Those problems come with my proof of question. I already found a better solution for this question, but there exists some confusion in the first proof occur to my head
Original Question
f(x) is derivable on [a, b], and $f^{'}(a)f^{'}(b)<0$. Prove that there exists at least one point $\xi$ that satisfies $f^{'}(\xi)=0$
Here is my proof,
Suppose that f(x) has no local extrema on (a, b).
Thus, f(x) is monotonic on (a, b), /* Question 1: Is this true? */
Therefore, $f^{'}(a)f^{'}(b)>0$, /* Question 2: Is this true? */
which contradicts the fact that $f^{'}(a)f^{'}(b)<0$.
Then, there exists at least one extremum on (a, b) and that point is $\xi$. Thus, $f^{'}(\xi)=0$.
Overall, my question is that if a function is derivable on [a, b] and monotonic on (a, b), can we say anything about the left or right hand derivatives at the end point?
Suppose f(x) is derivable on [a, b] and monotonically decreasing on (a, b).
Can we say that always f(x) < f(a) ($a\le x \le b$) and use the formula,
$\lim {\frac {f(x) - f(a)}{x-a}} = f^{'}(a) < 0$ ( x-> $a^{+}$ ).
If you know the intermediate value theorem, then the proof is very easy. Just observe that if $f'(a)*f'(b)<0$, at one endpoint the derivative is positive and on the other endpoint it is negative.
Concerning your proof:
If $f$ is differentiable and monotonic, then the derivative $f'$ does not change its sign. If $f$ is also stric monotonic, then also $f' \ne 0$ and thus $f'(a)*f'(b)>0$. Additionally, if $f$ has no local extremum, then $f$ is stric monotonic.
Now assume, that $f$ is differentable on $[a,b]$ and monotonically decreasing on $(a,b)$. Then $$ \lim_{x \searrow a} \frac {f(x)-f(a)}{x-a}\le0$$ because $f(x)<f(a)$ if $x>a$ and the limit exists per assumption.