Check the differentiability of \( f(x) \) at \( x = 1 \), where:
\[
f(x) =
\begin{cases}
x^2 + 1, & 0 \leq x < 1, \\
3 - x, & 1 \leq x \leq 2.
\end{cases}
\]
Show Hint
Solution and Explanation
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: \( \boxed{ {Not differentiable at } x = 1} \)
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