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Provide an appropriate response. Find the local extrema for \( f(x, y)=x^{3}-12 x+y^{2} \)

Ask by Pollard Chang. in the United States
Dec 19,2024

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Answer

The function \( f(x, y) = x^{3} - 12x + y^{2} \) has a local minimum at \( (2, 0) \) and a saddle point at \( (-2, 0) \).

Solution

To find the **local extrema** of the function \( f(x, y) = x^{3} - 12x + y^{2} \), we'll follow these steps: 1. **Find the Critical Points:** Critical points occur where the first partial derivatives are zero. - **Partial derivative with respect to \( x \):** \[ f_x = \frac{\partial f}{\partial x} = 3x^{2} - 12 \] Setting \( f_x = 0 \): \[ 3x^{2} - 12 = 0 \quad \Rightarrow \quad x^{2} = 4 \quad \Rightarrow \quad x = \pm 2 \] - **Partial derivative with respect to \( y \):** \[ f_y = \frac{\partial f}{\partial y} = 2y \] Setting \( f_y = 0 \): \[ 2y = 0 \quad \Rightarrow \quad y = 0 \] Therefore, the critical points are \( (2, 0) \) and \( (-2, 0) \). 2. **Classify the Critical Points Using the Second Derivative Test:** - **Second partial derivatives:** \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x, \quad f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2, \quad f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \] - **Determinant of the Hessian (D):** \[ D = f_{xx} \cdot f_{yy} - (f_{xy})^2 = (6x)(2) - 0 = 12x \] - **Evaluate \( D \) and \( f_{xx} \) at each critical point:** - **At \( (2, 0) \):** \[ D = 12 \times 2 = 24 > 0 \quad \text{and} \quad f_{xx} = 6 \times 2 = 12 > 0 \] Since \( D > 0 \) and \( f_{xx} > 0 \), \( (2, 0) \) is a **local minimum**. - **At \( (-2, 0) \):** \[ D = 12 \times (-2) = -24 < 0 \] Since \( D < 0 \), \( (-2, 0) \) is a **saddle point** (neither a local maximum nor a minimum). 3. **Conclusion:** - **Local Minimum:** \( (2, 0) \) - **Saddle Point:** \( (-2, 0) \) **Graphical Interpretation:** - At \( (2, 0) \), the function reaches its lowest nearby value, making it a local minimum. - At \( (-2, 0) \), the function has a saddle point, indicating a point of inflection where the function changes curvature. **Final Answer:** After simplifying, f  has a local minimum at (2 , 0) and a saddle point at ( –2, 0). Thus, the local extrema of f(x, y) = x³ – 12x + y² are: • A local minimum at (2, 0) • A saddle point at (−2, 0)

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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 \( (-2, 0) \).

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