<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] $f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$
$= x^4(x^4 - 8x^2 + 16) +19x^4 - 16x^4 - 12 x^3 + 14x^2 - 8x + 9$
$= x^4(x^2 - 4)^2 + 3x^2(x^2 - 4x + 4) + 14x^2 -12x^2 - 8x + 9$
$= x^4(x^2 - 4)^2 + 3x^2(x - 2)^2 + 2(x^2 - 4x + 4) + 1$
$= (x - 2)^2(x^4(x+2) + 3x^2 +2) + 1 \ge 1$ with equality holding iff $x = 2$.
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