From the following form of Taylor's Theorem and assuming that $|f(x)|\le 1$ and $|f''(x)|\le 1$ hold on $[0,2]$,
$$f(a+h) = f(a) + hf'(a) + (1/2)h^2f''(a+θh),$$
some application of Taylor's Theorem (and other linked theorems) concludes that $|f'(x)|\le 2$ on $[0,2].$
It is mentioned as a hint that writing the expansions of $f(0)$ and $f(2)$ about the point $x$ in $[0,2]$ might lead on the path, but I'm sure where to go or exactly what expansion that is... Can you help me? Thanks a lot
We make the substitution $g(x) = f(x+1)$ where $x \in [-1,1]$, for sake of simplicity,
The Taylor's theorem now asserts for $0 \le c \le 1$ and $x \in [-c,c]$,
$g(c) - g(x) = (c-x)g'(x) + \frac{(c-x)^2}{2}g''(c + \theta_1(c-x))$
and, $g(-c) - g(x) = (-c-x)g'(x) + \frac{(-c-x)^2}{2}g''(-c+\theta_2(-c-x))$
Subtracting, $g(c) - g(-c) = 2cg'(x) + \frac{(c-x)^2}{2}g''(c + \theta_1(c-x))-\frac{(-c-x)^2}{2}g''(-c+\theta_2(-c-x))$
or, $|g'(x)| < \left|\frac{g(c) - g(-c)}{2c}\right|+\frac{(c-x)^2}{4c}\left|g''(c + \theta_1(c-x))\right|+\frac{(-c-x)^2}{4c}\left|g''(-c+\theta_2(-c-x))\right| \le \frac{1}{c} + c$
We used $|g| \le 1$ and $|g''| \le 1$ in the last step.
Now, the above inequality holds when $c=1$ in particular.
We have $|g'(x)| \le 2$ for $x \in [-1,1]$; ie, $|f'(x)|\le 2$ for $x \in [0,2]$.