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 \( x = 1 \): The left-hand limit is \( f(1^-) = \lim_{x \to 1^-} f(x) = (1)^2 + 1 = 2 \). The right-hand limit is \( f(1^+) = \lim_{x \to 1^+} f(x) = 3 - 1 = 2 \). Since \( f(1^-) = f(1^+) = f(1) = 2 \), \( f(x) \) is continuous at \( x = 1 \).
2. Differentiability at \( x = 1 \): The left-hand derivative is calculated from \( f'(x) = \frac{d}{dx} (x^2 + 1) = 2x \), yielding \( f'(1^-) = 2(1) = 2 \). The right-hand derivative is calculated from \( f'(x) = \frac{d}{dx} (3 - x) = -1 \), yielding \( f'(1^+) = -1 \). Because \( f'(1^-) eq f'(1^+) \), the function is not differentiable at \( x = 1 \).
Final Answer: \( \boxed{{Not differentiable at } x = 1} \)