Question

Find the critical points and test for relative extrema. List the critical points for which the Second-Partials Test fails. $$ f(x, y)=x^{2 / 3}+y^{2 / 3} $$

Short answer

The critical point of the function is at (0,0) and the Second-Partials Test fails at this point.

Step-by-step solution

Calculate the first derivatives

The first step is to calculate the partial derivatives of the function with respect to \(x\) and \(y\). These are given by: \(f_x =\frac{2}{3}x^{-1/3}\) and \(f_y =\frac{2}{3}y^{-1/3}\).

Find the critical points

The critical points occur where both first derivatives are zero or undefined. From \(f_x =\frac{2}{3}x^{-1/3}\) and \(f_y =\frac{2}{3}y^{-1/3}\), we get that the critical point is at \((0,0)\).

Perform the Second-Partials Test

Calculate the second partial derivatives: \(f_{xx} =-\frac{2}{9}x^{-4/3}\), \(f_{yy} =-\frac{2}{9}y^{-4/3}\), and \(f_{xy} = 0\). The Hessian determinant is \(D(x,y) = f_{xx}f_{yy} - f_{xy}^2\). At the critical point (0, 0), all these second partial derivatives do not exist. Therefore, the Second-Partials Test fails at this critical point.