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 $\alpha+\beta=$

  • A
    $6$
  • B
    $3$
  • C
    $2$
  • D
    $0$

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