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

Concept: Functions involving modulus expressions are often non-differentiable at points where the quantity inside the modulus becomes zero. For a function involving absolute values: \[ |x| \] the graph has a sharp corner at the point where the expression changes sign. A function is differentiable at a point only if: \[ \text{LHD}=\text{RHD} \]

Step 1: Identify critical points. The function is: \[ f(x)=|x|+|x-1| \] The modulus expressions change sign at: \[ x=0 \quad \text{and} \quad x=1 \] Thus, these are the possible non-differentiable points.

Step 2: Write the function piecewise. For \(x<0\): \[ |x|=-x \] and \[ |x-1|=-(x-1)=1-x \] Thus, \[ f(x)=-x+1-x \] \[ f(x)=1-2x \] For \(0<x<1\): \[ |x|=x \] and \[ |x-1|=1-x \] Hence, \[ f(x)=x+1-x \] \[ f(x)=1 \] For \(x>1\): \[ |x|=x \] and \[ |x-1|=x-1 \] Thus, \[ f(x)=x+x-1 \] \[ f(x)=2x-1 \]

Step 3: Check differentiability at \(x=0\). For \(x<0\), \[ f(x)=1-2x \] Derivative: \[ f'(x)=-2 \] Thus, Left Hand Derivative: \[ LHD=-2 \] For \(0<x<1\), \[ f(x)=1 \] Derivative: \[ f'(x)=0 \] Thus, \[ RHD=0 \] Since, \[ LHD\neq RHD \] the function is not differentiable at \(x=0\).

Step 4: Check differentiability at \(x=1\). For \(0<x<1\), \[ f'(x)=0 \] Thus, \[ LHD=0 \] For \(x>1\), \[ f(x)=2x-1 \] Derivative: \[ f'(x)=2 \] Thus, \[ RHD=2 \] Since, \[ LHD\neq RHD \] the function is not differentiable at \(x=1\). Hence, the function is non-differentiable at both points. \[ \boxed{ \text{Neither differentiable at }x=0\text{ nor at }x=1 } \]