Finding the least possible value of f(a) given the value of f(b) and that f'(x) is less than a specified c. I've tried several ways of answering this type of question but I can't seem to figure it out. This is one variation of it:
Suppose that $f(x)$ is differentiable on [38,50]. Also suppose that $f(50)=21$ and $f'(x)<=10$ for all $x\in[38,50]$. What is the least possible value that $f(38)$ can take?
I have concluded that this is rooted in the Extreme-Value theorem but I'm not totally sure how to use that here.
My attempt to answer this:
$f'(x)<=10$ -- integrates to $f(x)<=10x$
this leads me to conclude that the highest value that f(38) can take is 380. 
Which is not what the question is asking me.  
If you can help that would be splendid.

Hint: By the Mean Value Theorem,
$$\frac{f(50)-f(38)}{50-38}=f'(c)$$
for some $c$ between $38$ and $50$.
Remark: We gave a formal solution, supported by the MVT.  But the intuition goes as follows. The rate of change of $f(x)$ is $\le 10$. So from $38$ to $50$, the function $f$ can grow by at most $(10)(12)$. Thus $f(38) \ge f(50)-120$.  We can easily produce an example where $f(38)$ is exactly equal to $21-120$ by letting $f'(x)=10$ for all $x$, that is, by using a function $f(x)$ of the shape $10x+b$.
