Question

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

• A. Continuous at x = 1 and not differentiable at x = 1.
• B. Continuous and differentiable at x = 1.
• C. Discontinuous and differentiable at x = 1.
• D. Neither continuous nor differentiable at x = 1.

Answer 1

The Correct Option is A

The function \( f(x) = |x - 1| \) is indeed continuous at \( x = 1 \). This can be understood as follows:

• Continuity at \( x = 1 \): A function \( f(x) \) is continuous at a point \( x = a \) if the limit of \( f(x) \) as \( x \) approaches \( a \) from the left is equal to the limit of \( f(x) \) as \( x \) approaches \( a \) from the right, and both are equal to \( f(a) \). For \( f(x) = |x - 1| \):
  1. As \( x \to 1^- \), \( f(x) = 1-x \rightarrow 0 \).
  2. As \( x \to 1^+ \), \( f(x) = x-1 \rightarrow 0 \).
  3. Thus, \(\lim_{x \to 1} f(x) = f(1) = 0\).
  The function is continuous at \( x = 1 \) because these values match.
• Differentiability at \( x = 1 \): A function is differentiable at a point if the derivative exists at that point. To determine if \( f(x) = |x - 1| \) is differentiable at \( x = 1 \), we consider the left-hand derivative and the right-hand derivative:
  1. From the left, \( f(x) = 1-x \), and the derivative \( f'(x) = -1 \).
  2. From the right, \( f(x) = x-1 \), and the derivative \( f'(x) = +1 \).
  Because the left-hand and right-hand derivatives are not equal, the derivative \( f'(x) \) does not exist at \( x = 1 \). Therefore, it is not differentiable at \( x = 1 \).

Thus, the correct description of \( f(x) = |x - 1| \) at \( x = 1 \) is that it is continuous at \( x = 1 \) and not differentiable at \( x = 1 \).