Example For \( f(x, y)=x^{2}+y^{2} \) at \( (0,0) \) critical point. \( |H|=4 \) while \( f_{x x}(0,0)=2 \). Hence, \( (0,0) \) has a local maximum of 0 .

To analyze the critical point \((0,0)\) for the function \( f(x, y) = x^2 + y^2 \), we can follow these steps: 1. **Find the first derivatives**: \[ f_x = \frac{\partial f}{\partial x} = 2x, \quad f_y = \frac{\partial f}{\partial y} = 2y \] Setting these equal to zero gives us the critical point: \[ 2x = 0 \quad \Rightarrow \quad x = 0 \] \[ 2y = 0 \quad \Rightarrow \quad y = 0 \] Thus, the only critical point is \((0,0)\). 2. **Find the second derivatives**: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2, \quad f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2, \quad f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \] 3. **Construct the Hessian matrix**: The Hessian matrix \( H \) is given by: \[ H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \] 4. **Calculate the determinant of the Hessian**: \[ |H| = f_{xx} f_{yy} - (f_{xy})^2 = 2 \cdot 2 - 0^2 = 4 \] 5. **Determine the nature of the critical point**: - Since \( |H| > 0 \) and \( f_{xx} > 0 \), the critical point \((0,0)\) is a local minimum. 6. **Evaluate the function at the critical point**: \[ f(0,0) = 0^2 + 0^2 = 0 \] Thus, the conclusion is that \((0,0)\) is a local minimum of the function \( f(x, y) = x^2 + y^2 \) with a minimum value of \( 0 \). In summary, the statement in your example is incorrect regarding the nature of the critical point. The correct conclusion is that \((0,0)\) is a local minimum, not a local maximum.