Key Concepts and Formulas

• Greatest Integer Function: $[x]$ denotes the greatest integer less than or equal to $x$.
• Fractional Part Function: $f(x) = x - [x]$ is the fractional part of $x$, and $0 \le f(x) < 1$.
• Continuity: A function $h(x)$ is continuous at a point $c$ if $\lim_{x \to c} h(x) = h(c)$. For a piecewise function, continuity needs to be checked at the points where the definition changes.
• Differentiability: A function $h(x)$ is differentiable at a point $c$ if the limit of the difference quotient exists. For a piecewise function, differentiability needs to be checked at the points where the definition changes. This often involves checking if the left-hand derivative equals the right-hand derivative.
• Minimum of Two Functions: $h(x) = \min\{f(x), g(x)\}$ means that at each point $x$, $h(x)$ takes the smaller value between $f(x)$ and $g(x)$. The graph of $h(x)$ will be the lower envelope of the graphs of $f(x)$ and $g(x)$.

Step-by-Step Solution

Step 1: Analyze the functions $f(x)$ and $g(x)$. We are given $f(x) = x - [x]$ and $g(x) = 1 - x + [x]$. The function $f(x)$ is the fractional part of $x$. It is periodic with period 1, and its graph is a sawtooth wave. The function $g(x)$ can be rewritten as $g(x) = 1 - (x - [x]) = 1 - f(x)$. The domain for $x$ is $[-2, 2]$.

Step 2: Determine the behavior of $f(x)$ and $g(x)$ over intervals. The greatest integer function $[x]$ changes its value at integer points. Within the interval $[-2, 2]$, the integer points are $-2, -1, 0, 1, 2$. Let's analyze the functions in the intervals defined by these integers.

• For $x \in [n, n+1)$, where $n$ is an integer, $[x] = n$.
  • $f(x) = x - n$. This is a line segment with slope 1.
  • $g(x) = 1 - x + n$. This is a line segment with slope -1.

Step 3: Find the points where $f(x) = g(x)$. The function $h(x) = \min\{f(x), g(x)\}$ will switch between $f(x)$ and $g(x)$ at the points where $f(x) = g(x)$. Set $f(x) = g(x)$: $x - [x] = 1 - x + [x]$ $2x = 1 + 2[x]$ $x = \frac{1}{2} + [x]$

Let $[x] = n$. Then $x = \frac{1}{2} + n$. We need to check if this value of $x$ is consistent with $[x] = n$. If $[x] = n$, then $n \le x < n+1$. Substituting $x = \frac{1}{2} + n$: $n \le \frac{1}{2} + n < n+1$ $0 \le \frac{1}{2} < 1$. This inequality is always true. So, the points where $f(x) = g(x)$ are of the form $x = n + \frac{1}{2}$, where $n$ is an integer.

Within the interval $[-2, 2]$, the points where $f(x) = g(x)$ are: For $n = -2$, $x = -2 + \frac{1}{2} = -1.5$ For $n = -1$, $x = -1 + \frac{1}{2} = -0.5$ For $n = 0$, $x = 0 + \frac{1}{2} = 0.5$ For $n = 1$, $x = 1 + \frac{1}{2} = 1.5$ For $n = 2$, $x = 2 + \frac{1}{2} = 2.5$ (This is outside our domain $[-2, 2]$).

So, the points where the definition of $h(x)$ might change are $x = -1.5, -0.5, 0.5, 1.5$.

Step 4: Define $h(x)$ piecewise in the relevant intervals. Consider an interval $[n, n+1)$. In this interval, $[x] = n$. $f(x) = x - n$ $g(x) = 1 - x + n$

We found that $f(x) = g(x)$ at $x = n + \frac{1}{2}$. Let's compare $f(x)$ and $g(x)$ in the subintervals $[n, n+\frac{1}{2})$ and $[n+\frac{1}{2}, n+1)$.

• For $x \in [n, n+\frac{1}{2})$: Let $x = n + \delta$, where $0 \le \delta < \frac{1}{2}$. $f(x) = (n+\delta) - n = \delta$ $g(x) = 1 - (n+\delta) + n = 1 - \delta$ Since $0 \le \delta < \frac{1}{2}$, we have $0 \le f(x) < \frac{1}{2}$ and $\frac{1}{2} < g(x) \le 1$. Thus, $f(x) < g(x)$ in this interval. So, $h(x) = f(x) = x - n$ for $x \in [n, n+\frac{1}{2})$.

• For $x \in [n+\frac{1}{2}, n+1)$: Let $x = n + \frac{1}{2} + \epsilon$, where $0 \le \epsilon < \frac{1}{2}$. $f(x) = (n + \frac{1}{2} + \epsilon) - n = \frac{1}{2} + \epsilon$ $g(x) = 1 - (n + \frac{1}{2} + \epsilon) + n = 1 - \frac{1}{2} - \epsilon = \frac{1}{2} - \epsilon$ Since $0 \le \epsilon < \frac{1}{2}$, we have $\frac{1}{2} \le f(x) < 1$ and $0 < g(x) \le \frac{1}{2}$. Thus, $g(x) \le f(x)$ in this interval. So, $h(x) = g(x) = 1 - x + n$ for $x \in [n+\frac{1}{2}, n+1)$.

Now, let's write $h(x)$ for the domain $[-2, 2]$:

• For $x \in [-2, -1.5)$: $[x] = -2$. $h(x) = x - (-2) = x+2$.
• For $x \in [-1.5, -1)$: $[x] = -1$. $h(x) = 1 - x + (-1) = -x$.
• For $x \in [-1, -0.5)$: $[x] = -1$. $h(x) = x - (-1) = x+1$.
• For $x \in [-0.5, 0)$: $[x] = -1$. $h(x) = 1 - x + (-1) = -x$.
• For $x \in [0, 0.5)$: $[x] = 0$. $h(x) = x - 0 = x$.
• For $x \in [0.5, 1)$: $[x] = 0$. $h(x) = 1 - x + 0 = 1-x$.
• For $x \in [1, 1.5)$: $[x] = 1$. $h(x) = x - 1$.
• For $x \in [1.5, 2)$: $[x] = 1$. $h(x) = 1 - x + 1 = 2-x$.
• At $x=2$: $[x]=2$. $f(2) = 2-2=0$. $g(2) = 1-2+2=1$. $h(2) = \min\{0, 1\} = 0$.

Let's check the endpoints and points where the definition of $[x]$ changes, and the points where $f(x)=g(x)$.

The critical points for continuity and differentiability are the integers and the half-integers within the domain. Integers in $[-2, 2]$ are $-2, -1, 0, 1, 2$. Half-integers in $(-2, 2)$ are $-1.5, -0.5, 0.5, 1.5$.

Let's re-evaluate $h(x)$ at the points where $[x]$ changes value (integers) and where $f(x)=g(x)$ (half-integers).

Step 5: Check for continuity. A function is continuous if it is continuous at every point in its domain. For piecewise functions, we need to check continuity at the points where the definition changes. These are the integer points and the half-integer points.

Let's examine continuity at the half-integer points ($x = n + 0.5$) first, as these are where the definition of $h(x)$ changes from $f(x)$ to $g(x)$ or vice versa. At $x = n + 0.5$: $f(n+0.5) = (n+0.5) - n = 0.5$ $g(n+0.5) = 1 - (n+0.5) + n = 1 - 0.5 = 0.5$ So, $f(n+0.5) = g(n+0.5) = 0.5$. Therefore, $h(n+0.5) = \min\{0.5, 0.5\} = 0.5$.

Now consider the limit from the left and right at $x = n + 0.5$. For $x \in [n, n+\frac{1}{2})$, $h(x) = x - n$. $\lim_{x \to (n+0.5)^-} h(x) = \lim_{x \to (n+0.5)^-} (x - n) = (n+0.5) - n = 0.5$.

For $x \in [n+\frac{1}{2}, n+1)$, $h(x) = 1 - x + n$. $\lim_{x \to (n+0.5)^+} h(x) = \lim_{x \to (n+0.5)^+} (1 - x + n) = 1 - (n+0.5) + n = 1 - 0.5 = 0.5$.

Since the left-hand limit, the right-hand limit, and the function value are all equal to 0.5 at $x = n + 0.5$, $h(x)$ is continuous at these half-integer points. The half-integer points in $(-2, 2)$ are $-1.5, -0.5, 0.5, 1.5$. So, $h(x)$ is continuous at these 4 points.

Now let's check continuity at the integer points ($x = n$). Consider $x = n$. The definition of $h(x)$ changes at $n$ based on the interval $[n, n+1)$ and $[n-1, n)$.

Let's consider the limit from the left and right at an integer $n$. For $x \in [n-1, n+\frac{1}{2})$: If $x \in [n-1, n)$, $[x] = n-1$. If $x \in [n-1, n-\frac{1}{2})$, $h(x) = x - (n-1) = x - n + 1$. If $x \in [n-\frac{1}{2}, n)$, $h(x) = 1 - x + (n-1) = -x + n$. For $x \in [n, n+1)$: If $x \in [n, n+\frac{1}{2})$, $h(x) = x - n$. If $x \in [n+\frac{1}{2}, n+1)$, $h(x) = 1 - x + n$.

Let's examine continuity at $x=0$: $\lim_{x \to 0^-} h(x)$: For $x \in [-0.5, 0)$, $[x] = -1$. $h(x) = 1 - x + (-1) = -x$. $\lim_{x \to 0^-} h(x) = \lim_{x \to 0^-} (-x) = 0$. $\lim_{x \to 0^+} h(x)$: For $x \in [0, 0.5)$, $[x] = 0$. $h(x) = x - 0 = x$. $\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} x = 0$. $h(0) = 0 - [0] = 0$. So, $h(x)$ is continuous at $x=0$.

Let's examine continuity at $x=1$: $\lim_{x \to 1^-} h(x)$: For $x \in [1, 1.5)$, $[x] = 1$. $h(x) = x - 1$. $\lim_{x \to 1^-} h(x) = \lim_{x \to 1^-} (x - 1) = 0$. $\lim_{x \to 1^+} h(x)$: For $x \in [1, 1.5)$, $[x] = 1$. (This is incorrect, need to consider the definition from the right side of 1). For $x \in [1, 1.5)$, $[x]=1$. $h(x) = x-1$. For $x \in [0.5, 1)$, $[x]=0$. $h(x) = 1-x$. $\lim_{x \to 1^-} h(x) = \lim_{x \to 1^-} (1-x) = 1-1=0$. $\lim_{x \to 1^+} h(x)$: For $x \in [1, 1.5)$, $[x]=1$. $h(x) = x-1$. $\lim_{x \to 1^+} h(x) = \lim_{x \to 1^+} (x-1) = 1-1=0$. $h(1) = 1 - [1] = 1-1 = 0$. So, $h(x)$ is continuous at $x=1$.

Let's examine continuity at $x=-1$: $\lim_{x \to -1^-} h(x)$: For $x \in [-1, -0.5)$, $[x] = -1$. $h(x) = x - (-1) = x+1$. $\lim_{x \to -1^-} h(x) = \lim_{x \to -1^-} (x+1) = -1+1=0$. $\lim_{x \to -1^+} h(x)$: For $x \in [-1, -0.5)$, $[x] = -1$. $h(x) = x+1$. (This is incorrect, need to consider the definition from the right side of -1). For $x \in [-1, -0.5)$, $[x]=-1$. $h(x) = x+1$. For $x \in [-1.5, -1)$, $[x]=-1$. $h(x) = 1 - x + (-1) = -x$. $\lim_{x \to -1^-} h(x) = \lim_{x \to -1^-} (-x) = -(-1) = 1$. $\lim_{x \to -1^+} h(x)$: For $x \in [-1, -0.5)$, $[x]=-1$. $h(x) = x+1$. $\lim_{x \to -1^+} h(x) = \lim_{x \to -1^+} (x+1) = -1+1=0$. Since $1 \ne 0$, $h(x)$ is not continuous at $x=-1$.

Let's examine continuity at $x=2$: $\lim_{x \to 2^-} h(x)$: For $x \in [1.5, 2)$, $[x]=1$. $h(x) = 2-x$. $\lim_{x \to 2^-} h(x) = \lim_{x \to 2^-} (2-x) = 2-2=0$. $h(2) = 2 - [2] = 2-2 = 0$. So, $h(x)$ is continuous at $x=2$.

Let's examine continuity at $x=-2$: $\lim_{x \to -2^+} h(x)$: For $x \in [-2, -1.5)$, $[x]=-2$. $h(x) = x+2$. $\lim_{x \to -2^+} h(x) = \lim_{x \to -2^+} (x+2) = -2+2=0$. $h(-2) = -2 - [-2] = -2 - (-2) = 0$. So, $h(x)$ is continuous at $x=-2$.

Summary of continuity: $h(x)$ is continuous at all half-integer points: $-1.5, -0.5, 0.5, 1.5$. (4 points) $h(x)$ is continuous at integer points $-2, 0, 1, 2$. $h(x)$ is NOT continuous at integer point $-1$.

So, $h(x)$ is continuous in $[-2, 2]$ except at $x=-1$.

Let's re-examine the definition of $h(x)$ and the points of discontinuity. The function $f(x) = x - [x]$ is discontinuous at every integer. The function $g(x) = 1 - x + [x]$ is also discontinuous at every integer. The minimum of two functions that are discontinuous at the same points can still be continuous.

Let's plot the graphs of $f(x)$ and $g(x)$. $f(x)$ is a sawtooth wave rising from 0 to 1 (not inclusive) in each unit interval. $g(x) = 1 - f(x)$. Its graph is the reflection of $f(x)$ about the line $y=0.5$.

Consider the interval $[0, 1)$: $f(x) = x$ $g(x) = 1 - x$ $f(x)=g(x)$ at $x=0.5$. $h(x) = x$ for $x \in [0, 0.5)$ $h(x) = 1-x$ for $x \in [0.5, 1)$ At $x=0$, $\lim_{x\to 0^-} h(x)$ needs to be checked. For $x \in [-1, 0)$, $[x]=-1$. $g(x) = 1-x+(-1) = -x$. So $\lim_{x\to 0^-} h(x) = \lim_{x\to 0^-} (-x) = 0$. $h(0) = 0 - [0] = 0$. At $x=1$, $\lim_{x\to 1^+} h(x)$ needs to be checked. For $x \in [1, 2)$, $[x]=1$. $f(x) = x-1$. So $\lim_{x\to 1^+} h(x) = \lim_{x\to 1^+} (x-1) = 0$. $h(1) = 1 - [1] = 0$.

The function $h(x)$ is continuous at all integer points. Let's recheck the continuity at $x=-1$. For $x \in [-1, 0)$: $[x] = -1$. For $x \in [-1, -0.5)$, $h(x) = x - (-1) = x+1$. For $x \in [-0.5, 0)$, $h(x) = 1 - x + (-1) = -x$. $\lim_{x \to -1^-} h(x)$: For $x \in [-2, -1)$, $[x] = -2$. For $x \in [-2, -1.5)$, $h(x) = x - (-2) = x+2$. For $x \in [-1.5, -1)$, $h(x) = 1 - x + (-2) = -x - 1$. $\lim_{x \to -1^-} h(x) = \lim_{x \to -1^-} (-x-1) = -(-1)-1 = 1-1 = 0$. $\lim_{x \to -1^+} h(x) = \lim_{x \to -1^+} (x+1) = -1+1 = 0$. $h(-1) = -1 - [-1] = -1 - (-1) = 0$. So $h(x)$ is continuous at $x=-1$.

It seems that $h(x)$ is continuous in $[-2, 2]$. Let's confirm this. The graph of $h(x)$ is the lower envelope of $y=x-[x]$ and $y=1-x+[x]$. The graph of $y=x-[x]$ is discontinuous at integers. The graph of $y=1-x+[x]$ is discontinuous at integers. However, their minimum can be continuous.

Consider the points $x=n$. $\lim_{x \to n^-} h(x)$. For $x$ slightly less than $n$, $[x] = n-1$. If $x \in [n-1, n - 0.5)$, $h(x) = x - (n-1) = x - n + 1$. If $x \in [n-0.5, n)$, $h(x) = 1 - x + (n-1) = -x + n$. So $\lim_{x \to n^-} h(x) = \lim_{x \to n^-} (-x+n) = -n+n = 0$. $h(n) = n - [n] = n - n = 0$. $\lim_{x \to n^+} h(x)$. For $x$ slightly greater than $n$, $[x] = n$. If $x \in [n, n+0.5)$, $h(x) = x - n$. So $\lim_{x \to n^+} h(x) = \lim_{x \to n^+} (x-n) = n-n = 0$. Thus, $h(x)$ is continuous at all integers. And we have already shown it is continuous at all half-integers. Therefore, $h(x)$ is continuous in $[-2, 2]$.

Step 6: Check for differentiability. A function is not differentiable at points where its graph has a sharp corner or a vertical tangent. For piecewise linear functions, this occurs at the points where the slope changes. The slope of $h(x)$ changes at:

1. The integer points ($x=n$) where the definition of $[x]$ changes.
2. The half-integer points ($x=n+0.5$) where $h(x)$ switches between $f(x)$ and $g(x)$.

Let's analyze the slopes in the intervals. For $x \in [n, n+0.5)$, $h(x) = x - n$. The slope is 1. For $x \in [n+0.5, n+1)$, $h(x) = 1 - x + n$. The slope is -1.

At the half-integer points $x = n+0.5$: The slope from the left is 1. The slope from the right is -1. Since the left-hand derivative (1) is not equal to the right-hand derivative (-1), $h(x)$ is not differentiable at the half-integer points. The half-integer points in $(-2, 2)$ are $-1.5, -0.5, 0.5, 1.5$. So, $h(x)$ is not differentiable at these 4 points.

Now let's check the integer points $x=n$. Consider $x=n$. If $n$ is an integer, the function definition changes. For $x \in [n-1, n)$: If $x \in [n-1, n-0.5)$, $h(x) = x - (n-1) = x - n + 1$. Slope = 1. If $x \in [n-0.5, n)$, $h(x) = 1 - x + (n-1) = -x + n$. Slope = -1. For $x \in [n, n+1)$: If $x \in [n, n+0.5)$, $h(x) = x - n$. Slope = 1. If $x \in [n+0.5, n+1)$, $h(x) = 1 - x + n$. Slope = -1.

Let's check differentiability at $x=n$. Left-hand derivative at $x=n$: We need to consider the behavior of $h(x)$ as $x$ approaches $n$ from the left. For $x$ in an interval just to the left of $n$, say $[n-1, n)$, the definition of $h(x)$ depends on whether $x$ is in $[n-1, n-0.5)$ or $[n-0.5, n)$. The "last" slope before $n$ is from the interval $[n-0.5, n)$, where $h(x) = -x + n$. The slope is -1. So, the left-hand derivative at $x=n$ is -1.

Right-hand derivative at $x=n$: For $x$ in an interval just to the right of $n$, say $[n, n+1)$, the definition of $h(x)$ depends on whether $x$ is in $[n, n+0.5)$ or $[n+0.5, n+1)$. The "first" slope after $n$ is from the interval $[n, n+0.5)$, where $h(x) = x - n$. The slope is 1. So, the right-hand derivative at $x=n$ is 1.

Since the left-hand derivative (-1) is not equal to the right-hand derivative (1) at any integer point $x=n$, $h(x)$ is not differentiable at all integer points. The integer points in $(-2, 2)$ are $-1, 0, 1$. So, $h(x)$ is not differentiable at these 3 points.

The domain is $[-2, 2]$. The points of non-differentiability in $(-2, 2)$ are the 4 half-integer points and the 3 integer points. Total points of non-differentiability in $(-2, 2)$ = 4 (half-integers) + 3 (integers) = 7 points.

These points are: $-1.5, -1, -0.5, 0, 0.5, 1, 1.5$.

Let's re-read the question carefully. "not differentiable at more than four points in $(-2, 2)$". We found 7 points of non-differentiability in $(-2, 2)$. This matches option (A).

Let's double check the intervals and slopes. $h(x) = \min\{x-[x], 1-x+[x]\}$. The graph of $h(x)$ is the lower envelope of the two lines. The points where the lines intersect are $x = n + 0.5$. The slope of $x-[x]$ is 1. The slope of $1-x+[x]$ is -1.

In the interval $[n, n+1)$: If $x \in [n, n+0.5)$, $h(x) = x-n$. Slope = 1. If $x \in [n+0.5, n+1)$, $h(x) = 1-x+n$. Slope = -1.

At $x = n+0.5$, the slope changes from 1 to -1. This is a sharp corner. So, not differentiable at $n+0.5$. Points: $-1.5, -0.5, 0.5, 1.5$ (4 points).

Now consider the integer points $x=n$. We need to check the left and right derivatives at $x=n$. Consider $x=0$. To the left of 0 (e.g., in $[-0.5, 0)$), $h(x) = 1-x+(-1) = -x$. Slope = -1. To the right of 0 (e.g., in $[0, 0.5)$), $h(x) = x-0 = x$. Slope = 1. So, at $x=0$, the left derivative is -1 and the right derivative is 1. Not differentiable.

Consider $x=1$. To the left of 1 (e.g., in $[0.5, 1)$), $h(x) = 1-x+0 = 1-x$. Slope = -1. To the right of 1 (e.g., in $[1, 1.5)$), $h(x) = x-1$. Slope = 1. So, at $x=1$, the left derivative is -1 and the right derivative is 1. Not differentiable.

Consider $x=-1$. To the left of -1 (e.g., in $[-1.5, -1)$), $h(x) = 1-x+(-2) = -x-1$. Slope = -1. To the right of -1 (e.g., in $[-1, -0.5)$), $h(x) = x-(-1) = x+1$. Slope = 1. So, at $x=-1$, the left derivative is -1 and the right derivative is 1. Not differentiable.

The integer points in $(-2, 2)$ are $-1, 0, 1$. So, $h(x)$ is not differentiable at these 3 points.

Total points of non-differentiability in $(-2, 2)$ = 4 (half-integers) + 3 (integers) = 7 points.

The function is continuous in $[-2, 2]$. The points of non-differentiability in $(-2, 2)$ are $-1.5, -1, -0.5, 0, 0.5, 1, 1.5$. There are exactly 7 such points.

Option (A) states: continuous in $[-2, 2]$ but not differentiable at more than four points in $(-2, 2)$. Our finding of 7 points of non-differentiability in $(-2, 2)$ satisfies "more than four points".

Let's check if any other option could be correct. (B) not continuous at exactly three points in $[-2, 2]$. We found it is continuous. (C) continuous in $[-2, 2]$ but not differentiable at exactly three points in $(-2, 2)$. We found 7 points. (D) not continuous at exactly four points in $[-2, 2]$. We found it is continuous.

So, option (A) is the correct answer.

The original solution stated "always continuous in [-2, 2] but not differentiable at 7 points." This matches our findings. The original solution also had a typo in the function definition $h(x) = \min\{x - [x], 1 - [x - [x])\}$ which should be $h(x) = \min\{x - [x], 1 - x + [x]\}$.

Let's ensure the wording of the question "more than four points" is correctly interpreted. Yes, 7 is more than 4.

The points of non-differentiability are indeed the points where the slope changes. The graph of $h(x)$ consists of line segments. The slope of $h(x)$ is either 1 or -1 in the open intervals. The slope changes at every integer and every half-integer.

Consider the interval $[-2, 2]$. The integer points are $-2, -1, 0, 1, 2$. The half-integer points are $-1.5, -0.5, 0.5, 1.5$.

Points where the slope could change:

• Integer points: $-1, 0, 1$. (Points $-2$ and $2$ are endpoints).
• Half-integer points: $-1.5, -0.5, 0.5, 1.5$.

Let's re-evaluate the derivatives at the integer points carefully. Consider $x=n$. The interval just to the left of $n$ is $[n-1, n)$. In this interval, the slope changes at $n-0.5$. So, for $x \in [n-1, n-0.5)$, slope is 1. For $x \in [n-0.5, n)$, slope is -1. The left derivative at $n$ is the slope of the segment ending at $n$, which is -1.

The interval just to the right of $n$ is $[n, n+1)$. In this interval, the slope changes at $n+0.5$. So, for $x \in [n, n+0.5)$, slope is 1. For $x \in [n+0.5, n+1)$, slope is -1. The right derivative at $n$ is the slope of the segment starting at $n$, which is 1.

So, at each integer $n$, the left derivative is -1 and the right derivative is 1. Thus, not differentiable at integers. Integers in $(-2, 2)$ are $-1, 0, 1$. (3 points).

At the half-integer points $x = n+0.5$. To the left of $n+0.5$, in $[n, n+0.5)$, the slope is 1. To the right of $n+0.5$, in $[n+0.5, n+1)$, the slope is -1. So, at each half-integer $n+0.5$, the left derivative is 1 and the right derivative is -1. Thus, not differentiable. Half-integers in $(-2, 2)$ are $-1.5, -0.5, 0.5, 1.5$. (4 points).

Total points of non-differentiability in $(-2, 2)$ = 3 (integers) + 4 (half-integers) = 7 points. The function is continuous in $[-2, 2]$.

Thus, the function is continuous in $[-2, 2]$ but not differentiable at exactly 7 points in $(-2, 2)$. This means it is not differentiable at more than four points in $(-2, 2)$.

Common Mistakes & Tips

• Confusing endpoints with interior points: Differentiability is usually checked in the open interval $(-2, 2)$. The behavior at the endpoints $-2$ and $2$ does not affect the number of points of non-differentiability in the open interval.
• Incorrectly determining the slope of $h(x)$: Always ensure you are taking the minimum of the two functions. Sketching the graphs of $f(x)$ and $g(x)$ and their minimum is very helpful.
• Overlooking integer points for non-differentiability: While the fractional part function $f(x)$ itself is discontinuous at integers, the minimum function $h(x)$ can be continuous at integers. However, the slope change at integers still leads to non-differentiability.

Summary

The function $h(x) = \min\{x-[x], 1-x+[x]\}$ is analyzed for continuity and differentiability in the interval $[-2, 2]$. We found that $h(x)$ is continuous at all points in $[-2, 2]$. The points where the differentiability might fail are the integer points (where $[x]$ changes) and the half-integer points (where $h(x)$ switches between $x-[x]$ and $1-x+[x]$). By examining the slopes of the piecewise linear segments of $h(x)$, we determined that the left and right derivatives do not match at all integer points ($-1, 0, 1$ in $(-2, 2)$) and all half-integer points ($-1.5, -0.5, 0.5, 1.5$ in $(-2, 2)$). This results in exactly 7 points of non-differentiability in the open interval $(-2, 2)$. Since 7 is more than 4, option (A) is the correct choice.

The final answer is \boxed{A}.