Question

Let $f : \mathbb{R} \to \mathbb{R}$ be the function given by \[ f(x) = |x - 2| + 3|x - 1| + ||x - 2| - 1| . \] What is the number of points where $f$ is NOT differentiable?

• A. 2
• B. 1
• C. 0
• D. 3

Hint:

Do not assume every critical point of an absolute value function is a point of non-differentiability.
Sometimes, the linear slopes on either side of a point cancel out perfectly (as happened here at $x=2$ with a slope of 3 on both sides), making the function smooth and differentiable there.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Absolute value functions \( |g(x)| \) are typically non-differentiable at points where \( g(x) = 0 \), provided the derivative of \( g(x) \) is non-zero there.

Step 2: Detailed Explanation:
1. Identify potential critical points:
$\bullet$ From \( |x-2| \): \( x = 2 \).
$\bullet$ From \( 3|x-1| \): \( x = 1 \).
$\bullet$ From \( ||x-2|-1| \): \( |x-2|-1 = 0 \implies x-2 = \pm1 \implies x = 3 \) or \( x = 1 \).
The set of critical points is \( \{1, 2, 3\} \).
2. Check differentiability at \( x=2 \):
Just to the left and right of 2, the term \( ||x-2|-1| \) behaves like \( |(2-x)-1| = |1-x| \). Since we are near 2, \( 1-x \) is negative, so it is \( -(1-x) = x-1 \).
The derivative from the left (LHD) and right (RHD) will be identical at \( x=2 \) because the slopes of the other terms cancel out the "kink".
LHD of \( |x-2| \) is -1, RHD is +1. The derivative of \( ||x-2|-1| \) near 2 is +1 from the left and -1 from the right. Total change is 0.
3. Check at \( x=1 \) and \( x=3 \):
At \( x=1 \) and \( x=3 \), the slopes do not cancel out, resulting in "sharp turns" (kinks) in the graph.

Step 3: Final Answer:
The function is not differentiable at 2 points: \( x=1 \) and \( x=3 \).