Question

Recall from calculus that ascritical point of a differentiable function f(x,y) is a point where the partial derivatives

∂f/∂x and ∂f/∂yvanish simultaneously. When f∈R[x,y], it follows that the critical points can be found by applying our techniques to the system of polynomial equations ∂f/∂x =
∂f/∂y =0. To see how this works, consider the function. f(x;y)=(x² +y² −4)(x² +y² −1)+(x−3/2)² +(y−3/2)²
a. Find all critical points of f(x,y).

Answer 1

To find all critical points of the given function
`\( f(x, y) = (x^2 + y^2 - 4)(x^2 + y^2 - 1) + (x - (3)/(2))^2 + (y - (3)/(2))^2 \)`, we need to determine where the partial derivatives
`\( (\partial f)/(\partial x) \)` and
`\( (\partial f)/(\partial y) \)`are simultaneously zero. Solving the system of polynomial equations
`\( (\partial f)/(\partial x) = (\partial f)/(\partial y) = 0 \)`will provide the critical points.

Step-by-step explanation:

The partial derivatives of f with respect to x and y are calculated as
`\( (\partial f)/(\partial x) \) and \( (\partial f)/(\partial y) \`. Setting these derivatives equal to zero and solving the resulting system of equations provides the critical points of the function. This involves finding values of x and y where both partial derivatives vanish simultaneously.

Applying this process to the given function f(x, y) we calculate
`\( (\partial f)/(\partial x) \) and \( (\partial f)/(\partial y) \)`, set them to zero, and solve the resulting system of polynomial equations. The solutions to this system represent the critical points of the function.

In summary, the critical points of the function are obtained by finding the values of x and y where both partial derivatives
`\( (\partial f)/(\partial x) \)`and
`\( (\partial f)/(\partial y) \)` simultaneously equal zero. Solving this system of equations provides insight into the critical points' locations.