Let f(x) = begin{cases} 3x, & x

Question

(D) For $0 \leq x \leq 2$,we analyze $f(x) = \min \{1+x+[x], x+2[x]\}$.Case $1$: $0 \leq x < 1$,then $[x] = 0$. So,$f(x) = \min \{1+x, x\} = x$.Case $

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$5$

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(D) For $0 \leq x \leq 2$,we analyze $f(x) = \min \{1+x+[x], x+2[x]\}$.Case $1$: $0 \leq x Case $2$: $1 \leq x Case $3$: At $x = 2$,$[x] = 2$. So,$f(2) = \min \{1+2+2, 2+2(2)\} = \min \{5, 6\} = 5$.Thus,the function is:$f(x) = \begin{cases} 3x, & x Checking continuity:At $x=0$: $\lim_{x 	o 0^-} f(x) = 0$,$\lim_{x 	o 0^+} f(x) = 0$,$f(0) = 0$. Continuous.At $x=1$: $\lim_{x 	o 1^-} f(x) = 1$,$\lim_{x 	o 1^+} f(x) = 3$. Discontinuous.At $x=2$: $\lim_{x 	o 2^-} f(x) = 4$,$\lim_{x 	o 2^+} f(x) = 5$. Discontinuous.So,$\alpha = 2$ (points are $x=1, 2$).Checking differentiability:At $x=0$: $f'(0^-) = 3$,$f'(0^+) = 1$. Not differentiable.At $x=1$: Discontinuous,so not differentiable.At $x=2$: Discontinuous,so not differentiable.So,$\beta = 3$ (points are $x=0, 1, 2$).Therefore,$\alpha + \beta = 2 + 3 = 5$.

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