Prove that $f(7) = 56$ given $f(1)$ and $f(9)$ and $f' \le 6$
- Assume $f(7)>56$. What does the mean value theorem say exists somewhere in the interval $(1,7)$?
- Assume $f(7)<56$. What does the mean value theorem say exists somewhere in the interval $(7,9)$?
Conclusion: $f(7)\not>56$ and $f(7)\not<56$, so we must have $f(7)=56$.
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.
Hint: Consider $g(x)=f(x)-6x$, values and derivative, or alternatively $h(x)=6x+14-f(x)$