Answer
$12$
Work Step by Step
Given $f’(x)\leq 2$ for all $x$ including the interval of $[0,6]$, the local increase for a starting point $x_i$ can be written as $\Delta f_i=f’(x_i)\Delta x_i$ and the total increase in the interval is the sum of all the local increases. We have $\Delta f_{[0,6]}=\sum \Delta f_i=\sum f’(x_i)\Delta x_i\leq 2\sum \Delta x_i=2(6-0)=12$. Thus the maximum increase of the function in the interval is $12$.
