Question

Check the differentiability of \( f(x) \) at \( x = 1 \), where: \[ f(x) = \begin{cases} x^2 + 1, & 0 \leq x < 1, \\ 3 - x, & 1 \leq x \leq 2. \end{cases} \]

Hint:

To check differentiability, verify continuity first and then compare left-hand and right-hand derivatives.

Answer 1

1. Continuity at \( x = 1 \): \[ f(1^-) = \lim_{x \to 1^-} f(x) = (1)^2 + 1 = 2, \] \[ f(1^+) = \lim_{x \to 1^+} f(x) = 3 - 1 = 2. \] Thus, \( f(1^-) = f(1^+) = f(1) = 2 \), so \( f(x) \) is continuous at \( x = 1 \). 
2. Differentiability at \( x = 1 \): Find the left-hand derivative: \[ f'(x) = \frac{d}{dx} (x^2 + 1) = 2x, \quad f'(1^-) = 2(1) = 2. \] Find the right-hand derivative: \[ f'(x) = \frac{d}{dx} (3 - x) = -1, \quad f'(1^+) = -1. \] Since \( f'(1^-) \neq f'(1^+) \), the function is not differentiable at \( x = 1 \).
Final Answer: \( \boxed{{Not differentiable at } x = 1} \)