Question

Let \( f(x) = |x| + |x-1| + |x+1| \) be a function defined on \( \mathbb{R} \), then \( f(x) \) is:

• A. differentiable for all \( x \in \mathbb{R} \)
• B. differentiable for all \( x \in \mathbb{R} \) other than \( x = -1, 0, 1 \)
• C. differentiable only for \( x = -1, 0, 1 \)
• D. not differentiable at any real point

Hint:

A function involving a sum of absolute values, \( \sum |x-a_i| \), will generally be non-differentiable at each point \(x=a_i\). To check, you don't always need to write out the full piecewise function. You can just check if the sum of the derivatives of the arguments changes sign at that point. For example at \(x=1\), the derivatives of \(x, x-1, x+1\) are \(1,1,1\). The signs of the absolute values are \(+, -, +\) just below 1, and \(+,+,+\) just above 1. The derivative just below 1 is \(+1 -1 +1 = 1\). The derivative just above 1 is \(+1+1+1=3\). Since \(1 \neq 3\), it's not differentiable.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The function \(f(x)\) is a sum of absolute value functions. The absolute value function \(|u|\) is not differentiable at points where its argument \(u\) is zero. The sum of differentiable functions is differentiable. The sum of a differentiable function and a non-differentiable function is non-differentiable. Therefore, we expect potential points of non-differentiability where the arguments of the absolute value functions become zero.

Step 2: Key Formula or Approach:
The points where the arguments of the absolute value functions are zero are: - \(|x|\) \(\implies\) \(x=0\) - \(|x-1|\) \(\implies\) \(x=1\) - \(|x+1|\) \(\implies\) \(x=-1\) These are the critical points where the definition of the function changes. Let's write \(f(x)\) as a piecewise function. The critical points divide the real line into four intervals: \( (-\infty, -1) \), \( [-1, 0) \), \( [0, 1) \), and \( [1, \infty) \).

Step 3: Detailed Explanation:
Let's define \(f(x)\) on each interval:
For \( x<-1 \): \(x\), \(x-1\), and \(x+1\) are all negative. \( f(x) = -x - (x-1) - (x+1) = -x - x + 1 - x - 1 = -3x \)
For \( -1 \le x<0 \): \(x\) and \(x-1\) are negative, \(x+1\) is non-negative. \( f(x) = -x - (x-1) + (x+1) = -x - x + 1 + x + 1 = -x + 2 \)
For \( 0 \le x<1 \): \(x\) and \(x+1\) are non-negative, \(x-1\) is negative. \( f(x) = x - (x-1) + (x+1) = x - x + 1 + x + 1 = x + 2 \)
For \( x \ge 1 \): \(x\), \(x-1\), and \(x+1\) are all non-negative. \( f(x) = x + (x-1) + (x+1) = x + x - 1 + x + 1 = 3x \) So, the piecewise function is: \[ f(x) = \begin{cases} -3x & \text{if } x<-1
-x+2 & \text{if } -1 \le x<0
x+2 & \text{if } 0 \le x<1
3x & \text{if } x \ge 1 \end{cases} \] Within each of these open intervals, the function is a simple polynomial, so it is differentiable. We only need to check for differentiability at the critical points \(x=-1, 0, 1\) by comparing the left-hand and right-hand derivatives. - At x = -1: - Left-hand derivative: \( \frac{d}{dx}(-3x) = -3 \) - Right-hand derivative: \( \frac{d}{dx}(-x+2) = -1 \) Since \( -3 \neq -1 \), \(f(x)\) is not differentiable at \(x=-1\). - At x = 0: - Left-hand derivative: \( \frac{d}{dx}(-x+2) = -1 \) - Right-hand derivative: \( \frac{d}{dx}(x+2) = 1 \) Since \( -1 \neq 1 \), \(f(x)\) is not differentiable at \(x=0\). - At x = 1: - Left-hand derivative: \( \frac{d}{dx}(x+2) = 1 \) - Right-hand derivative: \( \frac{d}{dx}(3x) = 3 \) Since \( 1 \neq 3 \), \(f(x)\) is not differentiable at \(x=1\).
Step 4: Final Answer:
The function \(f(x)\) is differentiable everywhere except at the points \(x = -1, 0, 1\).