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(D) Given $f(x) = |x-2| + x$. First,we check for continuity at $x=0$ and $x=2$. At $x=0$: $f(0) = |0-2| + 0 = 2$. $\lim_{x 	o 0^-} f(x) = |0-2| + 0 =

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continuous at both $x=2$ and $x=0$

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(D) Given $f(x) = |x-2| + x$. First,we check for continuity at $x=0$ and $x=2$. At $x=0$: $f(0) = |0-2| + 0 = 2$. $\lim_{x 	o 0^-} f(x) = |0-2| + 0 = 2$ and $\lim_{x 	o 0^+} f(x) = |0-2| + 0 = 2$. Since $\lim_{x 	o 0} f(x) = f(0)$,the function is continuous at $x=0$. At $x=2$: $f(2) = |2-2| + 2 = 2$. $\lim_{x 	o 2^-} f(x) = |2-2| + 2 = 2$ and $\lim_{x 	o 2^+} f(x) = |2-2| + 2 = 2$. Since $\lim_{x 	o 2} f(x) = f(2)$,the function is continuous at $x=2$. Now,we check for differentiability. We can write $f(x)$ as: $f(x) = \begin{cases} -(x-2) + x, & x For $x  2$,$f'(x) = 2$. At $x=2$,$LHD$ $= \lim_{h 	o 0^-} \frac{f(2+h)-f(2)}{h} = \lim_{h 	o 0^-} \frac{2-2}{h} = 0$. $RHD$ $= \lim_{h 	o 0^+} \frac{f(2+h)-f(2)}{h} = \lim_{h 	o 0^+} \frac{2(2+h)-2-2}{h} = \lim_{h 	o 0^+} \frac{2h}{h} = 2$. Since $LHD$ $
eq$ $RHD$,$f(x)$ is not differentiable at $x=2$. At $x=0$,$f(x) = 2$ in the neighborhood,so $f'(0) = 0$. Thus,$f(x)$ is differentiable at $x=0$. Therefore,the function is continuous at both $x=0$ and $x=2$.

The function $f(x) = |x-2| + x$ is

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