The minimum value of $x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$ is 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.
 A: This one is a bit tailor-made, I'm afraid.  The start of the polynomial naturally inclines to a perfect square trinomial (something that would obviously help in determining a minimum), and conveniently, it all falls out from there.
\begin{align}
f(x) & = x^8-8x^6+19x^4-12x^3+14x^2-8x+9 \\
     & = x^8-8x^6+16x^4 \\
     & \phantom{= x^8-8x^6\,\,}+\phantom{1}3x^4-12x^3+12x^2 \\
     & \phantom{= x^8-8x^6+19x^4-12x^3\,\,}+\phantom{1}2x^2-8x+8 \\
     & \phantom{= x^8-8x^6+19x^4-12x^3+14x^2-8x\,\,}+1
\end{align}
which allows us to rewrite $f(x)$ as
$$
f(x) = x^4(x^2-4)^2 + 3x^2(x-2)^2 + 2(x-2)^2 + 1
$$
The first three terms are clearly non-negative, and each reaches their minimum of $0$ at $x = 2$ (the first term also has a minimum at $x = -2$).  Thus, the minimum of $f(x)$ must be $1$.
This can't really be generalized.  (I mean, you can apply the approach generally, but it won't generally give you such a convenient result.)  I'm not sure I would have looked for this decomposition of $f(x)$ except for the presence of the question.
A: 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).
A: Let $a,b,c,d,e,k\in\Bbb R$,
$$(x^4-ax^2)^2+(bx^2-cx)^2+(dx-e)^2+k=\\ x^8- 2 a x^6+(a^2+ b^2) x^4 - 2 b c x^3 + (c^2 + d^2) x^2 - 2 d e x+e^2+k$$
By identification we easily obtain
$$a=4 \implies b = \sqrt{3} \implies c=\frac{6}{\sqrt{3}} \implies d= \sqrt{2} \implies e= \frac{4}{\sqrt{2}} \implies k=1$$
this implies that
$$x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9 = \\ \big(x^4-4x^2\big)^2+\Big(\sqrt{3}x^2-\frac{6}{\sqrt{3}}x\Big)^2+\Big(\sqrt{2}x-\frac{4}{\sqrt{2}}\Big)^2+1 \\ 
= \big(x^2(x^2-4)\big)^2+3\big(x(x-2)\big)^2+2\big(x-2\big)^2+1>0 $$
Hence, $-1$ can not be the minimal value.
As you have already noticed, it is easy to observe that $f(2)=1$ and since you know that exactly one answer is correct (see comments under OP), it must be the answer (d).
A: $f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$
$= x^4(x^4 - 8x^2 + 16) +19x^4 - 16x^4 - 12 x^3 + 14x^2 - 8x + 9$
$= x^4(x^2 - 4)^2 + 3x^2(x^2 - 4x + 4) + 14x^2 -12x^2 - 8x + 9$
$= x^4(x^2 - 4)^2 + 3x^2(x - 2)^2 + 2(x^2 - 4x + 4) + 1$
$= (x - 2)^2(x^4(x+2) + 3x^2 +2) + 1 \ge 1$ with equality holding iff $x = 2$.
