dy/dx= (2x-2xy)/(x^2+2y)

Find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of those points.

To find all the points on the curve where x = 2, we can substitute x = 2 into the given equation and solve for y.

Substituting x = 2 into the equation, we have:

dy/dx = (2(2) - 2(2)(y)) / (2^2 + 2y)

Simplifying, we get:

dy/dx = (4 - 4y) / (4 + 2y)

Now, to find the value of y, we need to set dy/dx equal to 0. This will give us the points on the curve where a horizontal tangent exists.

Setting dy/dx = 0, we have:

(4 - 4y) / (4 + 2y) = 0

To solve this equation, we set the numerator equal to zero:

4 - 4y = 0

Solving for y, we get y = 1.

So, when x = 2, we have a point (2, 1) on the curve.

To show that there is a horizontal tangent at (2, 1), we need to find the derivative of the given equation and substitute x = 2 and y = 1.

Differentiating the given equation with respect to x, we get:

d^2y/dx^2 = (2 - 2y - 2x(-y) - 2xy) / (x^2 + 2y)^2

Substituting x = 2 and y = 1 into the second derivative, we have:

d^2y/dx^2 = (2 - 2(1) - 2(2)(-1) - 2(2)(1)) / (2^2 + 2(1))^2

Simplifying, we get:

d^2y/dx^2 = (2 + 2 + 4 - 8) / (4 + 2)^2

d^2y/dx^2 = 0 / 36

Since the second derivative is equal to zero at (2, 1), it confirms that there is a horizontal tangent at that point.