<s>[INST] 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.
 [/INST] 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).
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