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) = 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

Well, isn't that a curve-y question! Let's put on our mathematical clown shoes and get to it!

To find the equation of the tangent line, we need to find the derivative of the curve at the point (0,3). The derivative will give us the slope of the curve at that point.

Taking the derivative of y = 6/(1+e^(-x)) using the chain rule, we get:

dy/dx = (6e^(-x))/((1+e^(-x))^2)

Plugging in x=0 into this derivative equation, we get:

dy/dx |(x=0) = (6e^0)/((1+e^0)^2)
= 6/4
= 3/2

So, at the point (0,3), the slope of the tangent line 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:

y - y₁ = m(x - x₁)

Plugging in the values, where (x₁, y₁) = (0, 3) and m = 3/2, we get:

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

Simplifying, we have:

y - 3 = (3/2)x

And voila! That's the equation of the tangent line to the curve y = 6/(1+e^(-x)) at the point (0,3).

To find the equation of the tangent line to the curve y = 6/(1+e^−x) at the point (0,3), we need to find the slope of the tangent line at that point and then use point-slope form to write the equation.

Step 1: Find the first derivative of the function y = 6/(1+e^−x) with respect to x.
Let's denote the function as f(x).
To find f'(x), we need to use the quotient rule.
The quotient rule states that if we have a function of the form f(x) = g(x)/h(x), then the derivative f'(x) is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2

In our case, g(x) = 6 and h(x) = (1+e^−x).

Using the quotient rule, we can find f'(x) as follows:

f'(x) = ((g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= ((0 * (1+e^−x)) - (6 * (-e^−x))) / ((1+e^−x)^2)
= (6e^−x) / ((1+e^−x)^2)

Step 2: Evaluate f'(x) at x = 0 to find the slope of the tangent line.
Substitute x = 0 into f'(x):

f'(0) = (6e^−0) / ((1+e^−0)^2)
= 6 / (1 + 1)^2
= 6 / 4
= 3/2

So the slope of the tangent line at the point (0,3) is 3/2.

Step 3: Use point-slope form to write the equation of the tangent line.
The point-slope form of a line is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope.

We know that the point (0,3) is on the tangent line and the slope of the tangent line is 3/2. So we can write the equation as:

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

Simplifying, we get:

y - 3 = (3/2)x

Finally, we can rearrange the equation to obtain the standard form:

2y - 6 = 3x

Therefore, the equation of the tangent line to the curve y = 6/(1+e^−x) at the point (0,3) is 2y - 6 = 3x.

To find the equation of the tangent line to the curve at the point (0,3), we need to find the derivative of the curve and then evaluate it at the point (0,3).

Step 1: Find the derivative of the curve y = 6/(1+e^(-x)) using the chain rule.

The derivative of y with respect to x is given by:
dy/dx = (d/dx)[6/(1+e^(-x))]

To find the derivative, we can use the quotient rule. Let f(x) = 6 and g(x) = (1+e^(-x)). Applying the quotient rule:
dy/dx = [g(x)*(d/dx)(f(x)) - f(x)*(d/dx)(g(x))]/[g(x)]^2

Since f(x) = 6, (d/dx)(f(x)) = 0. And since g(x) = (1+e^(-x)), (d/dx)(g(x)) = e^(-x) * (-1) = -e^(-x).

Substituting these values into the quotient rule expression, we get:
dy/dx = [g(x) * 0 - 6 * (-e^(-x))]/[g(x)]^2

Simplifying:
dy/dx = 6e^(-x)/[1+e^(-x)]^2

Step 2: Evaluate the derivative dy/dx at the point (0,3).
Substituting x = 0 into the derivative expression, we get:
dy/dx = 6e^0/(1+e^0)^2
dy/dx = 6/1^2
dy/dx = 6/1
dy/dx = 6

Step 3: Use the slope-intercept form of the equation of a line to write the equation of the tangent line.

The slope of the tangent line is the derivative dy/dx evaluated at the point (0,3). We found that dy/dx = 6 at (0,3).

Using the point-slope form of the equation of a line, we have:
y - y1 = m(x - x1)
where (x1, y1) is the given point (0,3) and m is the slope.

Substituting the values into the equation, we get:
y - 3 = 6(x - 0)

Simplifying, we get:
y - 3 = 6x

Rearranging the equation to the standard form, we have:
6x - y + 3 = 0

Therefore, the equation of the tangent line to the curve y = 6/(1+e^(-x)) at the point (0,3) is 6x - y + 3 = 0.