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Show that the equations $$2e^x + yu - 4v + 3 = 0$$ $$y \cos x - 6x + 2u - w = 0$$ can be solved for functions $$x = f_1(u,v,w)$$ $$y = f_2(u,v,w)$$ in a small ball with centre $(3,2,7)$ such that $f_1(3,2,7)=0$, $f_2(3,2,7)=1$.

Pedro
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Risa
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  • I know I have to use the implicit function theorem and jacobian matrix. but i dont know how to do the steps to get the answer – Risa Jun 09 '13 at 02:09
  • It would be good if you wrote where you're stuck, what ideas you have and what you have tried in the post. – Pedro Jun 09 '13 at 02:14
  • Define f:R^2 x R^2 -> R^2 by f(u,v,w) = (2e^x + yu - 4v + 3, y cos x - 6x + 2u - w). – Risa Jun 09 '13 at 02:25
  • (Continue) f is C^1 (R^2 x R^2) ie, f and its partial derivatives of order <= 1 are continuous (I'm not so sure about this, i just got it from my notes but I couldn't understand the C^1 for) f1(3,2,7) = 2e^x + yu - 4v + 3 f2(3,2,7) = y cos - 6x + 2u - w

    J(f) = det(df1/dx df1/dy df2/dx df2/dy) = det (2e^x u -y sin x - 6 cos x) = 2e^x cos x - [u(-y sin x - 6)] = 2e^x cos x + u (y sin x + 6) = 2e^x cos x + 3(y sin x +6)

    That's my answer but i doubt its correct...

    So can you tell me what's the answer now??

     – Risa Jun 09 '13 at 02:31

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