TITLE: The minimum value of $x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$ is
QUESTION [5 upvotes]: The minimum value of $$f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$$ is
(a)-1
(b)9
(c)6
(d)1
Apart from trying to obtain $1$, which in this case is simple and $f(2)=1$ is there a standard method to approach such problems.
Please keep in mind that this is an objective question in one of the competitive exam and you get around 5 mins to solve it.
Also this is asked in elementary section, so only knowledge of basic calculus and polynomials is assumed.

REPLY [2 votes]: Don't know that it counts as a heuristic, but if one assumes that the minimum is attained for an integer value of $x$, then either of the following observations leads to a plausible guess:

the only integer root of $f'(x)=0$ is $x=2\,$, which suggest that the answer is $f(2)=1$;
the only equations $f(x)=y \in \{-1,9,6,1\}$ with integer roots $x$ are in cases (b) and (d), which suggests that the answer is the lowest of the two i.e. (d).