Let f(x)= cases e^ x-1 , & x<0 x^2-5x+6, & x ≥ 0 cases and...

We are given the function:

$f(x) =$\begin{cases} e^{x-1}, & x < 0 \\ x^2 - 5x + 6, & x \geq 0 \end{cases}<br><br>We need to analyze the continuity and differentiability of $g(x) = f(|x|) + |f(x)|$.<br><br>For $x < 0$, $|x| = -x > 0$. Thus, $f(|x|) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6$.<br>Also, for $x < 0$, $f(x) = e^{x-1} > 0$, so $|f(x)| = e^{x-1}$.<br>Therefore, for $x < 0$, $g(x) = x^2 + 5x + 6 + e^{x-1}$.<br><br>For $x \geq 0$, $|x| = x$. Thus, $f(|x|) = f(x) = x^2 - 5x + 6$.<br>Therefore, for $x \geq 0$, $g(x) = x^2 - 5x + 6 + |x^2 - 5x + 6|$.<br><br>Let us check the continuity of $g(x)$ at $x = 0$:<br>$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (x^2 + 5x + 6 + e^{x-1}) = 6 + \dfrac{1}{e}$<br>$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (x^2 - 5x + 6 + |x^2 - 5x + 6|) = 6 + 6 = 12$<br>$g(0) = 12$<br><br>Since $\lim_{x \to 0^-} g(x) \neq \lim_{x \to 0^+} g(x)$, $g(x)$ is discontinuous at $x = 0$. For all other $x$, $g(x)$ is a sum of continuous functions and is therefore continuous. <br>Thus, the number of points of discontinuity is $\alpha = 1$.<br><br>Now, let us check the differentiability of $g(x)$.<br>Since $g(x)$ is discontinuous at $x = 0$, it is not differentiable at $x = 0$.<br><br>For $x < 0$, $g(x) = x^2 + 5x + 6 + e^{x-1}$, which is differentiable everywhere in its domain.<br><br>For $x > 0$, we can rewrite $g(x)$ by analyzing the sign of $x^2 - 5x + 6 = (x-2)(x-3)$:<br>$g(x) = $\begin{cases} 2(x^2 - 5x + 6), & x \in (0, 2] \cup [3, \infty) \\ 0, & x \in (2, 3) \end{cases}

Differentiating $g(x)$ for $x > 0, x \neq 2, 3$:
$g'(x) =$\begin{cases} 4x - 10, & x \in (0, 2) \cup (3, \infty) \\ 0, & x \in (2, 3) \end{cases}$$<br><br>Checking differentiability at$x = 2$:<br>$g'(2^-) = 4(2) - 10 = -2$<br>$g'(2^+) = 0$<br>Since$g'(2^-) \neq g'(2^+)$,$g(x)$is not differentiable at$x = 2$.<br><br>Checking differentiability at$x = 3$:<br>$g'(3^-) = 0$<br>$g'(3^+) = 4(3) - 10 = 2$<br>Since$g'(3^-) \neq g'(3^+)$,$g(x)$is not differentiable at$x = 3$.<br><br>Thus,$g(x)$is not differentiable at exactly three points:$x = 0, 2, 3$.<br>So, the number of points of non-differentiability is$\beta = 3$.<br><br>Finally,$\alpha + \beta = 1 + 3 = 4$.<br><br>Answer:$4$