Let $$f(x) = \begin{cases} \{4x^2 - 8x + 5\}, & \text{if } 8x^2 - 6x + 1 \ge 0 \\ [4x^2 - 8x + 5], & \text{if } 8x^2 - 6x + 1 < 0 \end{cases}$$, where $$[\alpha]$$ denotes the greatest integer less than or equal to $$\alpha$$ . Then the number of points in $$\mathbb{R}$$ where $$f$$ is not differentiable is ______.

We have $$f(x) = \begin{cases} \{4x^2 - 8x + 5\}, & \text{if } 8x^2 - 6x + 1 \ge 0 \\ [4x^2 - 8x + 5], & \text{if } 8x^2 - 6x + 1 < 0 \end{cases}$$ where $$\{\alpha\}$$ is the fractional part and $$[\alpha]$$ is the greatest integer function.

Observe that $$g(x) = 4x^2 - 8x + 5 = 4(x-1)^2 + 1$$.

Since $$g$$ attains its minimum at $$x = 1$$, this minimum value is $$g(1) = 1$$.

Moreover, $$g$$ is symmetric about $$x = 1$$.

To determine the intervals where each branch of $$f$$ applies, we solve $$8x^2 - 6x + 1 = 0$$.

$$8x^2 - 6x + 1 = (2x - 1)(4x - 1) = 0 \implies x = \dfrac{1}{4} \text{ or } x = \dfrac{1}{2}$$

It follows that $$8x^2 - 6x + 1 < 0$$ on $$\left(\dfrac{1}{4},\, \dfrac{1}{2}\right)$$ and is non-negative elsewhere.

Hence $$f(x) = \{g(x)\}$$ outside $$\left(\dfrac{1}{4},\, \dfrac{1}{2}\right)$$ and $$f(x) = [g(x)]$$ on $$\left(\dfrac{1}{4},\, \dfrac{1}{2}\right)$$.

In particular, evaluating $$g$$ at the endpoints of $$\left(\dfrac{1}{4},\, \dfrac{1}{2}\right)$$ yields:

$$g\!\left(\dfrac{1}{4}\right) = 4\!\left(\dfrac{3}{4}\right)^2 + 1 = 4 \cdot \dfrac{9}{16} + 1 = \dfrac{13}{4} = 3.25$$

$$g\!\left(\dfrac{1}{2}\right) = 4\!\left(\dfrac{1}{2}\right)^2 + 1 = 4 \cdot \dfrac{1}{4} + 1 = 2$$

Thus, on $$\left(\dfrac{1}{4},\, \dfrac{1}{2}\right)$$, $$g$$ decreases from $$3.25$$ to $$2$$.

Within this interval, the value of $$[g(x)]$$ depends on whether $$g(x)$$ lies in $$[3,3.25)$$ or $$[2,3)$$.

In particular, $$[g(x)]$$ equals $$3$$ when $$g(x)\in[3,3.25)$$ and equals $$2$$ when $$g(x)\in[2,3)$$.

Solving $$g(x)=3\implies (x-1)^2=\dfrac{1}{2}\implies x=1-\dfrac{1}{\sqrt{2}}\approx 0.293$$.

Since this value lies in $$\left(\dfrac{1}{4},\, \dfrac{1}{2}\right)$$, the floor function jumps from $$3$$ to $$2$$ at that point.

Next, examining the endpoints of $$\left(\dfrac{1}{4},\, \dfrac{1}{2}\right)$$, we find the following left and right limits of $$f(x)$$.

At $$x = \dfrac{1}{4}$$: from the left, $$f = \{g(1/4)\} = \{3.25\} = 0.25$$; from the right, $$f = [g(1/4^+)] = 3$$. There is a jump discontinuity.

At $$x = \dfrac{1}{2}$$: from the left, $$f = [g(1/2^-)] = 2$$; from the right, $$f = \{g(1/2)\} = \{2\} = 0$$. There is a jump discontinuity.

Therefore, the non-differentiable points occur at

(1) $$x = \dfrac{1}{4}$$: discontinuity between the two branches.

(2) $$x = 1 - \dfrac{1}{\sqrt{2}}$$: floor function jumps from $$3$$ to $$2$$.

(3) $$x = \dfrac{1}{2}$$: discontinuity between the two branches.

Thus, the answer is $$3$$.