Suppose $f(x)$ is a real valued polynomial function of degree $6$ satisfying the following conditions 
Suppose $f(x)$ is a real-valued polynomial function of degree $6$ satisfying the following conditions
(a) $f$ has minimum value at $x=0$ and $2$
(b) $f$ has maximum value at $x=1$
(c) $\displaystyle\lim_{x\to 0}\frac{1}{x}\ln\,$$\det\pmatrix{\frac{f(x)}{x}&1&0\\0&\frac{1}{x}&1\\1&0&\frac{1}{x}} = 2$
Determine $f(x)$.


I let
$$f(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$$
and I took $f'(0)=0$, $f'(2)=0$, $f'(1)=0$ and $$\lim_{x\to 0}\frac{1}{x}\ln(\frac{f(x)}{x^3}+1)=2$$ But I am not able to solve it further.

Finding such a polynomial is not terribly difficult. Starting with
$$p(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g,$$
$p'(0)=0$ implies $f=0$, and in order for $\log(1+p(x)/x^3)/x$ to tend to $2$ when $x\to 0$, we need $\log(1+p(x)/x^3)$ to behave like $2x$ when $x\to 0$, so we could ask $p(x)/x^3$ to behave like $2x$ when $x\to 0$, because we know that $\log(1+\xi)\approx \xi$ for small $\xi$. So if we set $c=2,d=e=f=0$, we get:
$$p(x)=ax^6+bx^5+2x^4$$
and so
$$\lim_{x\to 0}\frac{\log(1+\frac{ax^6+bx^5+2x^4}{x^3})}{x}=\lim_{x\to 0}\frac{\log(1+2x)}{x}=2$$
Now, the rest of the given information helps to determine the coefficients $a,b$. Namely:
$$p'(1)=0\Longrightarrow 8+6a+5b=0,\enspace p'(2)=0\Longrightarrow 64+192a+80b=0$$
solving these two equations gives $a=\frac{2}{3}, b=-\frac{12}{5}$, hence a polynominal satisfying the requirements is given by:
$$p(x)=\frac{2x^6}{3}-\frac{12x^5}{5}+2x^4$$
To prove that only this polynomial satisfies the requirements takes a little more effort.
