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 equal to ____.

Let the function f: [1, ∞) → R be defined by f(t) = (-1)ⁿ⁺¹2, if t = 2n-1, n ∈ N, ((2n+1-t))/2 f(2n-1) + ((t-(2n-1)))/2 f(2n+1), if 2n-1 < t < 2n+1, n ∈ N. Define g(x) = ∫₁ˣ f(t) dt, x ∈ (1, ∞). Let α denote the number of solutions of the equation g(x) = 0 in the interval (1, 8] and β = lim_(x → 1⁺) g(x)/(x-1). Then the value of α + β is equal to ____.

$f(2n-1)=(-1)^{n+1}2$

$f \text{ is linear on } [2n-1,2n+1]$

$g(x)=\displaystyle\int_1^x f(t)\,dt$

$\alpha = \#\{x\in(1,8]: g(x)=0\}$

$\beta = \lim_{x\to1^+}\dfrac{g(x)}{x-1}$

$\text{find } \alpha+\beta$

$1.\ f \text{ on each interval (linear)}$

$2.\ g(x)=\int_1^x f,\ \text{find its zeros}$

$3.\ \beta = g'(1) = f(1)$

$f(t)=4-2t,\quad t\in[1,3]$

$f(t)=2t-8,\quad t\in[3,5]$

$f(t+4)=f(t) \Rightarrow \text{period } 4$

$g(x)=\int_1^x(4-2t)\,dt = 4x-x^2-3$

$g(2)=1,\quad g(3)=0$

$g(x)=\int_3^x(2t-8)\,dt = x^2-8x+15$

$g(4)=-1,\quad g(5)=0$

$g(x+4)=g(x)$

$g=0 \text{ at } x=3,5,7 \in (1,8]$

$\alpha = 3$

$\beta = \lim_{x\to1^+}\dfrac{g(x)}{x-1} = g'(1) = f(1) = 2$

$\alpha+\beta = 3+2 = 5$

$\boxed{5}$