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(A) The absolute value function can be rewritten as a piecewise function: $f(x)= \begin{cases} 1-2x, & 	ext{if } x \leq \frac{1}{2} \ 2x-1, & 	ext{

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$f(x)$ is continuous but not differentiable at $x=\frac{1}{2}$

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(A) The absolute value function can be rewritten as a piecewise function: $f(x)= \begin{cases} 1-2x, & 	ext{if } x \leq \frac{1}{2} \ 2x-1, & 	ext{if } x > \frac{1}{2} \end{cases}$ Check for continuity at $x=\frac{1}{2}$: Left-hand limit: $\lim_{x ightarrow \frac{1}{2}^-} f(x) = \lim_{x ightarrow \frac{1}{2}} (1-2x) = 1-2(\frac{1}{2}) = 0$. Right-hand limit: $\lim_{x ightarrow \frac{1}{2}^+} f(x) = \lim_{x ightarrow \frac{1}{2}} (2x-1) = 2(\frac{1}{2})-1 = 0$. Function value: $f(\frac{1}{2}) = |1-2(\frac{1}{2})| = 0$. Since the left-hand limit, right-hand limit, and function value are all equal, the function is continuous at $x=\frac{1}{2}$. Check for differentiability at $x=\frac{1}{2}$: Left-hand derivative: $f'(x) = \frac{d}{dx}(1-2x) = -2$ for $x Right-hand derivative: $f'(x) = \frac{d}{dx}(2x-1) = 2$ for $x > \frac{1}{2}$. Since the left-hand derivative $(-2)$ and right-hand derivative $(2)$ are not equal, the function is not differentiable at $x=\frac{1}{2}$.

Let $f(x)=|1-2 x|$, then

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