Exercise 9.23

Question

Exercise 23: Define $f$ in $\R3$ by
    $$
        f(x,y_1,y_2)=x^2y_1+e^x+y_2.
    $$
    Show that $f(0,1,-1)=0$, $D_1f(0,1,-1)\ne0$, and that there exists therefore a differentiable function $g$ in some neighborhood of $(1,-1)$ in $\R2$, such that $g(1,-1)=0$ and
    $$
        f\big(g(y_1,y_2),y_1,y_2\big)=0.
    $$
    Find $D_1g(1,-1)$, and $D_2g(1,-1)$.

analambanomenos · Accepted Answer

\def\where{\,|\,}

Note that
    \begin{align*}
        D_1f(x,y_1,y_2) &= 2xy_1+e^x \
        D_2f(x,y_1,y_2) &= x^2 \
        D_3f(x,y_1,y_2) &= 1.
    \end{align*}
    Hence
    \begin{align*}
	f(0,1,-1) &= 0+1-1 = 0 \
	D_1f(0,1,-1) &= 0+1=1 \
	D_2f(0,1,-1) &= 0 \
	D_3f(0,1,-1) &= 1.
    \end{align*}
    We can apply the implicit function theorem with $m=1$, $n=2$, where
    \begin{align*}
        A_x &=
        \begin{pmatrix}
            1
        \end{pmatrix} \
        A_y &=
        \begin{pmatrix}
            0 & 1 
        \end{pmatrix}
    \end{align*}
    to conclude that there is a function $g$ in some neighborhood of $(1,-1)$ such that 
    \begin{align*}
        f\big(g(y_1,y_2),y_1,y_2\big) &= 0 \
        g'(1,-1) &= -A_x^{-1}A_y \
        &= -
        \begin{pmatrix}
            1
        \end{pmatrix}
        ^{-1}
        \begin{pmatrix}
            0 & 1 
        \end{pmatrix} \
        &=
        \begin{pmatrix}
            0 & -1 
        \end{pmatrix}
    \end{align*}
    so that $D_1g(1,-1)=0$ and $D_2g(1,-1)=-1$.

Exercise 9.23

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Exercise 9.23

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