Question

Find the critical points and test for relative extrema. List the critical points for which the Second-Partials Test fails. $$ f(x, y)=x^{3}+y^{3} $$

Short answer

The only critical point is (0, 0). The Second-Partials Test fails at this critical point.

Step-by-step solution

Calculate First-order Partial Derivatives

To find the critical points of the function \(f(x, y)=x^{3}+y^{3}\), first, calculate the first-order partial derivatives of the function with respect to \(x\) and \(y\). These are given as: \(f_{x} = 3x^{2}\) and \(f_{y} = 3y^{2}\).

Find the Critical Points

Critical points occur where the first-order partial derivatives are each equal to zero. From \(f_{x} = 3x^{2}\) and \(f_{y} = 3y^{2}\), it can be seen that the only common solution is (0, 0). Hence, (0, 0) is the only critical point of the function.

Calculate Second-order Partial Derivatives

Next, calculate the second-order partial derivatives. These are given as: \(f_{xx} = 6x\), \(f_{yy} = 6y\), \(f_{xy} = f_{yx} = 0\). (Here, \(f_{xx}\) and \(f_{yy}\) are the second-order partial derivatives with respect to \(x\) and \(y\) respectively, and \(f_{xy}\) and \(f_{yx}\) are the mixed second-order partial derivatives).

Perform the Second-Partials Test

The Second-Partials Test uses the discriminant \(D = f_{xx} * f_{yy} - (f_{xy})^{2} = (6x * 6y) - (0)^{2}\). Substituting the critical point (0, 0), we get \(D = 0\), so the Second-Partials Test fails at this critical point.