For the function: $$f(a,b,c,d,h)= \begin{pmatrix} 2e^{a}+ bc - 4d +3 \\ b\cos(a)-6a+2c-h \end{pmatrix}$$
Show that there is a continuously differentiable map $g$ defined in the neighborhood of $(3,2,7)$ with values in the neighborhood of $(0,1)$ so that $f(g(y),y)=0$ with all $y$ in the domain of $g$.
I get that the Jacobian can define the $(3,2,7)$ neighborhood, but how to proceed further? Or am I wrong. Help please.
plug in 0,1,3,2,7 to check you get o,o . Now calculate the Jacobian matrix of partial derivatives of the function f(a,b ,3.2.7) from R^2 to R^2 at a=0,b=1 and take its determinant to check it is not 0 .You say you did this .Now quote that the existence of the required neighborhoods follows from the implicit function theorem . This theorem does not give the size of the neighborhoods ,only that they exist . .Hope that helps .smn