Question

Find whether the function \( f(x) = \begin{cases x - 1, & x 2x - 3, & x \ge 2 \end{cases} \) at \( x = 2 \) is differentiable or not.}

Hint:

For piecewise linear functions, the LHD and RHD are simply the slopes of the line segments on either side of the point.
Slope for \( x < 2 \) is 1; slope for \( x \ge 2 \) is 2. Since they differ, non-differentiability is confirmed.

Answer 1

Concept:
• A function is differentiable at \( x = a \) if its Left Hand Derivative (LHD) and Right Hand Derivative (RHD) at that point exist and are equal.
• LHD at \( a = \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h} \).
• RHD at \( a = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h} \).
• First, check continuity; if a function is discontinuous, it is automatically non-differentiable.

Step 1: Check continuity at \( x = 2 \)
Functional value: \( f(2) = 2(2) - 3 = 1 \).
Left Hand Limit (LHL): \( \lim_{x \to 2^-} (x - 1) = 2 - 1 = 1 \).
Right Hand Limit (RHL): \( \lim_{x \to 2^+} (2x - 3) = 2(2) - 3 = 1 \).
Since \( \text{LHL} = \text{RHL} = f(2) \), the function is continuous.

Step 2: Calculate the Left Hand Derivative (LHD)
For \( x < 2 \), \( f(x) = x - 1 \).
\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h} \] Since \( 2+h < 2 \) for \( h < 0 \):
\[ \text{LHD} = \lim_{h \to 0} \frac{(2 + h - 1) - 1}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \]

Step 3: Calculate the Right Hand Derivative (RHD)
For \( x \ge 2 \), \( f(x) = 2x - 3 \).
\[ \text{RHD} = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h} \] Since \( 2+h > 2 \) for \( h > 0 \):
\[ \text{RHD} = \lim_{h \to 0} \frac{[2(2 + h) - 3] - 1}{h} = \lim_{h \to 0} \frac{4 + 2h - 3 - 1}{h} = \lim_{h \to 0} \frac{2h}{h} = 2 \]

Step 4: Compare LHD and RHD
We found \( \text{LHD} = 1 \) and \( \text{RHD} = 2 \).
Since \( \text{LHD} \neq \text{RHD} \), the function is not differentiable at \( x = 2 \).