Points of Discontinuity and Non Differentiability of Composite Absolute Function

To find the value of $\alpha + \beta$, we first analyze the given piecewise function $f(x)$ and then express $g(x) = f(|x|) + |f(x)|$ across different intervals.

The function $f(x)$ is defined as: $f(x) = \begin{cases} e^{x-1} &, x < 0 \\ x^2 - 5x + 6 &, x \ge 0 \end{cases}$

Step 1: Evaluating $f(|x|)$

Since $|x| \ge 0$ for all $x \in \mathbb{R}$, $f(|x|)$ always evaluates using the branch for $x \ge 0$: $f(|x|) = (|x|)^2 - 5|x| + 6 = x^2 - 5|x| + 6$

Writing this explicitly for $x < 0$ and $x \ge 0$: $f(|x|) = \begin{cases} x^2 + 5x + 6 &, x < 0 \\ x^2 - 5x + 6 &, x \ge 0 \end{cases}$

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Step 2: Evaluating $|f(x)|$

We analyze the sign of $f(x)$ in different regions:

1. For $x < 0$: $f(x) = e^{x-1} > 0 \implies |f(x)| = e^{x-1}$

2. For $x \ge 0$: $f(x) = x^2 - 5x + 6 = (x-2)(x-3)$

   • If $0 \le x < 2$, $f(x) > 0 \implies |f(x)| = x^2 - 5x + 6$
   • If $2 \le x \le 3$, $f(x) \le 0 \implies |f(x)| = -(x^2 - 5x + 6)$
   • If $x > 3$, $f(x) > 0 \implies |f(x)| = x^2 - 5x + 6$

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Step 3: Explicit Definition of $g(x) = f(|x|) + |f(x)|$

By adding $f(|x|)$ and $|f(x)|$ in their respective intervals, we get:

1. For $x < 0$: $g(x) = (x^2 + 5x + 6) + e^{x-1}$

2. For $0 \le x < 2$: $g(x) = (x^2 - 5x + 6) + (x^2 - 5x + 6) = 2(x^2 - 5x + 6)$

3. For $2 \le x \le 3$: $g(x) = (x^2 - 5x + 6) - (x^2 - 5x + 6) = 0$

4. For $x > 3$: $g(x) = (x^2 - 5x + 6) + (x^2 - 5x + 6) = 2(x^2 - 5x + 6)$

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Step 4: Points of Discontinuity ($\alpha$)

We test continuity of $g(x)$ at the boundary points $x = 0$, $x = 2$, and $x = 3$:

1. At $x = 0$: $\lim_{x \to 0^-} g(x) = 0^2 + 5(0) + 6 + e^{-1} = 6 + \frac{1}{e}$ $g(0) = \lim_{x \to 0^+} g(x) = 2(0^2 - 5(0) + 6) = 12$ Since $\lim_{x \to 0^-} g(x) \ne g(0)$, $g(x)$ is discontinuous at $x = 0$.

2. At $x = 2$: $\lim_{x \to 2^-} g(x) = 2(2^2 - 5(2) + 6) = 0$ $g(2) = \lim_{x \to 2^+} g(x) = 0$ $g(x)$ is continuous at $x = 2$.

3. At $x = 3$: $\lim_{x \to 3^-} g(x) = 0$ $g(3) = \lim_{x \to 3^+} g(x) = 2(3^2 - 5(3) + 6) = 0$ $g(x)$ is continuous at $x = 3$.

Thus, $g(x)$ is discontinuous at only 1 point ($x = 0$). $\alpha = 1$

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Step 5: Points of Non-Differentiability ($\beta$)

$g(x)$ is smooth within each open interval, so non-differentiability can only occur at boundary points $x = 0, 2, 3$:

1. At $x = 0$: Since $g(x)$ is discontinuous at $x = 0$, it is not differentiable at $x = 0$.

2. At $x = 2$:

   • Left-hand derivative: $g'(2^-) = \left.\frac{d}{dx}(2x^2 - 10x + 12)\right|_{x=2} = 4(2) - 10 = -2$
   • Right-hand derivative: $g'(2^+) = \left.\frac{d}{dx}(0)\right|_{x=2} = 0$ Since $g'(2^-) \ne g'(2^+)$, $g(x)$ is not differentiable at $x = 2$.

3. At $x = 3$:

   • Left-hand derivative: $g'(3^-) = \left.\frac{d}{dx}(0)\right|_{x=3} = 0$
   • Right-hand derivative: $g'(3^+) = \left.\frac{d}{dx}(2x^2 - 10x + 12)\right|_{x=3} = 4(3) - 10 = 2$ Since $g'(3^-) \ne g'(3^+)$, $g(x)$ is not differentiable at $x = 3$.

Thus, $g(x)$ is not differentiable at 3 points ($x = 0, 2, 3$). $\beta = 3$

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Conclusion

$\alpha + \beta = 1 + 3 = 4$