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

The absolute value function can be rewritten as a piecewise function: \(f(x)= \begin{cases}1-2 x, & \text { if } x \leq \frac{1}{2} \\ -(1-2 x), & \text { if } x>\frac{1}{2}\end{cases}\) 2. Check for continuity at \(x=\frac{1}{2}\) : Left-hand limit: \(\lim _{x \rightarrow \frac{1}{2}} f(x)=\lim _{x \rightarrow \frac{1}{2}}(1-2 x)=1-2\left(\frac{1}{2}\right)=0\) Right-hand limit: \(\lim _{x \rightarrow \frac{1}{2}} f(x)=\lim _{x \rightarrow \frac{1}{2}}-(1-2 x)=-\left(1-2\left(\frac{1}{2}\right)\right)=0\) Function value: \(f\left(\frac{1}{2}\right)=\left|1-2\left(\frac{1}{2}\right)\right|=|1-1|=0\) Since the left-hand limit, right-hand limit, and function value are all equal, the function is continuous at \(x=\frac{1}{2}\). 3. Check for differentiability at \(x=\frac{1}{2}\) : Left-hand derivative: \(f^{\prime}(x)=-2\) for \(x<\frac{1}{2}\), so \(\lim _{x \rightarrow \frac{1}{2}} f^{\prime}(x)=-2\). Right-hand derivative: \(f^{\prime}(x)=2\) for \(x>\frac{1}{2}\), so \(\lim _{x \rightarrow \frac{1}{2}^{+}} f^{\prime}(x)=2\). Since the left-hand derivative and right-hand derivative are not equal, the function is not differentiable at \(x=\frac{1}{2}\).