Let the function \( f : [1, \infty) \to \mathbb{R} \) be defined by:
\[
f(t) =
\begin{cases}
(-1)^{n+1}2, & \text{if } t = 2n-1, \, n \in \mathbb{N}, \\
\frac{(2n+1-t)}{2}f(2n-1) + \frac{(t-(2n-1))}{2}f(2n+1), & \text{if } 2n-1 < t < 2n+1, \, n \in \mathbb{N}.
\end{cases}
\]
Define \( g(x) = \int_1^x f(t) \, dt, \, x \in (1, \infty). \) Let \( \alpha \) denote the number of solutions of the equation \( g(x) = 0 \) in the interval \( (1, 8] \) and \( \beta = \lim_{x \to 1^+} \frac{g(x)}{x-1}. \) Then the value of \( \alpha + \beta \) is .........