Step-by-step Solution:

Given function: \[ f(x, y) = x^2 - 2x + y^2 + 2y - 2 \] Step 1: Finding Partial Derivatives \[ f_x = \frac{\partial f}{\partial x} = 2x - 2 \] \[ f_y = \frac{\partial f}{\partial y} = 2y + 2 \] Step 2: Finding Critical Points Setting \( f_x = 0 \) and \( f_y = 0 \): \[ 2x - 2 = 0 \Rightarrow x = 1 \] \[ 2y + 2 = 0 \Rightarrow y = -1 \] Thus, the critical point is \( (1, -1) \).
Step 3: Second-Order Partial Derivatives \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \] Step 4: Hessian Determinant
The Hessian determinant is given by: \[ D = (f_{xx} \cdot f_{yy}) - (f_{xy})^2 \] \[ D = (2 \times 2) - (0)^2 = 4 > 0 \] Since \( D > 0 \) and \( f_{xx} > 0 \), the function attains a minimum at \( (1, -1) \).
Conclusion: Thus, the function has a minimum at \( (1, -1) \)