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 Question:
We are given a function $f(x)$ containing absolute value terms.
Absolute value functions are continuous everywhere but typically have "corners" where they are not differentiable.
We need to find the exact number of points where $f(x)$ is not differentiable.

Step 2: Key Formula or Approach: The critical points where the arguments of the absolute values change sign are the potential points of non-differentiability.
These points are:
1. $x - 2 = 0 \implies x = 2$
2. $x - 1 = 0 \implies x = 1$
3. $|x - 2| - 1 = 0 \implies |x - 2| = 1 \implies x = 3$ or $x = 1$.
Thus, the critical points are $x = 1, 2,$ and $3$. We analyze the left-hand derivative (LHD) and right-hand derivative (RHD) at these three points.

Step 3: Detailed Explanation:

• Let us write the function without absolute values by dividing the domain into intervals:
• Interval 1: $x \le 1$
Here, $|x - 2| = 2 - x$, $|x - 1| = 1 - x$, and $||x - 2| - 1| = |2 - x - 1| = |1 - x| = 1 - x$.
\[ f(x) = (2 - x) + 3(1 - x) + (1 - x) = 6 - 5x \] Thus, $f'(x) = -5$.

• Interval 2: $1 < x \le 2$
Here, $|x - 2| = 2 - x$, $|x - 1| = x - 1$, and $||x - 2| - 1| = |1 - x| = x - 1$ (since $x > 1$).
\[ f(x) = (2 - x) + 3(x - 1) + (x - 1) = 3x - 2 \] What is the derivative? $f'(x) = 3$.

• Interval 3: $2 < x \le 3$
Here, $|x - 2| = x - 2$, $|x - 1| = x - 1$, and $||x - 2| - 1| = |x - 3| = 3 - x$ (since $x \le 3$).
\[ f(x) = (x - 2) + 3(x - 1) + (3 - x) = 3x - 2 \] What is the derivative? $f'(x) = 3$.

• Interval 4: $x > 3$
Here, $|x - 2| = x - 2$, $|x - 1| = x - 1$, and $||x - 2| - 1| = x - 3$.
\[ f(x) = (x - 2) + 3(x - 1) + (x - 3) = 5x - 8 \] What is the derivative? $f'(x) = 5$.

• Let us now check differentiability at the critical points: 1. At $x = 1$:
LHD $= -5$, RHD $= 3$. Since LHD $\neq$ RHD, the function is NOT differentiable at $x = 1$.
2. At $x = 2$:
LHD $= 3$, RHD $= 3$. Since LHD $=$ RHD, the function IS differentiable at $x = 2$.
3. At $x = 3$:
LHD $= 3$, RHD $= 5$. Since LHD $\neq$ RHD, the function is NOT differentiable at $x = 3$.

Step 4: Final Answer:
The points of non-differentiability are $x = 1$ and $x = 3$. Therefore, there are exactly 2 such points.