I read https://math.stackexchange.com/a/192440/85079. The intuition's in the final paragrah.
- I know how to prove this. But how can you presage to define $g(x) = f(x) – \alpha x$?
Unsanctioned: Officially integrals haven't been presented hence I think this is unwarranted,
but I know $\color{magenta}{\int} f'(c) \; dx = \color{magenta}{\int} \alpha \; dx \implies f(c)= \alpha + constant$.2. The answer to (a) mentions (Fermat = Interior Extremum) Theorem underneath? Does (a) follow from this theorem, ergo no need to reprove it? Why does Siminore prove (a) anew?
3. To prove $f'(a) \ge 0$ , Siminore takes $\color{darkred}{righthand}$ limit. To prove $f'(a) \ge 0$, $\color{seagreen}{\text{lefthand}}$.
Why? Can I swap? To prove $f'(a) \ge 0$, take $\color{seagreen}{\text{lefthand}}$ instead? To prove $f'(a) \ge 0$, $\color{darkred}{righthand}$?
I know $g'(x) = \lim _{h\rightarrow 0}\dfrac {g\left( x+h\right) -g\left( a\right) } {h}$ exists $\iff$ left-hand = right-hand limit.
Best Answer
I hope I understood your questions correctly.
The function $g$ is introduced just "to simplify matters" (as written above). It satisfies $g'(a)<0<g'(b)$, and the original problem is equivalent to finding som point $c$ such that $g'(c)=0$.
Siminore directly shows that the point $c$ does indeed exist. Part (a) of your exercise is intended to help you in finding where this point $c$ comes from. To show that $x$ and $y$ as above exist, use the definition of differentiability. You can write $g(x)=g(a)+(x-a)g'(a)+(x-a)\varepsilon (x)$, where $\varepsilon (x)\to 0$ as $x\to a^+$. Since $g'(a)<0$, you can find $\delta >0$ such that $g'(a)+\varepsilon(x)<0$ for every $x\in[0,\delta)$. Then $g(x)=g(a)+(x-a)(g'(a)+\varepsilon(x))<0$ for every $x\in (0,\delta)$. In the same way, you'll find that $g(y)<g(b)$ for every $y<b$ close enough to $b$.
Siminore does not talk about local min. His argument is that the function $g$ attains its maximum on $[a,b]$, and that it cannot be at $a$ or $b$. So it is attained at some $c\in (a,b)$ and hence Fermat's theorem can be used. In your setting, this is exactly the same except that you have to say that your function $g$ attains its minimum at some point $c\in [a,b]$. By part (a) of your exercise, $c$ can be equal neither to $a$ nor to $b$; so $c\in (a,b)$ and hence you can apply Fermat.