Let [t] denote the greatest integer less than or equal to t. Let f(x) = x - [x], g(x) = 1 - x + [x],

Q: Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

This isn't a physics problem involving energy, but a calculus problem about function behavior! We don't have 'roots' to reject; we have points of non-differentiability to identify. We look for sharp corners where the derivative is undefined, not solutions to an equation.

Q: Why are there sharp corners at the integers? Isn't the function continuous there?

Great observation! Yes, the function is continuous at the integers because the limit from the left and the right both approach zero. However, continuity does not guarantee differentiability. At $x=n$, the slope jumps from $1$ to $-1$, creating a sharp corner, which makes the derivative undefined.

Q: How do I know if a function is continuous just by looking at the graph?

A function is continuous if you can draw its graph without lifting your pen. In this case, the 'min' function connects the two pieces perfectly at the intersection points, so there are no jumps or holes. It's a continuous, albeit 'kinky', triangular wave!

Question

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f(x) = x - [x]$, $g(x) = 1 - x + [x]$, and $h(x) = \min\{f(x), g(x)\}$, $x \in [-2, 2]$. Then $h$ is :

Options:

(A) continuous in $[-2, 2]$ but not differentiable at more than four points in $(-2, 2)$

(B) not continuous at exactly three points in $[-2, 2]$

(C) continuous in $[-2, 2]$ but not differentiable at exactly three points in $(-2, 2)$

(D) not continuous at exactly four points in $[-2, 2]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The function is defined as $h(x) = \min(\{x\}, 1 - \{x\})$ over the interval $x \in [-2, 2]$. Here, $\{x\}$ denotes the fractional part of $x$, which is defined as $\{x\} = x - \lfloor x floor$.

To understand the behavior of $h(x)$, we must visualize the two components:
1. $f(x) = \{x\}$, which is a sawtooth wave increasing from $0$ to $1$ on every interval $[n, n+1)$.
2. $g(x) = 1 - \{x\}$, which is the reflection of the sawtooth wave, decreasing from $1$ to $0$ on every interval $[n, n+1)$.

### The Intersection: The Birth of the Triangular Wave

The function $h(x)$ selects the minimum of these two paths at any point $x$. The intersection occurs when:

$$\{x\} = 1 - \{x\}$$

Solving this yields $2\{x\} = 1$, or $\{x\} = 0.5$. At this value, both functions equal $0.5$.

For $0 \le \{x\} < 0.5$, the function $f(x)$ is smaller, so $h(x) = \{x\}$. For $0.5 < \{x\} < 1$, the function $g(x)$ is smaller, so $h(x) = 1 - \{x\}$. This creates a continuous triangular wave oscillating between $0$ and $0.5$.

### The Continuity Check

To verify continuity, we examine the transition points. At any integer $n$, $\{x\} = 0$, so $f(n) = 0$ and $g(n) = 1$. However, as $x 	o n^-$, $\{x\} 	o 1$, making $f(x) 	o 1$ and $g(x) 	o 0$.

Since $h(x) = \min(\{x\}, 1 - \{x\})$, at the integers, $h(n) = \min(0, 1) = 0$. Because the limits from both sides approach $0$, the function is **continuous** everywhere.

### The Differentiability Hunt

A function is non-differentiable at points where it exhibits "sharp corners" or "kinks." For $h(x)$, these occur at the troughs and the peaks of the triangular wave.

The troughs occur at the integers where $\{x\} = 0$. In the interval $(-2, 2)$, these points are:
$$x \in \{-1, 0, 1\}$$

The peaks occur at the half-integers where $\{x\} = 0.5$. In the interval $(-2, 2)$, these points are:
$$x \in \{-1.5, -0.5, 0.5, 1.5\}$$

### Final Conclusion

Summing these up, the total number of points of non-differentiability in the open interval $(-2, 2)$ is:
$$3 	ext{ (troughs)} + 4 	ext{ (peaks)} = 7 	ext{ points}$$

The function $h(x)$ is **continuous on $[-2, 2]$** but fails to be differentiable at **7 points** within the interval $(-2, 2)$.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[t]$ denote the greatest integer less than or equal to $t$ . Let $f (x) = x - [x]$ , $g (x) = 1 - x + [x]$ , and $h (x) = min {f (x), g (x)}$ , $x \in [- 2, 2]$ . Then $h$ is :

Select Answer:

Visualized Solution

Analyzing $f (x)$

Analyzing $g (x)$

Constructing $h (x)$

Behavior of $h (x)$ in $[0, 1]$

Checking Continuity

Differentiability of $h (x)$

Locating Sharp Corners

Final Answer

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Intersection

The Continuity Check

The Differentiability Hunt

Final Conclusion

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