Question

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

Hint:

For simple piecewise linear functions, you can find the derivatives of the individual branches directly. The derivative of the left branch ($x-1$) is 1, and the derivative of the right branch ($2x-3$) is 2. Since $1 \neq 2$, it is instantly proven non-differentiable!

Answer 1

Concept: A function $f(x)$ is differentiable at a given point $x = c$ if and only if its Left Hand Derivative (LHD) and Right Hand Derivative (RHD) both exist and are completely equal to each other at that point.
• Left Hand Derivative (LHD): $f'(c^-) = \lim_{h \to 0^+} \frac{f(c - h) - f(c)}{-h}$
• Right Hand Derivative (RHD): $f'(c^+) = \lim_{h \to 0^+} \frac{f(c + h) - f(c)}{h}$

Step 1: Calculate the functional value at $x = 2$ and evaluate the Left Hand Derivative (LHD).
First, identify the function value at $x = 2$. Looking at the conditional statements, for $x \ge 2$, we use $f(x) = 2x - 3$: \[ f(2) = 2(2) - 3 = 4 - 3 = 1 \] Now, let us calculate the LHD at $x = 2$ by approaching from values less than 2 ($f(x) = x - 1$): \[ \text{LHD} = \lim_{h \to 0^+} \frac{f(2 - h) - f(2)}{-h} \] Substitute $f(2-h) = (2 - h) - 1 = 1 - h$ and $f(2) = 1$: \[ \text{LHD} = \lim_{h \to 0^+} \frac{(1 - h) - 1}{-h} = \lim_{h \to 0^+} \frac{-h}{-h} = 1 \]

Step 2: Evaluate the Right Hand Derivative (RHD) and check for equality.
Next, let us calculate the RHD at $x = 2$ by approaching from values greater than or equal to 2 ($f(x) = 2x - 3$): \[ \text{RHD} = \lim_{h \to 0^+} \frac{f(2 + h) - f(2)}{h} \] Substitute $f(2+h) = 2(2 + h) - 3 = 4 + 2h - 3 = 1 + 2h$ and $f(2) = 1$: \[ \text{RHD} = \lim_{h \to 0^+} \frac{(1 + 2h) - 1}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2 \] Comparing the two directional derivatives: \[ \text{LHD} = 1, \quad \text{RHD} = 2 \quad \implies \quad \text{LHD} \neq \text{RHD} \] Since the Left Hand Derivative does not match the Right Hand Derivative at $x = 2$, the function is not differentiable at $x = 2$.