Apply the second partials test to $f$ at $(0, 0)$. For which values of $k$ is the test inconclusive

Question

This is what i done, not entirely sure it is correct. first found the derivative of $f(x,y)$ with respect to $x$ it was $2x+ky$ then found the derivative of $f(x,y)$ with respect to y it was $2y+kx$ i then let both of them equal $0$

solved $2x+ky=0$ looking for a value of x and i got $x=-ky/2$ i then put this value for x back into $2y+kx=0$ i got y to cancel and i got a value for k. i got $k=2$.

so then i rewrote my derivatives as $2x+2y=0$ and $2y+2x=0$ went ahead and tried to solve one of them for a value of x so i could sub it back and i got $x=-y$ then i went and subbed this new value for x back into the equation and i'm getting $0=0$ This means that for k=2 the test is inclusive correct? considering the question is stating "values" meaning plural? how would find more values?

Answer 1

You don't need to solve for $x,y$ since the critical point is already given. Plugging $(0,0)$ into first partials of $f$. If they are both evaluated $0$, then this verifies $(0,0)$ is a critical point.

Now you need to find the Hessian matrix $$\begin{pmatrix}\frac{\partial^2 f}{\partial x^2}&\frac{\partial^2 f}{\partial x\partial y}\\ \frac{\partial^2 f}{\partial y\partial x}&\frac{\partial^2 f}{\partial y^2}\end{pmatrix}.$$

Then see here for the second derivative test by Hessain matrix.