The function f(x) = begin{cases} x, & text{if | vedclass

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(A) Given $f(x) = \begin{cases} x, & 0 \le x \le 1 \ 1, & 1 < x \le 2 \end{cases}$Check continuity at $x = 1$:Left-hand limit: $\lim_{x 	o 1^-} f(x)

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Continuous at all $x$,$0 \le x \le 2$ and differentiable at all $x$,except $1$ in the interval $(0, 2)$

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(A) Given $f(x) = \begin{cases} x, & 0 \le x \le 1 \ 1, & 1 Check continuity at $x = 1$:Left-hand limit: $\lim_{x 	o 1^-} f(x) = \lim_{h 	o 0} f(1 - h) = \lim_{h 	o 0} (1 - h) = 1$Right-hand limit: $\lim_{x 	o 1^+} f(x) = \lim_{h 	o 0} f(1 + h) = 1$Since $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1) = 1$,the function is continuous at $x = 1$.Check differentiability at $x = 1$:Left-hand derivative: $LHD = \lim_{h 	o 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h 	o 0} \frac{(1 - h) - 1}{-h} = \lim_{h 	o 0} \frac{-h}{-h} = 1$Right-hand derivative: $RHD = \lim_{h 	o 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h 	o 0} \frac{1 - 1}{h} = 0$Since $LHD 
eq RHD$,the function is not differentiable at $x = 1$.Thus,the function is continuous for all $x \in [0, 2]$ and differentiable for all $x \in [0, 2]$ except at $x = 1$.

The function $f(x) = \begin{cases} x, & \text{if } 0 \le x \le 1 \\ 1, & \text{if } 1 < x \le 2 \end{cases}$ is

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