Question

In the interval \([0, 3]\), the function \(f(x) = |x - 1| + |x - 2|\) is:

• A. Discontinuous
• B. Differentiable
• C. Continuous but not differentiable at \(x = 2\) only
• D. Continuous but not differentiable at \(x = 1\) and \(x = 2\)

Hint:

When analyzing functions involving absolute values, identify points where the function's formula changes, as these are potential locations of non-differentiability. Always check both continuity and differentiability at these points.

Answer 1

The Correct Option is D

The function \(f(x) = |x - 1| + |x - 2|\) is defined as the sum of two absolute value functions. To analyze its continuity and differentiability, let's break down the function into different intervals based on the points where the arguments inside the absolute values change sign, namely at \(x = 1\) and \(x = 2\). 
Step 1: Analyze the behavior of \(f(x)\) in different intervals: - For \(x \in [0, 1]\), \(x - 1 \leq 0\) and \(x - 2 \leq 0\), so \(f(x) = -(x - 1) - (x - 2) = -2x + 3\). - For \(x \in [1, 2]\), \(x - 1 \geq 0\) and \(x - 2 \leq 0\), so \(f(x) = (x - 1) - (x - 2) = 1\). - For \(x \in [2, 3]\), \(x - 1 \geq 0\) and \(x - 2 \geq 0\), so \(f(x) = (x - 1) + (x - 2) = 2x - 3\).
Step 2: Check the continuity of \(f(x)\): - The function is continuous at all points within the interval \([0, 3]\) because the absolute value functions are continuous, and the piecewise expressions match at the endpoints of the intervals: \[ \lim_{x \to 1^-} f(x) = 1 \quad {and} \quad \lim_{x \to 1^+} f(x) = 1, \] \[ \lim_{x \to 2^-} f(x) = 1 \quad {and} \quad \lim_{x \to 2^+} f(x) = 1. \] Hence, \(f(x)\) is continuous at \(x = 1\) and \(x = 2\).
Step 3: Check the differentiability of \(f(x)\): - At \(x = 1\), the left-hand derivative is the derivative of \(f(x) = -2x + 3\), which is \(-2\), and the right-hand derivative is the derivative of \(f(x) = 1\), which is \(0\). Since these derivatives do not match, \(f(x)\) is not differentiable at \(x = 1\). - Similarly, at \(x = 2\), the left-hand derivative is the derivative of \(f(x) = 1\), which is \(0\), and the right-hand derivative is the derivative of \(f(x) = 2x - 3\), which is \(2\). Again, the derivatives do not match, so \(f(x)\) is not differentiable at \(x = 2\). Thus, \(f(x)\) is continuous but not differentiable at \(x = 1\) and \(x = 2\).