Obviosly $ f_{1}(x)=a_{1}(x-x_{0})^{2},f_{2}(x)=a_{2}(x-x_{0})^{2}.$

You're right obviously. First observation: $ a1\cdot a2>0$. From the condition $ f'_{1}(x)f'_{2}(x)\ge 0$ we get that $ f'_{1}$ and $ f'_{2}$ have the same root $ x_{0}$. But if we put $ x=x_{0}$ we get that it is a root for $ f_{1}$ and $ f_{2}$ too. So $ f_{1}(x)=a_{1}(x-x_{0})^{2}$ $ f_{2}(x)=a_{2}(x-x_{0})^{2}$ $ g(x)=\sqrt{a_{1}a_{2}}(x-x_{0})^{2}$