Consider the function f(x), a fourth degree polynomial such that $\underset{x\to 1}{\lim }\frac{f\left(x\right)}{{\left(x-1\right)}^2}=1,f^{\text{'}}\left(0\right)=-6,f^{\text{'}}\left(2\right)=6$

The minimum value of f(x) is

$\begin{array}{l}\underset{x\to 1}{lim} \frac{f\left(x\right)}{(x-1{)}^2}=1\\ \Rightarrow f\left(1\right)=0,f^{\mathrm{\prime }}\left(1\right)=0,f^{\prime \prime }\left(1\right)=2\\ \therefore f\left(x\right)=A(x-1{)}^4+B(x-1{)}^3+(x-1{)}^2\\ f^{\mathrm{\prime }}\left(x\right)=4A(x-1{)}^3+3B(x-1{)}^2+2(x-1)\\ f^{\mathrm{\prime }}\left(0\right)=-6\Rightarrow -4A+3B-2=-6 -----\left(\text{i}\right)\\ f^{\mathrm{\prime }}\left(2\right)=6\Rightarrow 4A+3B+2=6-----\left(\text{ii}\right)\\ \text{ (i), (ii) }\Rightarrow \mathrm{A}=1,\mathrm{B}=0\\ \therefore f\left(x\right)=(x-1{)}^4+(x-1{)}^2\ge 0\\ f\left(1\right)=0\text{ is the minimum }\end{array}$