Question

Identify where the function x³y³ - 130xy = 0 has a horizontal tangent.

Answer 1

We solve the partial derivative with respect to x for zero. This results in y(3x²y² - 130) = 0, which gives us y=0 and y = ±√(130/3x²). Hence, horizontal tangents occur at those points.

Step-by-step explanation:

To identify where the function x³y³ - 130xy = 0 has a horizontal tangent, we must understand the notion of a horizontal tangent in the context of functions of two variables. A horizontal tangent occurs when the partial derivative with respect to y, represented as dy/dx, is zero. In other words, we need to find where the change in y in relation to x is zero.

Let's first find the partial derivative with respect to x, which is given by:

f'_x(x, y) = ∂3x²y³ - 130y

For a horizontal tangent, f'_x(x, y) must equal zero, so we need to solve the equation:

3x²y³ - 130y = 0

To solve this equation, we can factor out y:

y(3x²y² - 130) = 0

So, y = 0 or 3x²y² - 130 = 0. For y = 0, this is trivially true for all x. To find other points, we solve 3x²y² - 130 = 0 for y:

y² = ⅔130/(3x²)

y = ±√(130/3x²)

So the function has horizontal tangents at y = 0, and y = ±√(130/3x²) for nonzero values of x such that this expression is real.