Question

If $ f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2x - 1, & x > 1 \end{cases} $, then:

• A. \( f \) is not continuous but differentiable at \( x = 1 \)
• B. \( f \) is differentiable at \( x = 1 \)
• C. \( f \) is continuous but not differentiable at \( x = 1 \)
• D. \( f \) is discontinuous at \( x = 1 \)

Hint:

When checking for continuity and differentiability, ensure the left-hand and right-hand limits as well as the function values and derivatives are consistent at the point in question.

Answer 1

The Correct Option is C

To determine whether the function is continuous and differentiable at \( x = 1 \), we need to check the following:
Step 1: Continuity at \( x = 1 \)
A function is continuous at a point if: \[ \lim_{x \to c} f(x) = f(c) \] For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit and the right-hand limit must be equal to the value of the function at \( x = 1 \). - The function is defined as \( f(x) = x \) for \( 0 \leq x \leq 1 \), so at \( x = 1 \), we have \( f(1) = 1 \). - The function is also defined as \( f(x) = 2x - 1 \) for \( x>1 \), and the right-hand limit as \( x \to 1^+ \) gives: \[ \lim_{x \to 1^+} f(x) = 2(1) - 1 = 1 \] Thus, the left-hand limit, right-hand limit, and \( f(1) \) are all equal to 1. Therefore, \( f(x) \) is continuous at \( x = 1 \).
Step 2: Differentiability at \( x = 1 \)
A function is differentiable at a point if the derivative from the left equals the derivative from the right at that point. - The derivative of \( f(x) = x \) for \( 0 \leq x \leq 1 \) is \( f'(x) = 1 \). - The derivative of \( f(x) = 2x - 1 \) for \( x>1 \) is \( f'(x) = 2 \). At \( x = 1 \), the left-hand derivative is 1, and the right-hand derivative is 2. Since these are not equal, \( f(x) \) is not differentiable at \( x = 1 \). Thus, the function is continuous but not differentiable at \( x = 1 \).