Question

If \( f(x) = \begin{cases} 2x - 3, & -3 \leq x \leq -2 \\x + 1, & -2<x \leq 0 \end{cases} \), check the differentiability of \( f(x) \) at \( x = -2 \).

Hint:

A function must be continuous at a point to be differentiable there. If discontinuity exists, differentiability fails.

Answer 1

To determine if the piecewise function \( f(x) \) is differentiable at \( x = -2 \), we first examine its continuity at that point. The function is defined as:
\[ f(x) = \begin{cases} 2x - 3, & -3 \leq x \leq -2 \\ x + 1, & -2 < x \leq 0 \end{cases} \]

Step 1: Evaluate Continuity at \( x = -2 \)

Left-Hand Limit (LHL):
Using the first part of the function for \( x \to -2^- \):
\[ \lim_{x \to -2^-} f(x) = 2(-2) - 3 = -4 - 3 = -7 \]

Right-Hand Limit (RHL):
Using the second part of the function for \( x \to -2^+ \):
\[ \lim_{x \to -2^+} f(x) = -2 + 1 = -1 \]

Since the LHL (\(-7\)) does not equal the RHL (\(-1\)), the function is not continuous at \( x = -2 \).

Step 2: Conclusion on Differentiability
A function must be continuous at a point to be differentiable there. As \( f(x) \) is not continuous at \( x = -2 \), it cannot be differentiable at \( x = -2 \).

Final Answer:
The function \( f(x) \) is not differentiable at \( x = -2 \).