Provide an appropriate response. Find the local extrema for \( f(x, y)=x^{3}-12 x+y^{2} \)
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To find the local extrema of the function \( f(x, y) = x^{3} - 12x + y^{2} \), we first need to compute the partial derivatives and set them to zero. The partial derivative with respect to \( x \) is \( f_x = 3x^{2} - 12 \), and with respect to \( y \) it is \( f_y = 2y \). Setting \( f_x = 0 \) gives us \( 3x^{2} - 12 = 0 \), leading to \( x^{2} = 4 \), so \( x = 2 \) or \( x = -2 \). Setting \( f_y = 0 \) gives \( 2y = 0 \), leading to \( y = 0 \). Thus, the critical points are \( (2, 0) \) and \( (-2, 0) \). Next, we use the second derivative test: 1. Compute \( f_{xx} = 6x \), \( f_{yy} = 2 \), and \( f_{xy} = 0 \). 2. The discriminant \( D = f_{xx} f_{yy} - (f_{xy})^2 \). - At \( (2, 0) \): - \( f_{xx} = 12 \), \( f_{yy} = 2 \), \( D = 12 \cdot 2 - 0^2 = 24 \) (positive, and \( f_{xx} > 0 \)): local minimum. - At \( (-2, 0) \): - \( f_{xx} = -12 \), \( f_{yy} = 2 \), \( D = -12 \cdot 2 - 0^2 = -24 \) (negative): saddle point. Thus, \( f \) has a local minimum at \( (2, 0) \) and a saddle point at \( (-2, 0) \).