The Textbook Question
For $F_1(x,y) = \frac{1}{4}x^4 + x^2y + y^2$ and $F_2(x,y) = x^3 + xy - x$ find the second derivative matrices $H_1$ and $H_2$:
\begin{bmatrix}\frac{\partial^2 F}{\partial x^2}&\frac{\partial^2 F}{\partial x \partial y}\\\frac{\partial^2 F}{\partial x \partial y}&\frac{\partial^2 F}{\partial y^2}\end{bmatrix}
$H_1$ is positive definite so $F_1$ is concave up (= convex). Find the minimum point of $F_1$ and the saddle point of $F_2$ (look only where first derivatives are zero).
The Textbook Answer For $F_1$
$F_1 = \begin{bmatrix}6x^2&2x\\2x&2\end{bmatrix} $ is positive definite if $x \ne 0$;
$F_1 = (\frac{1}{2}x^2+y)^2=0$ on the curve $x^2+y=0$
My Answer For $F_1$
$\frac{\partial F_1}{\partial x}= x^3+2xy = x(x^2+2y)$
$\frac{\partial F_1}{\partial x}= 0 $
when
$x =0$
or when
$y = -\frac{1}{2}x^2$
so
$\frac{\partial^2 F_1}{\partial x^2}= 3x^2+2y=3x^2-x^2=2x^2$
$\frac{\partial^2 F_1}{\partial x \partial y}=2x$
$\frac{\partial^2 F_1}{\partial y^2}= 2$
$H_1 = \begin{bmatrix}2x^2&2x\\2x&2\end{bmatrix} $
A singular matrix
The minimum is the parabola $y = -\frac{1}{2}x^2$
Could someone please verify whether the textbook answer is wrong, the author is cleverer than I am so my calculations should be treated with more suspicion.