Prove that there is such A, such that for each $x \in [0,2]:$ $|f(x)-x| \le A(x-1)^2 $. Let $f$ have continuous derivatives until second order in the interval $[0,2]$, s.t. $f(1)=f^\prime (1)=1$.
Prove that there is such A, such that for each $x \in [0,2]:$
$$|f(x)-x| \le A(x-1)^2 $$ 
SOLUTION ATTEMPT:
I know that:


*

*$\lim_{x\to x_0} f^\prime  (x)=f^\prime (x_0)$

*$\lim_{x\to x_0} f^{\prime \prime}   (x)=f^{\prime \prime} (x_0)$


I don't have any idea how to continue from here, any hints? 

Hint: Expand to a first order Taylor polynomial about the point $x_0 = 1$ with the (Lagrange) mean value remainder form.
