\[ f(x) = \begin{cases} 3x, & \text{if } x < 0 \\ \min(1 + x + \lfloor x \rfloor, 2 + x \lfloor x \rfloor), & \text{if } 0 \leq x \leq 2 \\ 5, & \text{if } x > 2 \end{cases} \]
The function changes definition at \(x = 0\) and \(x = 2\). Evaluate limits from left and right at these points:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 3x = 0 \]
\[ \lim_{x \to 0^+} f(x) = \min(1 + 0 + 0, 2 + 0 \times 0) = 1 \]
\[ \lim_{x \to 2^-} f(x) = \min(1 + 2 + 1, 2 + 2 \times 1) = 4 \]
\[ \lim_{x \to 2^+} f(x) = 5 \]
Discontinuity at \(x = 0\) and \(x = 2\).
Check for differentiability at integer points within \([0, 2]\) and at \(x = 2\), as \(f(x)\) involves the floor function, which is non-differentiable at integers:
\[ f'(x) \text{ is not defined at } x = 1, 2 \]
\[ \alpha = 2 \quad (\text{discontinuity at 0 and 2}) \]
\[ \beta = 3 \quad (\text{non-differentiability at 0, 1, and 2}) \]
\[ \alpha + \beta = 5 \]
Let \( f(x) = -3x^2(1 - x) - 3x(1 - x)^2 - (1 - x)^3 \). Then, \( \frac{df(x)}{dx} = \)
Let the area of the bounded region $ \{(x, y) : 0 \leq 9x \leq y^2, y \geq 3x - 6 \ be $ A $. Then 6A is equal to: