Second derivative Test becomes zero

Question

Consider the domain $D = \{ (,) ∈ ℝ^2: ≤ \}$ and the function $ℎ: → ℝ$ defined by $ℎ((,)) = ( −2)^4 +(−1)^4$, $(,) ∈ $. Find the minimum value of $h$ in the domain $D$:
a) $1/2$
b) $1/4$
c) $1/8$
d) $1/16$

My work:

$f'(x) = 4(x-2)^3$
$f'(y) = 4(y-1)^3$
$f''(x) = 12(x-2)^2$
f_{xy} = $4(y-1)^3$ (partial derivative of $x,y$)
Critical points are $x = 2$ and $y = 1$,
Performing second derivative test with $x = 2$ and $y = 1$ I get zero , How do I proceed and where have I gone wrong? kindly help . The answer is Option C: 1/8

Answer 1

Since $(2,1)$ is not in the domain of $D$ of $h$, we have $\operatorname{grad}h(x,y) \ne (0,0)$ for all $(x,y) \in D$. Thus investigate the function $h$ on $ \partial D=\{(x,y): x=y\}$.

Hence consider $f(x):=(x-2)^4+(x-1)^4$. Can you proceed?