Determine $f(x)$ which satisfies given condition Suppose $f(x)$ is real valued function of degree $6$ satisfying the following conditions:
$1.$ $f(x)$ has minimum at $x=0$ and $x=2$
$2.$ $f(x)$ has maximum at x=1
$3.$ $lim(x \to 0)$ $\frac{ln(\Delta)}{x}=2$ where $\Delta$ is
\begin{vmatrix}
\frac{f(x)}{x} & 1 & 0 &\\ 
0 & \frac{1}{x} & 1 &\\ 
1 & 0 & \frac{1}{x} &\\ 
 \notag
\end{vmatrix}
Determine $f(x)$
If we assume $f(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$, set $f'(x)=0$ at given critical points even then we will get only three equations.
One equation can be formed by using the fact that $\Delta$ must be equal to $1$ at $x=0$ as limit is finite.
One more equation can be formed by using $lim(x \to 0)$ $\frac{ln(\Delta)}{x}=2$ but still the solution is short of two equations (As we have to find 7 variables) and method it too length as well.
Could someone suggest a better approach?
 A: The determinant is $\Delta=1+f(x)/x^3$. Condition 3 imposes that $\ln(1+f(x)/x^3)\sim 2x$ for $x\rightarrow 0$, then necessarily $g=f=e=d=0$, and a condition arises on $c$ to get the value 2. The condition at $x=0$ is automatically fulfilled.
If I am not wrong,
$$
f(x)=-(12/5)x^5+(2/3)x^6+2x^4
$$
A: For the limit to exist we need 
\begin{align*}
&\lim_{x\rightarrow 0} \ln \Delta(0) = 0\\
\implies &\lim_{x\rightarrow 0} \Delta(0) = 1\\
\implies &\lim_{x\rightarrow 0} f(x)/x^3+1 = 1\\
\implies &\lim_{x\rightarrow 0} f(x)/x^3 = 0.
\end{align*}
Since $f(x)$ is continuous this is only possible if $f(0) = 0$.  By L'hopital you also get $f'(0) = f''(0) = f'''(0) = 0$, so $f(x)$ is divisible by $x^4$.
The derivative has zeros at $x=0,1,2$, so $f'(x) = g(x)x(x-1)(x-2)$ with $g(x)$ having degree 2.  Since $f(x)$ is divisible by $x^4$, $f'(x)$ will be divisible by $x^3$, so $g(x)=cx^2$.
So 
\begin{align*}
f'(x) &= cx^3(x-1)(x-2) = c(x^5-3x^4+2x^3)\\
f(x) &= c((1/6)x^6 - (3/5)x^5 + (1/2)x^4)
\end{align*}
Now solve for $c$ to make the limit work.
