Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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How do I get fxx, fyy, and fxy?
**Review: Find and classify CPs (Critical Points) of \( f(x, y) = e^{x^2 + y^2 - 2x} \)** 1. Compute the partial derivative with respect to \( x \): \[ f_x = (2x - 2) e^{x^2 + y^2 - 2x} \] Setting \( f_x = 0 \), gives \( x = 1 \). *Note: \( e^{\text{any #}} > 0 \)* 2. Compute the partial derivative with respect to \( y \): \[ f_y = (2y) e^{x^2 + y^2 - 2x} \] Setting \( f_y = 0 \), gives \( y = 0 \). 3. The critical point is \( CP = (1, 0) \). 4. Second partial derivatives: \[ f_{xx} = 2e^{x^2 + y^2 - 2x} + (2x - 2)^2 e^{x^2 + y^2 - 2x} \] \[ f_{yy} = 2e^{x^2 + y^2 - 2x} + 4y^2 e^{x^2 + y^2 - 2x} \] \[ f_{xy} = 2y(2x - 2) e^{x^2 + y^2 - 2x} \] 5. Evaluate the Hessian determinant \( H \) at the critical point \( (1, 0) \): \[ H(1, 0) = 2e^{-1}(2e^{-1}) - 0^2 > 0 \] Additionally, check \( f_{xx}(1, 0) > 0 \): \[ f_{xx}(1, 0) = 2e^{-1} > 0 \] 6. Conclusion: Relative minimum at \( (1, 0) \).
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Transcribed Image Text:**Review: Find and classify CPs (Critical Points) of \( f(x, y) = e^{x^2 + y^2 - 2x} \)** 1. Compute the partial derivative with respect to \( x \): \[ f_x = (2x - 2) e^{x^2 + y^2 - 2x} \] Setting \( f_x = 0 \), gives \( x = 1 \). *Note: \( e^{\text{any #}} > 0 \)* 2. Compute the partial derivative with respect to \( y \): \[ f_y = (2y) e^{x^2 + y^2 - 2x} \] Setting \( f_y = 0 \), gives \( y = 0 \). 3. The critical point is \( CP = (1, 0) \). 4. Second partial derivatives: \[ f_{xx} = 2e^{x^2 + y^2 - 2x} + (2x - 2)^2 e^{x^2 + y^2 - 2x} \] \[ f_{yy} = 2e^{x^2 + y^2 - 2x} + 4y^2 e^{x^2 + y^2 - 2x} \] \[ f_{xy} = 2y(2x - 2) e^{x^2 + y^2 - 2x} \] 5. Evaluate the Hessian determinant \( H \) at the critical point \( (1, 0) \): \[ H(1, 0) = 2e^{-1}(2e^{-1}) - 0^2 > 0 \] Additionally, check \( f_{xx}(1, 0) > 0 \): \[ f_{xx}(1, 0) = 2e^{-1} > 0 \] 6. Conclusion: Relative minimum at \( (1, 0) \).
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