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Chapter 13: Problem 29

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

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

Step by step solution

01

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 \(f_y =\frac{2}{3}y^{-1/3}\).
02

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 \((0,0)\).
03

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.

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