You should think geometrically. The function $$f(x,y)=-x^{2012}-y^{2012}$$ can be written $$f(x,y)=-(x^{2012}+y^{2012})$$ But because 2012 is even, $x^{2012}\geq0$ and $y^{2012}\geq 0$. Therefore $x^{2012}+y^{2012}\geq0$ and so $f(x,y)\leq 0$ for all points $(x,y)$. Therefore your critical point $(0,0)$ is a global maximum.
Whoever assigned this problem probably wanted you to notice that your function $f$ is similar to
$$g(x,y)=x^2+y^2$$which is an upward-opening elliptic paraboloid that has a global minimum at the origin. Similarly, $$h(x,y)=-x^2-y^2$$ is a downward-opening elliptic paraboloid that has a global maximum at the origin.
The only difference between these functions and your function $f$ is that $f$ rises much, much faster when $|x|>1$ and $|y|>1$ and is much, much flatter when $|x|<1$ and $|y|<1$, because the exponent is so much higher. In fact, the graph would look more like a box than a paraboloid because the walls of the surface are so steep.
In general, there's no surefire method for analyzing the local behavior of functions where the second derivative test comes back inconclusive. In practice, you should think geometrically or look at higher order derivatives to get a sense of what's going on.
To use the latter approach, consider taking the 2012th partial derivatives of your function. Both the $x$ and $y$ partial at the origin will be $-2012!$, and combining this with the multivariate Taylor expansion (and reflecting on the mixed partials) gives another way of seeing that your function is negative definite near the origin. (The reason the second derivative test fails for this function is that it is too flat near its critical point. This extreme flatness is what makes so many of the higher-order derivatives zero.)
But your function is so simple to understand that its global properties are obvious if you think geometrically. Other functions might require you to think harder about what's happening graphically, in which case the analytic approach with higher derivatives could be helpful.
This webpage states and proves a version of the higher-order derivative test that applies not only to functions defined on $\mathbb{R}^2$ or $\mathbb{R}^N$, but functions defined on arbitrary Banach spaces. First there is this theorem:
Theorem 38 (Higher derivative test). Let $A\subseteq E$ be an open set and let f$:A\to\mathbb{R}$. Assume that $f$ is $(p-1)$ times continuously differentiable and that $D^p f(x)$ exists for some $p\ge 2$ and $x\in A$. Also assume that $f'(x),\dots,f^{(p-1)}(x)=0$ and $f^{(p)}(x)\ne 0$. Write $h^{(p)}$ for the $p$-tuple $(h,\dots,h)$.
- If $f$ has an extreme value at $x$, then $p$ is even and the form $f^{(p)}(x)h^{(p)}$ is semidefinite.
- If there is a constant $c$ such that $f^{(p)}(x)h^{(p)}\ge c > 0$ for all $|h|=1$, then $f$ has a strict local minimum at $x$ and (1) applies.
- If there is a constant $c$ such that $f^{(p)}(x)h^{(p)}\le c < 0$ for all $|h|=1$, then $f$ has a strict local maximum at $x$ and (1) applies.
Then there is this corollary for the finite dimensional case, which is what we’re interested in:
Corollary 39 (Higher derivative test, finite-dimensional case). In Theorem 38, further assume that $E$ is finite-dimensional. Then $h\mapsto f^{(p)}(x)h^{(p)}$ has both a minimum and maximum value on the set $\{h\in E:|h|=1\}$, and:
- If the form $f^{(p)}(x)h^{(p)}$ is indefinite, then $f$ does not have an extreme value at $x$.
- If the form $f^{(p)}(x)h^{(p)}$ is positive definite, then $f$ has a strict local minimum at $x$.
- If the form $f^{(p)}(x)h^{(p)}$ is negative definite, then $f$ has a strict local maximum at $x$.
Here $f^{(p)}(x)$ denotes a tensor containing all the pure and mixed partial derivatives of $f$ of order $p$, evaluated at $x$.
Best Answer
No, you have it backwards. Assuming $\epsilon >0$, then subtracting $\epsilon$ from $x$ produces a value less than (i.e., to the left of) $x$, and adding $\epsilon$ to $x$ produces a value greater than (i.e., to the right of) $x$. In other words,
$$x-\epsilon < x$$ and $$x < x+\epsilon.$$