Proof with theorems like mean value theorem

Question

This must be a very elementary problem, someone may asked before, I am not much exposed to math, I don't know what keyword to search it.

I'm learning from Spivak's calculus, in Chapter 11, question 28, he asked to prove:

if $f'(x)\le M$, for all $x$ in $[a,b]$, then $f(b)-f(a)\le M(b-a)$.

The answer given in the answer book is:

We have

$$\frac{f(b)-f(a)}{b-a}=f'(x) \quad for\,some\,x\,in\,(a,b) $$ $$\le M,\quad\quad\quad\quad\,\,\,$$ so $f(b)-f(a)\le M(b-a)$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Since the mean value theorem states only "for some", then for some function I can choose$\frac{f(b)-f(a)}{b-a}>f'(x)$, even though, I can't prove such case must exist, but would the possibility that such case exist weaker the proof? And more I want to know is how to make the most use of theorems contain "for some $x$"? What kind of stuff should I read as a math beginner?

Answer 1

Of course that you can have$$\frac{f(b)-f(a)}{b-a}>f'(x)\tag1$$for some other $x$. If, for instance, $f(x)=x^2$, $a=0$, and $b=1$, then we are comparing $\frac{f(b)-f(a)}{b-a}=1$ with $f'(x)=2x$. And, when $x$ goes from $0$ to $1$, we have cases in which, in fact, $(1)$ holds: any $x\in\left[0,\frac12\right)$ will do.

That doesn't change two facts:

1. There is some $x\in[a,b]$ such that $\frac{f(b)-f(a)}{b-a}=f'(x)$;
2. For each $x\in[a,b]$, $f'(x)\leqslant M$.

Puting all this together, we get that, in fact,$$\frac{f(b)-f(a)}{b-a}\leqslant M.$$