Calculate ${f^{\ast}}'(0)$ if $f^{\ast}$ is an odd expansion of the function f which is defined on the interval $0\leq x\leq L$

Question

Imagine that the function $f(x)$ is defined on the interval $0\leq x\leq L$. $f$ is defined and continuous on $[0,L]$ and differentiable on $(0,L)$. also $f(0)=f(L)=0$ (The function $f$ is the initial displacement of a string of length $L$ whose endpoints are fastened relative to the $x-$ axis ) If we denote the odd expansion of this function by $f^{\ast}(x)$ then $f^{\ast}\left({x}\right)=f(x)$ for $0\leq x\leq L$ and $f^{\ast}\left({x}\right)=-f(-x)$ for $-L\leq x<0$. I know that ${f^{\ast}}'\left({x}\right)=f'(x)$ for ${0 < x < L}$ and ${f^{\ast}}'\left({x}\right)=f'(-x)$ for ${-L < x < 0}$ but what is ${f^{\ast}}'\left({0}\right)$ ?is function ${f^{\ast}}'$ continuous?