Hey I was wondering if I'm missing something in my proof or if there are any gaps of logic.. Specifically I'm not sure about my justification of the chain rule and overall accuracy..
b.
Claim: $g(x) = -g(-x)$ ($g$ is odd)
Proof:
$-g\left(-x\right)\ =\ f\left(-x\right)-f\left(x\right)\ =>g\left(-x\right)=f\left(x\right)-f\left(-x\right)$
Then we have
$g\left(x\right)=f\left(x\right)-f\left(-x\right) = -g(-x)$
Therefore $g$ is odd.
Now we show there exists a point $x_{0} ∈ R $ such that $g'(x_{0}) = 0$.
Note that by the algebra of differentiable functions $g$ is differentiable.
By (3a) we know that since $g$ is odd, we have
$\lim _{x\to \infty }\left(g\left(x\right)\right)=0 =>\:\lim _{x\to -\infty }\left(g\left(x\right)\right)=-0 = 0$
Therefore by HW10 question (5b) we know that $g$ has an extreme point in $R$, we shall denote it $x_{0}$.
Since $g$ is differentiable at $x_{0}$, and $x_{0}$ is an extreme point of $g$, by Fermat's theorem we have $g'(x_{0}) = 0$.
c.
Note that since $f$ is differentiable at $x_{0}$, and $g'$ is differetiable at $f(x_{0})$ by the chain rule we have
$g'\left(x_{0}\right)=f'\left(x_{0}\right)-\left(-f'\left(-x_{0}\right)\right)\ =f'\left(x_{0}\right)+f'\left(-x_{0}\right)=0$
$=>f'\left(x_{0}\right)+f'\left(-x_{0}\right)=0$
Which mean $f'\left(x_{0}\right)$ and $f'\left(-x_{0}\right)$ are additive inverses of each other.
Then either $f'\left(x_{0}\right)=f'\left(-x_{0}\right)=0$, in which case there exist $x_{1}∈R$ such that $f'\left(x_{1}\right)=0$.
or we have $f'\left(x_{0}\right)\cdot f'\left(-x_{0}\right)<0$, in which case by Darboux's theorem there exist a point $c∈R$ such that
$x_{0}<c<-x_{0}$ or $-x_{0}<c<x_{0}$ and $f'(c)=0$.
Therefore there must exist an $x_{1}∈R$ such that $f'\left(x_{1}\right)=0$.