Step-by-step Solution:

1. Analyze the function f(x)

The function $f(x)$ is defined as $f(x) = |2x - 1| + |2x + 1|$ on the interval $(-1, 1)$. We can write this as a piecewise function by considering the points where the arguments of the absolute value functions are zero, which are $x = 1/2$ and $x = -1/2$.

• For $x \in (-1, -1/2)$: $2x - 1 < 0$ and $2x + 1 < 0$. $f(x) = -(2x - 1) - (2x + 1) = -2x + 1 - 2x - 1 = -4x$.
• For $x \in [-1/2, 1/2]$: $2x - 1 \le 0$ and $2x + 1 \ge 0$. $f(x) = -(2x - 1) + (2x + 1) = -2x + 1 + 2x + 1 = 2$.
• For $x \in (1/2, 1)$: $2x - 1 > 0$ and $2x + 1 > 0$. $f(x) = (2x - 1) + (2x + 1) = 4x$.

So, the piecewise definition of $f(x)$ is: $f(x) = \begin{cases} -4x & \text{if } -1 < x < -1/2 \\ 2 & \text{if } -1/2 \le x \le 1/2 \\ 4x & \text{if } 1/2 < x < 1 \end{cases}$

The function $f(x)$ is continuous on $(-1, 1)$. It is not differentiable at $x = -1/2$ and $x = 1/2$ because the left and right derivatives do not match at these points.

2. Analyze the function g(x)

The function $g(x)$ is defined as $g(x) = x - [x]$, which is the fractional part function, on the interval $(-1, 1)$. We can write this as a piecewise function.

• For $x \in (-1, 0)$: $[x] = -1$, so $g(x) = x - (-1) = x + 1$.
• For $x \in [0, 1)$: $[x] = 0$, so $g(x) = x - 0 = x$.

So, the piecewise definition of $g(x)$ is: $g(x) = \begin{cases} x + 1 & \text{if } -1 < x < 0 \\ x & \text{if } 0 \le x < 1 \end{cases}$ The range of $g(x)$ on $(-1, 1)$ is $[0, 1)$. The function $g(x)$ is discontinuous at $x = 0$ because $\lim_{x \to 0^-} g(x) = 1$ while $g(0) = 0$.

3. Construct the composite function $(f \circ g)(x)$

Let $h(x) = (f \circ g)(x) = f(g(x))$. The domain of $h(x)$ is $(-1, 1)$. Since the range of $g(x)$ is $[0, 1)$, we only need the definition of $f(y)$ for $y \in [0, 1)$. From Step 1, for $y \in [0, 1)$: $f(y) = \begin{cases} 2 & \text{if } 0 \le y \le 1/2 \\ 4y & \text{if } 1/2 < y < 1 \end{cases}$ Now we substitute $y = g(x)$:

• For $x \in (-1, 0)$: $g(x) = x + 1$. The range of $g(x)$ is $(0, 1)$.
  • If $0 < g(x) \le 1/2$, i.e., $0 < x+1 \le 1/2$, which means $-1 < x \le -1/2$. In this case, $h(x) = f(g(x)) = 2$.
  • If $1/2 < g(x) < 1$, i.e., $1/2 < x+1 < 1$, which means $-1/2 < x < 0$. In this case, $h(x) = f(g(x)) = 4g(x) = 4(x+1)$.
• For $x \in [0, 1)$: $g(x) = x$. The range of $g(x)$ is $[0, 1)$.
  • If $0 \le g(x) \le 1/2$, i.e., $0 \le x \le 1/2$. In this case, $h(x) = f(g(x)) = 2$.
  • If $1/2 < g(x) < 1$, i.e., $1/2 < x < 1$. In this case, $h(x) = f(g(x)) = 4g(x) = 4x$.

Combining these, we get the piecewise definition for $h(x) = (f \circ g)(x)$: $h(x) = \begin{cases} 2 & \text{if } -1 < x \le -1/2 \\ 4(x+1) & \text{if } -1/2 < x < 0 \\ 2 & \text{if } 0 \le x \le 1/2 \\ 4x & \text{if } 1/2 < x < 1 \end{cases}$

4. Find the number of points of non-continuity (c)

We check for continuity at the points where the definition of $h(x)$ changes: $x = -1/2, 0, 1/2$.

• At $x = -1/2$: $\lim_{x \to -1/2^-} h(x) = \lim_{x \to -1/2^-} 2 = 2$. $\lim_{x \to -1/2^+} h(x) = \lim_{x \to -1/2^+} 4(x+1) = 4(-1/2 + 1) = 2$. $h(-1/2) = 2$. Since the limits and the function value are equal, $h(x)$ is continuous at $x = -1/2$.
• At $x = 0$: $\lim_{x \to 0^-} h(x) = \lim_{x \to 0^-} 4(x+1) = 4(0+1) = 4$. $\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} 2 = 2$. Since the left-hand limit and right-hand limit are not equal, $h(x)$ is NOT continuous at $x = 0$.
• At $x = 1/2$: $\lim_{x \to 1/2^-} h(x) = \lim_{x \to 1/2^-} 2 = 2$. $\lim_{x \to 1/2^+} h(x) = \lim_{x \to 1/2^+} 4x = 4(1/2) = 2$. $h(1/2) = 2$. So, $h(x)$ is continuous at $x = 1/2$.

There is only one point of discontinuity in $(-1, 1)$, which is $x = 0$. Therefore, $c = 1$.

5. Find the number of points of non-differentiability (d)

A function is not differentiable at points where it is not continuous, or where the left-hand and right-hand derivatives are not equal (sharp corners).

• At $x = 0$: Since $h(x)$ is not continuous at $x=0$, it is not differentiable at $x=0$.
• At $x = -1/2$: We check the left-hand derivative (LHD) and right-hand derivative (RHD). LHD at $x=-1/2$: For $x < -1/2$, $h(x) = 2$, so $h'(x) = 0$. LHD = 0. RHD at $x=-1/2$: For $x > -1/2$, $h(x) = 4(x+1)$, so $h'(x) = 4$. RHD = 4. Since LHD $\neq$ RHD, $h(x)$ is NOT differentiable at $x = -1/2$.
• At $x = 1/2$: We check the LHD and RHD. LHD at $x=1/2$: For $x < 1/2$ (and $x>0$), $h(x) = 2$, so $h'(x) = 0$. LHD = 0. RHD at $x=1/2$: For $x > 1/2$, $h(x) = 4x$, so $h'(x) = 4$. RHD = 4. Since LHD $\neq$ RHD, $h(x)$ is NOT differentiable at $x = 1/2$.

The points in $(-1, 1)$ where $h(x)$ is not differentiable are $x=0, x=-1/2,$ and $x=1/2$. There are 3 such points. Therefore, $d = 3$.

6. Calculate c + d

We have found that $c=1$ and $d=3$. The value of $c + d = 1 + 3 = 4$.

Final Answer: The value of $c+d$ is 4.