Question

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 \( x = 1 \).

Hint:

For piecewise functions, compute LHD and RHD separately at the point of interest to check differentiability.

Answer 1

Step 1: Calculate the Left-Hand Derivative (LHD)
The LHD at \( x = 1 \) is defined as: \[ {LHD} = \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{-h}. \] Using \( f(x) = x^2 + 1 \) for \( x<1 \): \[ {LHD} = \lim_{h \to 0^-} \frac{(1 - h)^2 + 1 - 2}{-h}. \] After simplification: \[ {LHD} = \lim_{h \to 0^-} \frac{1 - 2h + h^2 - 1}{-h} = \lim_{h \to 0^-} \frac{-2h + h^2}{-h}. \] Factoring the expression: \[ {LHD} = \lim_{h \to 0^-} (2 - h) = 2. \]
Step 2: Calculate the Right-Hand Derivative (RHD)
The RHD at \( x = 1 \) is defined as: \[ {RHD} = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h}. \] Using \( f(x) = 3 - x \) for \( x>1 \): \[ {RHD} = \lim_{h \to 0^+} \frac{[3 - (1 + h)] - 2}{h}. \] After simplification: \[ {RHD} = \lim_{h \to 0^+} \frac{-h}{h} = -1. \] 
Step 3: Assess Differentiability
As the \( {LHD} eq {RHD} \), the function \( f(x) \) is not differentiable at \( x = 1 \).