Show to which $A$ and $B$ the Fourier series converges uniformly I need to show that the function $f(x)$ below is continuous:
$$f(x) = \left\{\begin{matrix}
Ax + b  : -\pi \leq x<0,\\ 
cos(x) : 0\leq x \leq \pi.
\end{matrix}\right.$$
The main thing I need to show for the function to converge uniformly is that it is continuous in $[-\pi,\pi]$.
The book I am reading tells me I need to find $A$ and $B$, so $f(0^-)$ = $f(0^+)$, which I understand well. However, it also says that I need $A$,$B$ so $f(-\pi^+)$ = $f(\pi^-)$. This one I don't understand, it's in a way says the function needs to be continuous with it's next phase, like the $(\pi,3\pi]$ repetition but the sentence I know of only talks about the boundaries of $[-\pi,\pi]$.
So my question is why
$f(-\pi^+)$ = $f(\pi^-)$ is a requirement? Or why the periodic extension of $f$ needs to be continuous?

Because you need the periodic extension of $f$ to be continuous (and piecewise $C^1$.) Take into a account that the Fourier series is periodic of period $2\,\pi$, and (if convergent) the sum at $\pi$ and $-\pi$ will coincide. If the Fourier series is to converge to $f$ on $[-\pi,\pi]$, it is necessary then that $f(-\pi)=f(\pi)$.
