Question

The points of non differentiability of \(f(x) = |x-2| + |x - 3|\)
A. 1
B. 2
C. 3
D. 4
E. 5
Choose the correct answer from the options given below:

• A. A and B only
• B. B and C only
• C. A and C only
• D. A and D only

Answer 1

The Correct Option is B

To find the points of non-differentiability of the function \(f(x) = |x-2| + |x-3|\), we need to determine where its derivative does not exist. The function is composed of two absolute value expressions, each of which is non-differentiable at its respective point of change for the sign, which occurs at the inputs where the expression inside the absolute value is zero.

The points of interest are:

• \(|x-2| = 0\) when \(x = 2\). This is a point where the function might be non-differentiable.

• \(|x-3| = 0\) when \(x = 3\). This is another point where the function might be non-differentiable.

Checking Differentiability:

The general approach is to consider the derivative of \(f(x)\) and evaluate it around these points.

• For \(x < 2\): Both terms are negative inside the absolute values, so:

  \(f(x) = -(x-2) - (x-3) = -x+2 - x + 3 = -2x + 5\)

  The derivative in this range is \(f'(x) = -2\).

• For \(2 \leq x < 3\): The first term is positive, the second still negative, so:

  \(f(x) = (x-2) - (x-3) = x - 2 - x + 3 = 1\)

  The derivative here is \(f'(x) = 0\).

• For \(x \geq 3\): Both terms are positive, so:

  \(f(x) = (x-2) + (x-3) = x - 2 + x - 3 = 2x - 5\)

  The derivative in this range is \(f'(x) = 2\).

Conclusion:

The piecewise derivatives have different values around the points \(x = 2\) and \(x = 3\). Therefore, \(f(x)\) is not differentiable at \(x = 2\) and \(x = 3\).

The correct answer is: B and C only