Hi I was wondering if my proof makes sense or if I have some flaws in logic or mistakes I need to take care of..
Question 5.
a.
$f\:'\left(T\right)=\lim \:_{h\to \:o}\frac{f\left(T+h\right)-f\left(T\right)}{h}=\lim\: _{h\to o}\frac{f\left(0+h\right)-f\left(0\right)}{h}=f\:'\left(0\right)$
b.
Note that since $f$ is differentiable on $[0,T]$, it is also continuous on $[0,T]$.
We know that $ f(0)=f(T)$ for every $x∈R$, therefore by Rolle's theorem there exist a point $0 < x_{1} < T $ such that $f'(x_{1}) = 0$.
Note that $f$ is also differential and continuous on $[x_{1},x_{1}+T]$, therefore since we have $f(x_{1}) = f(x_{1}+T)$
by Rolle's theorem there exist a point $x_{1} < x_{2} < (x_{1}+T)$ such that $f'(x_{2})=0$.
We have two cases, $0 < x_{1} < x_{2} ≤ T < (x_{1}+T)$ or $0 < x_{1} < T < x_{2} < (x_{1}+T)$
Case 1: $x_{1} < x_{2} ≤ T$
In this case $x_{2}∈[0,T]$ so we have $f'(x_{2})=f'(x_{1})=0$ such that $x_{1} ≠ x_{2}$
Case 2: $T < x_{2} < (x_{1}+T)$
In this case $f(x_{2}) = f(x_{2}-T) = 0$, and $0 < (x_{2}-T) < x_{1} < T$
Therefore we have $(x_{2}-T) ∈ [0,T]$ and $f(x_{2}-T) = f(x_{1}) = 0$ such that $(x_{2}-T)≠x_{1}$.