= Solution
Substitution gives
$$
f(1)=-6,\qquad f'(1)=f''(1)=f^{(3)}(1)=0,\qquad f^{(4)}(1)=24.
$$
In particular, $1$ is a <stationary point>. For $x$ sufficiently close to $1$, <Taylor's theorem> with fifth-order remainder gives
$$
\boxed{f(x)=-6+(x-1)^4+\frac{f^{(5)}(\xi)}{120}(x-1)^5,}
$$
where $\xi$ lies between $1$ and $x$. The fifth <derivative> is bounded near $1$, so the remainder is $O((x-1)^5)$ and is smaller in magnitude than half the positive quartic term for sufficiently small nonzero $x-1$. Therefore \b[$x=1$ is a strict <local minimum>], despite the vanishing <second derivative>.
An exact algebraic check gives the stronger global conclusion:
$$
f(x)+6=\frac{(x-1)^4}{x^2}\geq0\qquad(x\ne0).
$$
Equality holds only at $x=1$, so it is the unique <global minimum> on the actual domain.
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