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Chapter 13: Problem 3

Find any critical points and relative extrema of the function. $$ f(x, y)=\sqrt{x^{2}+y^{2}+1} $$

Short Answer

Expert verified
The function \(f(x, y)=\sqrt{x^{2}+y^{2}+1}\) has one critical point at the origin (0,0) and this is a local minimum.

Step by step solution

01

Calculate the partial derivatives.

The partial derivatives are calculated using the following formulas: \n \( f_x = \frac{x}{\sqrt{x^2 + y^2 + 1}} \) \n \( f_y = \frac{y}{\sqrt{x^2 + y^2 + 1}} \) \n These are the derivatives of the function with respect to x and y, respectively.
02

Find the critical points.

Critical points occur when either \(f_x = 0\) or \(f_y = 0\) or both. Setting \(f_x = 0\) gives \(x = 0\) and setting \(f_y = 0\) gives \(y = 0\). So, the only critical point is at the origin \( (0,0) \).
03

Use the second-partial derivative test.

Calculate the second partial derivatives and use them to form the matrix H. Now, the second partial derivatives are: \n \(f_{xx} = \frac{y^2 + 1}{(x^2 + y^2 + 1)^{3/2}}\), \(f_{yy} = \frac{x^2 + 1}{(x^2 + y^2 + 1)^{3/2}}\) and \(f_{xy} = f_{yx} = -\frac{xy}{(x^2 + y^2 + 1)^{3/2}}\). Thus, the determinant of H at (0,0) is: \(D = f_{xx}(0,0)f_{yy}(0,0) - (f_{xy}(0,0))^2 = 0\). Since D=0, the test is inconclusive.
04

Analyze the function near the origin.

Since the second-derivative test was inconclusive, it's necessary to analyze the function near the origin. Notice that \(f(x,y) = \sqrt{x^2 + y^2 + 1}\) is always positive for any (x, y). This means the origin (0,0) will be a local minimum since there aren't any points (x, y) for which \(f(x, y)\) is less than \(f(0,0)\).

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