Let f(x) = begin{cases} e^{x-1}; x

Question

(D) Given $f(x) = e^{x-1}$ for $x < 0$ and $f(x) = x^2 - 5x + 6$ for $x \ge 0$. At $x = 0$, $f(0^-) = e^{-1} \approx 0.368$ and $f(0^+) = 0^2 - 5(0) +

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$4$

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(D) Given $f(x) = e^{x-1}$ for $x At $x = 0$, $f(0^-) = e^{-1} \approx 0.368$ and $f(0^+) = 0^2 - 5(0) + 6 = 6$. Since $f(0^-) 
eq f(0^+)$, $f(x)$ is discontinuous at $x = 0$. Now, $g(x) = f(|x|) + |f(x)|$. For $x For $x \ge 0$, $g(x) = f(x) + |f(x)|$. If $f(x) \ge 0$ (i.e.,$x \in [0, 2] \cup [3, \infty)$), $g(x) = 2f(x) = 2(x^2 - 5x + 6)$. If $f(x) Checking continuity: At $x = 0$, $g(0^-) = 0^2 + 5(0) + 6 + e^{-1} = 6 + e^{-1}$ and $g(0^+) = 2(6) = 12$. Since $6 + e^{-1} 
eq 12$, $g(x)$ is discontinuous at $x = 0$. Thus, $\alpha = 1$. Checking differentiability: $g(x)$ is not differentiable at $x = 0$ (discontinuity). For $x > 0$, $g(x)$ is defined as $2(x^2 - 5x + 6)$ for $x \in [0, 2] \cup [3, \infty)$ and $0$ for $x \in (2, 3)$. At $x = 2$, $g(2^-) = 2(4 - 10 + 6) = 0$ and $g(2^+) = 0$. $g'(2^-) = 2(2x - 5)|_{x=2} = 2(4-5) = -2$, while $g'(2^+) = 0$. Not differentiable at $x = 2$. At $x = 3$, $g(3^-) = 0$ and $g(3^+) = 2(9 - 15 + 6) = 0$. $g'(3^-) = 0$, while $g'(3^+) = 2(2x - 5)|_{x=3} = 2(6-5) = 2$. Not differentiable at $x = 3$. Thus, $g(x)$ is not differentiable at $x = 0, 2, 3$. So $\beta = 3$. Therefore, $\alpha + \beta = 1 + 3 = 4$.

Let $f(x) = \begin{cases} e^{x-1}; x < 0 \\ x^2-5x+6; x \ge 0 \end{cases}$ and $g(x) = f(|x|) + |f(x)|$. If the number of points where $g$ is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha + \beta$ is equal to ————

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