Check the differentiability of \( f(x) = \begin{cases}
x^2 + 1, & 0 \leq x < 1 \\
3 - x, & 1 \leq x \leq 2
\end{cases} \) at \( x = 1 \).
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Solution and Explanation
Step 1: Find the left-hand derivative (LHD)
The LHD at \( x = 1 \) is: \[ {LHD} = \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{-h}. \] Substituting \( f(x) = x^2 + 1 \) for \( x<1 \): \[ {LHD} = \lim_{h \to 0^-} \frac{(1 - h)^2 + 1 - 2}{-h}. \] Simplify: \[ {LHD} = \lim_{h \to 0^-} \frac{1 - 2h + h^2 - 1}{-h} = \lim_{h \to 0^-} \frac{-2h + h^2}{-h}. \] Factorize: \[ {LHD} = \lim_{h \to 0^-} (2 - h) = 2. \]
Step 2: Find the right-hand derivative (RHD)
The RHD at \( x = 1 \) is: \[ {RHD} = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h}. \] Substituting \( f(x) = 3 - x \) for \( x>1 \): \[ {RHD} = \lim_{h \to 0^+} \frac{[3 - (1 + h)] - 2}{h}. \] Simplify: \[ {RHD} = \lim_{h \to 0^+} \frac{-h}{h} = -1. \]
Step 3: Check differentiability
Since \( {LHD} \neq {RHD} \), \( f(x) \) is not differentiable at \( x = 1 \).
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