Let f(x) = begin{cases} 3-x & text{if } x

Question

(C) To check for continuity at $x = -3$: $\lim_{x 	o -3^-} f(x) = \lim_{x 	o -3^-} (3-x) = 3 - (-3) = 6$. $\lim_{x 	o -3^+} f(x) = 6$. $f(-3) = 6$.

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) To check for continuity at $x = -3$: $\lim_{x 	o -3^-} f(x) = \lim_{x 	o -3^-} (3-x) = 3 - (-3) = 6$. $\lim_{x 	o -3^+} f(x) = 6$. $f(-3) = 6$. Since $\lim_{x 	o -3^-} f(x) = \lim_{x 	o -3^+} f(x) = f(-3)$, the function is continuous at $x = -3$. To check for continuity at $x = 3$: $\lim_{x 	o 3^-} f(x) = 6$. $\lim_{x 	o 3^+} f(x) = \lim_{x 	o 3^+} (3+x) = 3+3 = 6$. $f(3) = 6$. Since $\lim_{x 	o 3^-} f(x) = \lim_{x 	o 3^+} f(x) = f(3)$, the function is continuous at $x = 3$. Thus, the function is continuous everywhere, so the number of points of discontinuity is $\alpha = 0$. To check for differentiability at $x = -3$: Left-hand derivative $(LHD)$ at $x = -3$: $\frac{d}{dx}(3-x) = -1$. Right-hand derivative $(RHD)$ at $x = -3$: $\frac{d}{dx}(6) = 0$. Since $LHD 
eq RHD$, $f$ is not differentiable at $x = -3$. To check for differentiability at $x = 3$: $LHD$ at $x = 3$: $\frac{d}{dx}(6) = 0$. $RHD$ at $x = 3$: $\frac{d}{dx}(3+x) = 1$. Since $LHD 
eq RHD$, $f$ is not differentiable at $x = 3$. Thus, the number of points of non-differentiability is $\beta = 2$. Therefore, $\alpha + \beta = 0 + 2 = 2$.

Let $f(x) = \begin{cases} 3-x & \text{if } x < -3 \\ 6 & \text{if } -3 \leq x \leq 3 \\ 3+x & \text{if } x > 3 \end{cases}$. Let $\alpha$ be the number of points of discontinuity of $f$ and $\beta$ be the number of points where $f$ is not differentiable. Then $\alpha+\beta=$

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