We are given $f:$$R\rightarrow$$R$ differentiable in $[a,b]$ such that $f'(a)<f'(b)$. We need to prove that $\forall\beta\in$$[f'(a),f'(b)]$$\exists$$x\in$$[a,b]$ such that $f'(x)=\beta$.
This seems trivial. I have defined another function $g(x)=f(x)-\beta$$x$. It gets a minimum on $[a,b]$, but f.o.c. may not apply. Right?
The problem's guidelines tell that one must first show the minimum point cannot be on the endpoints of $[a,b]$ and I don't know why.
If you know the Weierstrass theorem and conclude the existence of minimum to $g(x)=f(x)-\beta x$, you can proceed as the guideline recommends. The left end point cannot be minimum since it is not a local minimum ($g'(a)<0$ the function is strictly decreasing). Similarly, the right end point is not the minimum by the same reason - not a local minimum ($g'(b)>0$ so strictly increasing). Therefore, the minimum must be an inner point and you can apply the Fermat theorem $g'(c)=0$. Note that you can apply Fermat only if the extremum is not on the boundary.
The cases $\beta=f'(a)$ and $\beta)=f'(b)$ are trivial, so we may suppose $f'(a)<\beta<f'(b)$.
The function $g(x)=f(x)-\beta x$ is continuous since it is differentiable, so that it attains a minimum on $[a,b]$. Actually, it can't be attained at $a$: if it were, we would have $\;f(x)-\beta x>f(a)-\beta a$ for all $x\in (a,b]$ (by definition of a minimum), so that $$\frac{f(x)-f(a)}{x-a}>\beta,\enspace\text{whence}\enspace f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\ge \beta,$$ which contradicts the hypothesis on $\beta$. Similarly the minimum can't be attained at $b$.
So the minimum of $g$ is attained at an interior point. At such a point, we have $g'(x)=0$, i.e. $f'(x)=\beta$.
Another proof, based on topology:
Let $T$ be the triangle $\;\bigl\{(x,y)\in [a,b]\times[a,b]\mid x<y\bigr\}$ and $h$ be the function defined on $T$ by $$h(x,y)=\frac{f(x)-f(y)}{x-y}.$$ $T$ is connected, since it is convex, and $h$ is continuous. Hence $\;h(T)$ is a connected subspace of $\mathbf R$, i.e. an interval.
By definition of the derivative, $f'([a,b])\subset \overline{h(T)\strut}$, and by the Mean Value Theorem, $\;h(T)\subset f'([a,b])$. Thus $$h(T)\subset f'([a,b])\subset \overline{h(T)\strut}$$ As $h(T)$ and $\overline{h(T)\strut}$ are intervals, this proves $f'(I)$ is an interval.
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f.o.c.? – Bernard Jul 18 '15 at 12:11