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Given f(x,412) = x+ 4y? +1622 9(2,412) = = =) g(X x2-1=0 F(x, y12, 1) = frayi2) + %3D f(2,412,x) = (2?4 4y +162) + d (2y2-1) Fa = Fy = 8y +) Fz 327 +) 242-1 Upon using the Lagrange, why did Fx equals 2x + A if the basis function is f(2,412,x) = (2?4 442+162 + (242-1) Shouldn't it be Ex = 2x + Ayz since we take the partial with respect to x where y and z are treated constants and x as a variable.

Given f(x,412) = x+ 4y? +1622 9(2,412) = = =) g(X x2-1=0 F(x, y12, 1) = frayi2) + %3D f(2,412,x) = (2?4 4y +162) + d (2y2-1) Fa = Fy = 8y +) Fz 327 +) 242-1 Upon using the Lagrange, why did Fx equals 2x + A if the basis function is f(2,412,x) = (2?4 442+162 + (242-1) Shouldn't it be Ex = 2x + Ayz since we take the partial with respect to x where y and z are treated constants and x as a variable.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Given f(x,412) = x+ 4y? +1622 9(2,412) = -) g(x4) - x42-1=0 FIx, y12, 1) = F(2,412,x) = (2?4 4y +162) + d (2y2-1) Fa = Fy = 8y +) Fz 327 +) FX 242-1 Upon using the Lagrange, why did Fx equals 2x + A if the basis function is f(2,412,x) = (2?t4y+162 + x (2Y2-1) Shouldn't it be Fx = 2x + Ayz since we take the partial with respect to x where y and z are treated constants and x as a variable.
Transcribed Image Text:Given f(x,412) = x+ 4y? +1622 9(2,412) = -) g(x4) - x42-1=0 FIx, y12, 1) = F(2,412,x) = (2?4 4y +162) + d (2y2-1) Fa = Fy = 8y +) Fz 327 +) FX 242-1 Upon using the Lagrange, why did Fx equals 2x + A if the basis function is f(2,412,x) = (2?t4y+162 + x (2Y2-1) Shouldn't it be Fx = 2x + Ayz since we take the partial with respect to x where y and z are treated constants and x as a variable.
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