Points of Discontinuity and Non Differentiability of Composite Function

To find the value of $c + d$, we first determine the explicit mathematical expressions for the functions $g(x)$, $f(x)$, and their composite function $(f \circ g)(x)$ on the interval $(-1, 1)$.

Step 1: Analyze the function $g(x)$

The function $g: (-1, 1) \to (-1, 1)$ is defined as: $g(x) = x - [x]$ where $[x]$ is the greatest integer less than or equal to $x$.

For $x \in (-1, 1)$, we split the domain into two intervals:

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

Thus,

$$g(x) = \begin{cases} x + 1, & -1 < x < 0 \\ x, & 0 \le x < 1 \end{cases}$$

Note that $g(x) \in [0, 1)$ for all $x \in (-1, 1)$.

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Step 2: Analyze the function $f(t)$

The function $f: (-1, 1) \to \mathbb{R}$ is defined as: $f(t) = |2t - 1| + |2t + 1|$

Since the range of $g(x)$ is contained in $[0, 1)$, we only need to evaluate $f(t)$ for $t \in [0, 1)$:

• For $t \ge 0$, we have $2t + 1 > 0$, so $|2t + 1| = 2t + 1$.
• For $0 \le t < \frac{1}{2}$, $2t - 1 < 0 \implies |2t - 1| = 1 - 2t$. $f(t) = (1 - 2t) + (2t + 1) = 2$
• For $\frac{1}{2} \le t < 1$, $2t - 1 \ge 0 \implies |2t - 1| = 2t - 1$. $f(t) = (2t - 1) + (2t + 1) = 4t$

Thus, for $t \in [0, 1)$:

$$f(t) = \begin{cases} 2, & 0 \le t < \frac{1}{2} \\ 4t, & \frac{1}{2} \le t < 1 \end{cases}$$

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

We evaluate $(f \circ g)(x) = f(g(x))$ by substituting $g(x)$ into $f$:

1. For $x \in \left(-1, -\frac{1}{2}\right)$: $g(x) = x + 1 \in \left(0, \frac{1}{2}\right) \implies (f \circ g)(x) = 2$

2. For $x \in \left[-\frac{1}{2}, 0\right)$: $g(x) = x + 1 \in \left[\frac{1}{2}, 1\right) \implies (f \circ g)(x) = 4(x + 1) = 4x + 4$

3. For $x \in \left[0, \frac{1}{2}\right)$: $g(x) = x \in \left[0, \frac{1}{2}\right) \implies (f \circ g)(x) = 2$

4. For $x \in \left[\frac{1}{2}, 1\right)$: $g(x) = x \in \left[\frac{1}{2}, 1\right) \implies (f \circ g)(x) = 4x$

Writing $(f \circ g)(x)$ piecewise over $(-1, 1)$:

$$(f \circ g)(x) = \begin{cases} 2, & -1 < x < -\frac{1}{2} \\ 4x + 4, & -\frac{1}{2} \le x < 0 \\ 2, & 0 \le x < \frac{1}{2} \\ 4x, & \frac{1}{2} \le x < 1 \end{cases}$$

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Step 4: Determine points of discontinuity ($c$)

The potential points of discontinuity in $(-1, 1)$ are $x = -\frac{1}{2}$, $x = 0$, and $x = \frac{1}{2}$.

1. At $x = -\frac{1}{2}$: $\lim_{x \to -\frac{1}{2}^-} (f \circ g)(x) = 2$ $\lim_{x \to -\frac{1}{2}^+} (f \circ g)(x) = 4\left(-\frac{1}{2}\right) + 4 = 2$ $(f \circ g)\left(-\frac{1}{2}\right) = 2$ Since $\text{LHL} = \text{RHL} = (f \circ g)\left(-\frac{1}{2}\right)$, $(f \circ g)$ is continuous at $x = -\frac{1}{2}$.

2. At $x = 0$: $\lim_{x \to 0^-} (f \circ g)(x) = 4(0) + 4 = 4$ $\lim_{x \to 0^+} (f \circ g)(x) = 2$ Since $\text{LHL} \neq \text{RHL}$, $(f \circ g)$ is discontinuous at $x = 0$.

3. At $x = \frac{1}{2}$: $\lim_{x \to \frac{1}{2}^-} (f \circ g)(x) = 2$ $\lim_{x \to \frac{1}{2}^+} (f \circ g)(x) = 4\left(\frac{1}{2}\right) = 2$ $(f \circ g)\left(\frac{1}{2}\right) = 2$ Since $\text{LHL} = \text{RHL} = (f \circ g)\left(\frac{1}{2}\right)$, $(f \circ g)$ is continuous at $x = \frac{1}{2}$.

Therefore, the number of points of discontinuity is: $c = 1 \quad (\text{at } x = 0)$

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Step 5: Determine points of non-differentiability ($d$)

Since $(f \circ g)(x)$ is discontinuous at $x = 0$, it is automatically not differentiable at $x = 0$.

Now, check the derivatives at the continuous transition points $x = -\frac{1}{2}$ and $x = \frac{1}{2}$:

1. At $x = -\frac{1}{2}$: $\text{LHD} = \left.\frac{d}{dx}(2)\right|_{x \to -\frac{1}{2}^-} = 0$ $\text{RHD} = \left.\frac{d}{dx}(4x + 4)\right|_{x \to -\frac{1}{2}^+} = 4$ Since $\text{LHD} \neq \text{RHD}$, $(f \circ g)$ is not differentiable at $x = -\frac{1}{2}$.

2. At $x = \frac{1}{2}$: $\text{LHD} = \left.\frac{d}{dx}(2)\right|_{x \to \frac{1}{2}^-} = 0$ $\text{RHD} = \left.\frac{d}{dx}(4x)\right|_{x \to \frac{1}{2}^+} = 4$ Since $\text{LHD} \neq \text{RHD}$, $(f \circ g)$ is not differentiable at $x = \frac{1}{2}$.

All other points inside the open intervals $\left(-1, -\frac{1}{2}\right)$, $\left(-\frac{1}{2}, 0\right)$, $\left(0, \frac{1}{2}\right)$, and $\left(\frac{1}{2}, 1\right)$ are differentiable since the function is linear on those open intervals.

Therefore, the points of non-differentiability are $x \in \left\{-\frac{1}{2}, 0, \frac{1}{2}\right\}$, giving: $d = 3$

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Conclusion

$c + d = 1 + 3 = 4$