Let be $f:\mathbb{R}\to\mathbb{R}$ given by an even and $2\pi$-periodic continuation of$$f(x):=\begin{cases}4x^2-\pi^2,&x\in\left[0,\frac{\pi}{2}\right]\\8\pi x-3\pi^2-4x^2,&x\in\left[\frac{\pi}{2},\pi\right].\end{cases}$$Show that $f$ is continously differentiable.
My approach:
As $f(-x)=f(x)$ the even and $2\pi$-periodic continuation of $f$ on $[-\pi,\pi]$ attains the form of$$f(x):=\begin{cases}-8\pi x-3\pi^2-4x^2,&x\in\left[-\pi,-\frac{\pi}{2}\right]\\4x^2-\pi^2,&x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\\8\pi x-3\pi^2-4x^2,&x\in\left[\frac{\pi}{2},\pi\right].\end{cases}$$
We define $I_1:=\left[-\pi,-\frac{\pi}{2}\right]$ , $I_2:=\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and $I_3:=\left[\frac{\pi}{2},\pi\right]$ then each restriction $f_{|I_1}$, $f_{|I_2}$, $f_{|I_3}$is obviously continously differentiable.
We see that $f'_{|I_1}(x)=-8\pi-8x$, $f'_{|I_2}(x)=8x$ and $f'_{|I_3}(x)=8\pi-8x$. Since, \begin{align*}&f'_{|I_1}\left(-\frac{\pi}{2}\right)=-4\pi=f'_{|I_2}\left(-\frac{\pi}{2}\right)\\&f'_{|I_2}\left(\frac{\pi}{2}\right)=4\pi=f'_{|I_3}\left(\frac{\pi}{2}\right)\\&f'_{|I_3}\left(\pi\right)=0=f'_{|I_1}\left(-\pi\right)\\\end{align*}we get\begin{align*}&f'(x):=\begin{cases}-8\pi-8x,&x\in\left[-\pi,-\frac{\pi}{2}\right]\\8x,&x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\\8\pi-8x,&x\in\left[\frac{\pi}{2},\pi\right].\end{cases}\end{align*}S0 $f$ is continously differentiable on $[-\pi,\pi]$ and due to $2\pi$-periodicity on $\mathbb{R}$.
I am not sure if my continuation of $f$ is correct and if I can show the differentiability of $f$ this way?
