Let the function $$f : [1, \infty) \to \mathbb{R}$$ be defined by

$$f(t) = \begin{cases} (-1)^{n+1} \cdot 2, & \text{if } t = 2n-1, \, n \in \mathbb{N}, \\ \frac{(2n+1-t)}{2} f(2n-1) + \frac{(t-(2n-1))}{2} f(2n+1), & \text{if } 2n-1 < t < 2n+1, \, n \in \mathbb{N}. \end{cases}$$

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 ______.

The values of $$f(t)$$ at the odd integers are obtained directly from the first line of the definition.

$$f(2n-1)=(-1)^{\,n+1}\,2 \quad\Rightarrow\quad f(1)=2,\;f(3)=-2,\;f(5)=2,\;f(7)=-2,\dots$$

Between two successive odd integers $$2n-1$$ and $$2n+1$$ the function is a straight-line joining these end-values, i.e. it is linear on every interval $$[2n-1,\,2n+1]$$.

The integral required is $$g(x)=\displaystyle\int_{1}^{x}f(t)\,dt,\;x\gt 1.$$ Because $$g'(x)=f(x)$$, the right-hand derivative of $$g$$ at $$x=1$$ equals $$f(1)$$: $$\beta=\lim_{x\to1^{+}}\frac{g(x)}{x-1}=g'(1^{+})=f(1)=2.\tag{1}$$

Next we look for the zeros of $$g(x)$$ in $$(1,8].$$ Observe first that on each interval $$[2n-1,\,2n+1]$$ the end-values of $$f$$ are equal in magnitude and opposite in sign, so the average value of $$f$$ over the full length $$2$$ is zero. Hence

$$\int_{2n-1}^{2n+1}f(t)\,dt=0 \;\Longrightarrow\; g(2n+1)=g(2n-1).$$

Starting from $$g(1)=0$$, this gives successively

$$g(3)=0,\;g(5)=0,\;g(7)=0,\dots$$

Thus $$x=3,5,7$$ are certainly zeros lying in $$(1,8].$$ To see whether any additional zeros occur inside the open sub-intervals, evaluate $$g(x)$$ on each of them.

The straight-line expression for $$f$$ here is $$f(t)=2-2(t-1).$$ Integrating,

$$g(x)=\int_{1}^{x}\!\bigl[2-2(t-1)\bigr]dt =(2x-(x-1)^2)-2 =-x^2+4x-3.$$ This quadratic satisfies $$g(1)=g(3)=0$$ and $$g(2)=1\gt0,$$ so no other root occurs in $$(1,3).$$

Here $$f(t)=-2+2(t-3).$$ Put $$y=x-3\;(0\le y\le2).$$ Then

$$g(x)=\int_{3}^{x}\!\bigl[-2+2(t-3)\bigr]dt =-2y+y^{2}=y(y-2).$$ Hence $$g(3)=g(5)=0$$ and $$g(x)\lt0$$ for $$x\in(3,5).$$ Again, no extra root.

With $$u=x-5\;(0\le u\le2)$$ one obtains $$g(x)=2u-u^{2}=u(2-u)\gt0$$ in $$(5,7),$$ and only the end-points give zeros.

Let $$v=x-7$$. The integral gives $$g(x)=-2v+v^{2}=v(v-2)\lt0$$ for $$v\in(0,1],$$ so no zero arises in $$(7,8]$$ except the already counted one at $$x=7.$$

Collecting all solutions in $$(1,8]$$:

$$x=3,\;5,\;7\quad\Rightarrow\quad\alpha=3.\tag{2}$$

Finally, from $$\eqref{1}$$ and $$\eqref{2}$$

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

Hence the required value equals 5.