If $f\left(x\right)=x^{\alpha }\log x$ and $f\left(0\right)=0$, then the value of $\alpha$ for which Rolle's theorem can be applied in $[0, 1]$ is:

Let $g\left(x\right)$ be a polynomial of degree one and $f\left(x\right)$ be a function defined by $f\left(x\right)=\left\{\begin{array}{ll}g\left(x\right),& x\le 0\\ {\left(\frac{1+x}{2+x}\right)}^{\frac{1}{x}},& x>0\end{array}\right.$.

If $f\left(x\right)$ is continuous at $x=0$ and $f(-1)=f\text{'}\left(1\right)$, then $g\left(x\right)$ is equal to