1. **Problem Statement:** Given the functions $f(x) = \{x\}$ (the fractional part of $x$), $g(x) = 1 - \{x\}$, and $h(x) = \min(f(x), g(x))$, analyze the continuity and differentiability of $h(x)$ on the interval $[-2,2]$. 2. **Recall definitions:** - The fractional part function $\{x\}$ is defined as $x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. - $f(x) = \{x\}$ is continuous everywhere but not differentiable at integers because of the jump in the floor function. - $g(x) = 1 - \{x\}$ is also continuous everywhere and has the same differentiability properties as $f(x)$. 3. **Analyze $h(x) = \min(f(x), g(x))$:** - Since $f(x)$ and $g(x)$ are continuous, $h(x)$, being the minimum of two continuous functions, is also continuous everywhere on $[-2,2]$. 4. **Where does $f(x) = g(x)$?** - Solve $\{x\} = 1 - \{x\} \implies 2\{x\} = 1 \implies \{x\} = \frac{1}{2}$. - At points where the fractional part is $0.5$, $f(x)$ and $g(x)$ intersect. 5. **Behavior of $h(x)$:** - For $\{x\} < 0.5$, $f(x) < g(x)$ so $h(x) = f(x)$. - For $\{x\} > 0.5$, $g(x) < f(x)$ so $h(x) = g(x)$. - At $\{x\} = 0.5$, $h(x) = f(x) = g(x) = 0.5$. 6. **Differentiability analysis:** - Both $f(x)$ and $g(x)$ are linear with slope 1 and -1 respectively between integers. - $h(x)$ switches from $f(x)$ to $g(x)$ at points where $\{x\} = 0.5$. - At these points, the left-hand derivative of $h(x)$ is $1$ (from $f(x)$) and the right-hand derivative is $-1$ (from $g(x)$). - Since the left and right derivatives differ, $h(x)$ is not differentiable at points where $\{x\} = 0.5$. 7. **At integers:** - $f(x)$ and $g(x)$ have jump discontinuities in their derivatives but are continuous. - $h(x)$ inherits the same continuity and differentiability properties at integers as $f(x)$ and $g(x)$. - Since $f(x)$ and $g(x)$ are not differentiable at integers, $h(x)$ is also not differentiable at integers. **Final conclusion:** - $h(x)$ is continuous everywhere on $[-2,2]$. - $h(x)$ is not differentiable at integers and at points where $\{x\} = 0.5$ within $[-2,2]$. **Summary:** $$\text{Non-differentiable points} = \{x \in [-2,2] : x \in \mathbb{Z} \text{ or } \{x\} = 0.5\}$$