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? 

For each $t\in[0,2]$, the Mean Value Theorem says that there is a $\xi$ between $t$ and $1$ so that
$$
\frac{f'(t)-1}{t-1}=f''(\xi)
$$
Since $f''$ is continuous in $[0,2]$, there is an $A$ so that
$$
\left|f''(\xi)\right|\le 2A
$$
for $\xi\in[0,2]$. Therefore,
$$
\left|f'(t)-1\right|\le 2A\left|t-1\right|
$$
For each $x\in[0,2]$, integration in $t$ from $1$ to $x$ yields that
$$
\left|f(x)-x\right|\le A(x-1)^2
$$
