In the interval [0, 3],the function f(x) = |x | vedclass

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(D) The function is defined as $f(x) = |x - 1| + |x - 2|$ for $x \in [0, 3]$.By breaking the absolute values,we get:$f(x) = \begin{cases} -(x-1) - (x-

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Continuous but not differentiable at $x = 1$ and $x = 2$

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(D) The function is defined as $f(x) = |x - 1| + |x - 2|$ for $x \in [0, 3]$.By breaking the absolute values,we get:$f(x) = \begin{cases} -(x-1) - (x-2) = -2x + 3, & 0 \leq x \leq 1 \ (x-1) - (x-2) = 1, & 1 Since $f(x)$ is a sum of continuous functions,it is continuous everywhere in $[0, 3]$.At $x = 1$,the left-hand derivative is $\frac{d}{dx}(-2x+3) = -2$ and the right-hand derivative is $\frac{d}{dx}(1) = 0$. Since $-2 
eq 0$,$f(x)$ is not differentiable at $x = 1$.At $x = 2$,the left-hand derivative is $\frac{d}{dx}(1) = 0$ and the right-hand derivative is $\frac{d}{dx}(2x-3) = 2$. Since $0 
eq 2$,$f(x)$ is not differentiable at $x = 2$.Thus,$f(x)$ is continuous but not differentiable at $x = 1$ and $x = 2$.

In the interval $[0, 3]$,the function $f(x) = |x - 1| + |x - 2|$ is

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