Question

The set of all values of x for which $f(x) = ||x|-1|$ is differentiable is

• A. $\{-1,1\}$
• B. $\mathbb{R} - \{-1,1,0\}$
• C. $\mathbb{R}$
• D. $(0, \infty)$

Hint:

To find the points where a function with nested absolute values like $f(x) = |g(|h(x)|)|$ is not differentiable, you must check all points where the arguments of any of the absolute value functions become zero. In this case, check where $h(x)=0$ and where $g(|h(x)|)=0$.

Answer 1

The Correct Option is B

Step 1: Identify the potential points of non-differentiability.
The function $g(u) = |u|$ is not differentiable at $u=0$. In our function $f(x) = ||x|-1|$, non-differentiability can occur at points where the argument of either absolute value function is zero.

Step 2: Check the inner absolute value function.
The inner function is $|x|$. Its argument is $x$. The function $|x|$ is not differentiable at the point where its argument is zero, which is $x=0$. Therefore, $f(x)$ is not differentiable at $x=0$.

Step 3: Check the outer absolute value function.
The outer function is $|u|$ where the argument is $u = |x|-1$. This function will not be differentiable at points where its argument is zero. We set the argument to zero: \[ |x|-1 = 0 \implies |x|=1. \] This equation has two solutions: $x=1$ and $x=-1$. At these two points, the graph of $f(x)$ will have sharp corners, indicating non-differentiability.

Step 4: Combine the points of non-differentiability.
From the analysis, the function $f(x)$ fails to be differentiable at three points: $x=0$, $x=1$, and $x=-1$. The function is differentiable for all other real numbers. Therefore, the set of values where $f(x)$ is differentiable is the set of all real numbers excluding these three points. \[ \text{Set} = \mathbb{R} - \{-1, 0, 1\}. \]