Let the function $f:[1,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathbb{R}$ be defined by

$f(t)=\{\begin{array}{cc}(-<!--\; -\; -->1{)}^{n+1}2,& \text{ if }t=2n-<!--\; -\; -->1,n\in <!--\; \in \; -->\mathbb{N}\\ \frac{(2n+1-<!--\; -\; -->t)}{2}f(2n-<!--\; -\; -->1)+\frac{(t-<!--\; -\; -->(2n-<!--\; -\; -->1))}{2}f(2n+1),& \text{ if }2n-<!--\; -\; -->1<t<2n+1,n\in <!--\; \in \; -->\mathbb{N}\end{array}$

Define $g(x)=\underset{1}{\overset{x}{\int <!--\; \int \; -->}}f(t)dt,x\in <!--\; \in \; -->(1,\mathrm{\infty <!--\; \infty \; -->})$. Let α denote the number of solutions of the equation g(x) = 0 in the interval $(1,8]$ and $\mathrm{\beta <!--\; \beta \; -->}=\underset{x\to <!--\; \to \; -->1+}{lim}\frac{g(x)}{x-<!--\; -\; -->1}$. Then the value of α + β is equal to ______.

First, let us analyze the definition of the function $f:[1,\mathrm{\infty })\to \mathbb{R}$.

For odd integers of the form $t=2n-1$ where $n\in \mathbb{N}$, we have:
$f(2n-1)=(-1{)}^{n+1}2$

Evaluating this at consecutive values of $n$ gives:
- For $n=1$: $t=1\Rightarrow f(1)=(-1{)}^2\cdot 2=2$
- For $n=2$: $t=3\Rightarrow f(3)=(-1{)}^3\cdot 2=-2$
- For $n=3$: $t=5\Rightarrow f(5)=(-1{)}^4\cdot 2=2$
- For $n=4$: $t=7\Rightarrow f(7)=(���1{)}^5\cdot 2=-2$

For open intervals $2n-1<t<2n+1$, the function is defined as:
$f(t)=\frac{2n+1-t}{2}f(2n-1)+\frac{t-(2n-1)}{2}f(2n+1)$

This expression represents the linear interpolation between the points $(2n-1,f(2n-1))$ and $(2n+1,f(2n+1))$. Thus, the function $f(t)$ is a continuous, piecewise linear (sawtooth-like) function connecting the vertices:
$(1,2),(3,-2),(5,2),(7,-2),(9,2),\dots$

Let us determine the value of $\beta$:
$\beta =\underset{x\to 1^{+}}{lim}\frac{g(x)}{x-1}$

Since $g(1)={\int }_1^{1}f(t)dt=0$, by the definition of the derivative and the Fundamental Theorem of Calculus:
$\beta =g^{\prime }(1^{+})=f(1)=2$

Next, let us analyze the equation $g(x)=0$ for $x\in (1,8]$.

1. For the interval $t\in [1,3]$:
The line segment connecting $(1,2)$ and $(3,-2)$ is given by:
$f(t)=2-2(t-1)=4-2t$
Integrating to find $g(x)$ on this interval:
$g(x)={\int }_1^x(4-2t)dt={[4t-t^2]}_1^x=-x^2+4x-3$
Setting $g(x)=0\Rightarrow -(x-1)(x-3)=0$.
Since $x\in (1,8]$, this yields one solution: $x=3$.
Note that $g(3)=0$.

2. For the interval $t\in [3,5]$:
The line segment connecting $(3,-2)$ and $(5,2)$ is:
$f(t)=-2+2(t-3)=2t-8$
Integrating for $x\in [3,5]$:
$g(x)=g(3)+{\int }_3^x(2t-8)dt=0+{[t^2-8)]}_3^x=x^2-8x+15$
Setting $g(x)=0\Rightarrow (x-3)(x-5)=0$.
Since $x>3$ in this sub-interval, this yields the solution: $x=5$.
Note that $g(5)=0$.

3. For the interval $t\in [5,7]$:
Due to symmetry and periodicity of the sawtooth function (shifted by 4 units), the behavior of $g(x)$ on $[5,7]$ is identical to $[1,3]$ with $g(5)=0$:
$g(x)=0\Rightarrow x=7$.
Note that $g(7)=0$.

4. For the interval $t\in [7,8]$:
Since $f(t)$ is linear from $-2$ at $t=7$ to $2$ at $t=9$, we have $f(t)<0$ for all $t\in [7,8)$ and $f(8)=0$.
Thus, the integral $g(x)=g(7)+{\int }_7^{x}f(t)dt={\int }_7^{x}f(t)dt$ is strictly decreasing and negative for $x\in (7,8]$.
Therefore, there are no solutions to $g(x)=0$ in the interval $(7,8]$.

Combining the solutions in the interval $(1,8]$, we find exactly three values: $x=3$, $x=5$, and $x=7$.
Thus, the number of solutions is:
$\alpha =3$

Finally, we calculate the sum $\alpha +\beta$:
$\alpha +\beta =3+2=5$