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linearization there. Then use it to approximate f(1.1, –0.1). SOLUTION The partial derivatives are fx(x, y) = fy(x, y) = fx(1, 0) = 5 fy(1, 0) = 5. Both fy and fy are continuous functions, so f is differentiable. The linearization is L(x, у) f(1, 0) + fx(1, 0)(x – 1) + fy(1, 0)(y – 0) = 5 + + 5. y %3D The corresponding linear approximation is 5xexy = So f(1.1, -0.1) - L(1.1, -0.1) = Compare this with the actual value. (Round your answer to five decimal places.) f(1.1, –0.1) = 5.5e¬0.11 Read It

linearization there. Then use it to approximate f(1.1, –0.1). SOLUTION The partial derivatives are fx(x, y) = fy(x, y) = fx(1, 0) = 5 fy(1, 0) = 5. Both fy and fy are continuous functions, so f is differentiable. The linearization is L(x, у) f(1, 0) + fx(1, 0)(x – 1) + fy(1, 0)(y – 0) = 5 + + 5. y %3D The corresponding linear approximation is 5xexy = So f(1.1, -0.1) - L(1.1, -0.1) = Compare this with the actual value. (Round your answer to five decimal places.) f(1.1, –0.1) = 5.5e¬0.11 Read It

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**EXAMPLE 2**: Show that \( f(x, y) = 5xe^{xy} \) is differentiable at (1, 0) and find its linearization there. Then use it to approximate \( f(1.1, -0.1) \). **SOLUTION**: The partial derivatives are \[ f_x(x, y) = \_\_\_\_\_ \quad f_y(x, y) = \_\_\_\_\_ \] \[ f_x(1, 0) = 5 \quad f_y(1, 0) = 5. \] Both \( f_x \) and \( f_y \) are continuous functions, so \( f \) is differentiable. The linearization is \[ L(x, y) = f(1, 0) + f_x(1, 0)(x - 1) + f_y(1, 0)(y - 0) \] \[ = 5 + (\_\_\_\_\_)(x - 1) + 5 \cdot y \] \[ = \_\_\_\_\_ .\] The corresponding linear approximation is \[ 5xe^{xy} \approx \_\_\_\_\_ \] so \[ f(1.1, -0.1) \approx L(1.1, -0.1) = \_\_\_\_\_ .\] Compare this with the actual value. (Round your answer to five decimal places.) \[ f(1.1, -0.1) = 5.5e^{-0.11} \approx \_\_\_\_\_ .\]
Transcribed Image Text:**EXAMPLE 2**: Show that \( f(x, y) = 5xe^{xy} \) is differentiable at (1, 0) and find its linearization there. Then use it to approximate \( f(1.1, -0.1) \). **SOLUTION**: The partial derivatives are \[ f_x(x, y) = \_\_\_\_\_ \quad f_y(x, y) = \_\_\_\_\_ \] \[ f_x(1, 0) = 5 \quad f_y(1, 0) = 5. \] Both \( f_x \) and \( f_y \) are continuous functions, so \( f \) is differentiable. The linearization is \[ L(x, y) = f(1, 0) + f_x(1, 0)(x - 1) + f_y(1, 0)(y - 0) \] \[ = 5 + (\_\_\_\_\_)(x - 1) + 5 \cdot y \] \[ = \_\_\_\_\_ .\] The corresponding linear approximation is \[ 5xe^{xy} \approx \_\_\_\_\_ \] so \[ f(1.1, -0.1) \approx L(1.1, -0.1) = \_\_\_\_\_ .\] Compare this with the actual value. (Round your answer to five decimal places.) \[ f(1.1, -0.1) = 5.5e^{-0.11} \approx \_\_\_\_\_ .\]
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