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Step-by-Step Solution
Step 1: Understand the Definitions
• Let $[x]$ denote the greatest integer less than or equal to $x$ (the floor function).
• Define $f(x) = x - [x]$, which represents the fractional part of $x$ (i.e., $f(x) \in [0,1)$).
• Define $g(x) = 1 - x + [x]$. Observe that $g(x) = 1 - (x - [x]) = 1 - f(x)$.
• Define $h(x) = \min \{ f(x), g(x) \}$ for $x \in [-2, 2]$.
Step 2: Express the Function in Terms of the Fractional Part
Since $f(x)$ is the fractional part of $x$, it satisfies $0 \leq f(x) < 1$. Then:
\[
g(x) = 1 - f(x).
\]
Hence:
\[
h(x) = \min \{ f(x), 1 - f(x) \}.
\]
The function $\min \{ t, 1 - t \}$ for $t \in [0,1]$ is symmetric about $t = 0.5$. Specifically,
• If $0 \leq t \leq 0.5$, then $\min \{ t, 1 - t \} = t$.
• If $0.5 < t < 1$, then $\min \{ t, 1 - t \} = 1 - t$.
So $h(x)$ switches between $f(x)$ and $1 - f(x)$ at those points where $f(x) = 0.5$.
Step 3: Analyze Continuity of $h(x)$
1. The fractional part function $f(x)$ itself has jump discontinuities at integer points of $x$. That is, at every integer $n$, $f(x)$ jumps from a value close to 1 (when $x \to n^-$) to 0 (when $x \to n^+$).
2. However, $g(x) = 1 - f(x)$ also has a corresponding jump at every integer, but in an exactly opposite manner.
3. At each integer $x = n$, $f(n) = 0$ and $g(n) = 1$. Thus, $h(n) = \min \{ 0, 1 \} = 0$. Just as $x$ approaches $n$ from the left, $f(x)$ approaches 1 and $g(x)$ approaches 0, which also makes the minimum approach 0. Consequently, there is no jump in $h(x)$ itself and it remains continuous at every integer.
4. Between integers, $f(x)$ is continuous and so is $1 - f(x)$. Thus, the minimum of these two continuous pieces remains continuous. Therefore, $h(x)$ is continuous for all $x \in [-2, 2]$.
Step 4: Identify Points of Non-Differentiability
1. $h(x)$ can fail to be differentiable where either $f(x)$ or $g(x)$ has a corner (or cusp), or where $h(x)$ switches from $f(x)$ to $g(x)$ or vice versa.
2. The floor function $[x]$ creates “corners” in $f(x)$ at each integer $x$. Since $h(x)$ is the minimum, we must check these integer points for potential non-differentiability.
3. Additionally, $h(x)$ switches its definition whenever $f(x) = 0.5$. That happens when $x - [x] = 0.5$. These are the half-integer points (e.g., $n + 0.5$) in the interval $[-2, 2]$.
4. Each such point can be a point of “kink” or corner in the graph of $h(x)$, resulting in non-differentiability. In the interval $[-2, 2]$, one finds multiple integers ($-2, -1, 0, 1, 2$) and the midpoints in between ($-1.5, -0.5, 0.5, 1.5$). Overall, these cause more than four points of non-differentiability (commonly counted as seven within $[-2,2]$, as the solution suggests).
5. However, at each of these points, $h(x)$ remains continuous, but the left-hand derivative and right-hand derivative do not match, leading to a corner.
Step 5: Conclusion
1. $h(x)$ is continuous on the closed interval $[-2, 2]$.
2. $h(x)$ is not differentiable at more than four (specifically, seven) points where the floor function and the switching between $f(x)$ and $g(x)$ create corners.
Hence, the correct statement is:
“continuous in $[-2, 2]$ but not differentiable at more than four points in $(-2, 2)$.”
Reference to Provided Image
This image conceptually illustrates the behavior of $f(x), g(x)$, and their minimum, confirming continuity but highlighting the points of non-differentiability.
