Question

The function \(f(x)=|x|+|x-1|\) is:

• A. Differentiable at \(x=0\) but not at \(x=1\)
• B. Neither differentiable at \(x=0\) nor at \(x=1\)
• C. Differentiable at \(x=1\) but not at \(x=0\)
• D. Differentiable at \(x=0\) and \(x=1\)

Hint:

Whenever modulus functions appear, first identify where the expressions inside modulus become zero. These points are candidates for non-differentiability.

Answer 1

The Correct Option is B

To determine the differentiability of the function \(f(x) = |x| + |x - 1|\) at \(x = 0\) and \(x = 1\), we need to analyze the behavior of the function around these points.

The absolute value function \( |x| \) is defined as:

1. \(|x| = x\) if \(x \ge 0\)
2. \(|x| = -x\) if \(x < 0\)

Thus, we can express the function \(f(x)\) piecewise based on the intervals defined by the expressions inside the absolute values:

• For \(x < 0\), \(f(x) = -x - (x-1) = -2x + 1\)
• For \(0 \le x < 1\), \(f(x) = x - (x-1) = 1\)
• For \(x \ge 1\), \(f(x) = x + (x-1) = 2x - 1\)

We will now check the differentiability at \(x = 0\) and \(x = 1\).

Checking differentiability at \(x = 0\):

The left-hand derivative (LHD) at \(x = 0\) is computed as:

\(\lim_{{h \to 0^-}} \frac{f(0 + h) - f(0)}{h} = \lim_{{h \to 0^-}} \frac{-2h + 1 - 1}{h} = \lim_{{h \to 0^-}} \frac{-2h}{h} = -2\)

The right-hand derivative (RHD) at \(x = 0\) is computed as:

\(\lim_{{h \to 0^+}} \frac{f(0 + h) - f(0)}{h} = \lim_{{h \to 0^+}} \frac{1 - 1}{h} = 0\)

Since LHD \(\neq\) RHD at \(x = 0\), \(f(x)\) is not differentiable at \(x = 0\).

Checking differentiability at \(x = 1\):

The left-hand derivative (LHD) at \(x = 1\) is computed as:

\(\lim_{{h \to 0^-}} \frac{f(1 + h) - f(1)}{h} = \lim_{{h \to 0^-}} \frac{1 - (2-1)}{h} = \lim_{{h \to 0^-}} \frac{-h}{h} = -1\)

The right-hand derivative (RHD) at \(x = 1\) is computed as:

\(\lim_{{h \to 0^+}} \frac{f(1 + h) - f(1)}{h} = \lim_{{h \to 0^+}} \frac{2h + 1 - 1}{h} = \lim_{{h \to 0^+}} \frac{2h}{h} = 2\)

Since LHD \(\neq\) RHD at \(x = 1\), \(f(x)\) is not differentiable at \(x = 1\).

Thus, the function \(f(x) = |x| + |x - 1|\) is neither differentiable at \(x=0\) nor at \(x=1\).

Correct Answer: Neither differentiable at \(x=0\) nor at \(x=1\)