The function f(x) = |x| + |x - 1| is | vedclass

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(A) We have $f(x) = |x| + |x - 1|$.By definition of absolute value,we can write $f(x)$ as:$f(x) = \begin{cases} -2x + 1, & x < 0 \ 1, & 0 \le x < 1 \

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Continuous at $x = 1$,but not differentiable at $x = 1$

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(A) We have $f(x) = |x| + |x - 1|$.By definition of absolute value,we can write $f(x)$ as:$f(x) = \begin{cases} -2x + 1, & x For continuity at $x = 1$:$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (1) = 1$$\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} (2x - 1) = 2(1) - 1 = 1$$f(1) = 2(1) - 1 = 1$Since $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$,the function is continuous at $x = 1$.For differentiability at $x = 1$:Left-hand derivative $(LHD)$ at $x = 1$ is $\lim_{h 	o 0^-} \frac{f(1+h) - f(1)}{h} = \frac{d}{dx}(1) = 0$.Right-hand derivative $(RHD)$ at $x = 1$ is $\lim_{h 	o 0^+} \frac{f(1+h) - f(1)}{h} = \frac{d}{dx}(2x - 1) = 2$.Since $LHD 
eq RHD$,the function is not differentiable at $x = 1$.

The function $f(x) = |x| + |x - 1|$ is

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