TITLE: Prove that $f(7) = 56$ given $f(1)$ and $f(9)$ and $f' \le 6$
QUESTION [6 upvotes]: Let $f(x)$ be continues function at $[1,9]$ and differentiable at $(1,9)$ and also $f(1) = 20 , f(9) = 68 $ and $ |f'(x)| \le 6$ for every $x \in (1,9)$.
I need to prove that $f(7) = 56$.
I started by using the Lagrange theorem and found that there exist $ 1<c<9$ such that $f'(c) = 6$ but I'm not sure how is this relevant and how to proceed.

REPLY [3 votes]: I would like to say that, with the given conditions, the only possible function is a line, connecting $(1,20)$ and $(9,68)$.
Let's say the curve goes off below the line, at some point $(x1,y1)$. Then, it has to return to the line eventually, as it needs to get to $(9,68)$, at some point $(x2,y2)$. The derivative of $f(x)$, in the interval $(x1,x2)$, should at some point become equal to $\frac{y2−y1}{x2−x1}>6$. 
If it goes off, above the line, then we may apply a rather same reasoning.