. The equation of a line in point-slope form is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope.
Using the point (0,3) and the slope -3/2, we can plug these values into the point-slope form to get the equation of the tangent line:
y - 3 = (-3/2)(x - 0)
Simplifying:
y - 3 = (-3/2)x
y = (-3/2)x + 3
Therefore, the equation of the tangent line to the curve y = 6/(1 + e^(-x)) at point (0,3) is y = (-3/2)x + 3.